MOTION SAMPLE BOOKLET CLASS XI - Motion IIT- · PDF fileNo part of this work herein should be reproduced or used either ... STOICHIOMETRY-I ... (denoted by I) is an important quantity

MOTION SAMPLE BOOKLET CLASS XI

• ROTATIONAL• STOICHIOMETRY - I

• TRIGONOMETRIC RATIOS & IDENTITIES

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ROTATIONAL

S.NO. TOPIC PAGE NO.

THEORY WITH SOLVED EXAMPLES ..................................................... 5 – 24

EXERCISE - I ............................................................................................ 25 – 37

EXERCISE - II ........................................................................................... 38 – 54

EXERCISE - III .......................................................................................... 55 – 68

EXERCISE -IV ........................................................................................... 69 - 86

ANSWER KEY.......................................................................................... 87 - 89

CONTENTS

THEORY AND EXERCISE BOOKLET

STOICHIOMETRY-I



EXERCISE - I ............................................................................................ 120 – 126

EXERCISE - II ........................................................................................... 127 – 135

EXERCISE - III .......................................................................................... 136 – 143

EXERCISE - IV .......................................................................................... 144 – 145

ANSWER KEY.......................................................................................... 146 – 148

TRIGONOMETRIC RATIOS & IDENTITIES (PHASE–I)



EXERCISE - I ............................................................................................ 166 – 172

EXERCISE - II ........................................................................................... 173 – 179

EXERCISE - III .......................................................................................... 180 – 192

EXERCISE - IV .......................................................................................... 193 – 198

ANSWER KEY.......................................................................................... 199 - 200

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)

SYLLABUS

• ROTATIONAL

Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment ofinertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque;Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation;Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies;Collision of point masses with rigid bodies.

• STOICHIOMETRY-I

Mole concept; Chemical formulae; Balanced chemical equations; Calculations (based

on mole concept) involving common oxidation-reduction, neutralisation, and displacement

reactions; Concentration in terms of mole fraction, molarity, molality and normality.

• TRIGONOMETRIC RATIOS & IDENTITIES (PHASE–I)

Trigonometric functions, their periodicity and graphs, addition and subtraction formulae,

formulae involving multiple and sub-multiple angles

TOTAL NO. OF QUESTIONS

• PHYSICS

No. of Unsolved Example : 1000 (Approx)

No. of Solved Example : 5000 (Approx)

• CHEMISTRY



• MATHEMATICS



ROTATIONAL DYNAMICS Page # 5

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1. RIGID BODY :Rigid body is defined as a system of particles in which distance between each pair of particlesremains constant (with respect to time) that means the shape and size do not change,during the motion. Eg. Fan, Pen, Table, stone and so on.Our body is not a rigid body, two blocks with a spring attached between them is also not arigid body. For every pair of particles in a rigid body, there is no velocity of seperation orapproach between the particles. In the figure shown velocities of A and B with respect to

ground are VA and VB respectively

1B

AVA

VB2

B

VB cos 2VB sin 2

VA sin 1

VA cos 1

A

B

A

VBA

If the above body is rigidVA cos 1 = VB cos 2

Note : With respect to any particle of rigid body the motion of any other particle of that rigid body iscircular.

VBA = relative velocity of B with respect to A.

Types of Motion of rigid body

Pure TranslationalMotion

Pure RotationalMotion

Combined Translational and Rotational Motion

1.1. Pure Translational Motion :A body is said to be in pure translational motion if the displacement of each particle is sameduring any time interval however small or large. In this motion all the particles have same s v,& a at an instant.

example.A box is being pushed on a horizontal surface.

1610

6 610

V Vcm of any particle, a acm of any particle

S Scm of any particle

For pure translational motion :-vv

vv

v

v vv

m1 m1

m2m2

m3 m3

m4 m4 m5m5 m6

m8m7 m7m8

m6

ROTATIONAL DYNAMICS

Page # 6 ROTATIONAL DYNAMICS


F m a m a m aext 1 1 2 2 3 3 .............

Where m1, m2, m3, ......... are the masses of different particles of the body having accelerations

a a a1 2 3, , ,............... respectively..

But acceleration of all the particles are same So, a.........aaa 321

F Maext

Where M = Total mass of the body

a = acceleration of any particle or of centre of mass of body

P m v m v m v1 1 2 2 3 3 .............Where m1, m2, m3 ...... are the masses of different particles of the body having velocitiesv v v1 2 3, , ............. respectively

But velocities of all the particles are same so v v v v1 2 3 ..........

P MvWhere v = velocity of any particle or of centre of mass of the body..

Total Kinetic Energy of body = 2222

211 Mv

21...........vm

21vm

21

1.2. Pure Rotational Motion :A body is said to be in pure rotational motion if the perpendicular distance of each particleremains constant from a fixed line or point and do not move parallel to the line, and that line

is known as axis of rotation. In this motion all the particles have same , and at aninstant. Eg. : - a rotating ceiling fan, arms of a clock.For pure rotation motion :-

sr

Where = angle rotated by the particle

s = length of arc traced by the particle.r = distance of particle from the axis of rotation.

m2 m1

m3m4

m5 m6ddt Where = angular speed of the body..

ddt

Where = angular acceleration of the body..

All the parameters , and are same for all the particles. Axis of rotation is perpendicular tothe plane of rotation of particles.Special case : If = constant, = 0 + t Where 0 = initial angular speed

021

2t t t = time interval

2 = 02 + 2

Total Kinetic Energy 12

121 1

22 2

2m v m v .................

12 1 1

22 2

2 2[ ................]m r m r

12

2I Where I = Moment of Inertia = m r m r1 12

2 22 .......

= angular speed of body.



1.3 Combined translation and rotational MotionA body is said to be in translation and rotational motion if all the particles rotates about anaxis of rotation and the axis of rotation moves with respect to the ground.

2. MOMENT OF INERTIALike the centre of mass, the moment of inertia is a property of an object that is related to itsmass distribut ion. The moment of inert ia (denoted by I) is an important quantity in the studyof system of particles that are rotating. The role of the moment of inertia in the study ofrotational motion is analogous to that of mass in the study of linear motion. Moment of inertiagives a measurement of the resistance of a body to a change in its rotaional motion. If a bodyis at rest, the larger the moment of inertia of a body the more difficuilt it is to put that bodyinto rotational motion. Similarly, the larger the moment of inertia of a body, the more difficultto stop its rotational motion. The moment of inertia is calculated about some axis (usually therotational axis).Moment of inertia depends on :(i) density of the material of body(ii) shape & size of body(iii) axis of rotationIn totality we can say that it depends upon distribution of mass relative to axis of rotation.

Note :Moment of inertia does not change if the mass :(i) is shifted parallel to the axis of the rotation(ii) is rotated with constant radius about axis of rotation

2.1 Moment of Inertia of a Single ParticleFor a very simple case the moment of inertia of asingle particle about an axis is given by,

r

I = mr2 ...(i)Here, m is the mass of the particle and r its distance from the axis under consideration.

2.2 Moment of Inertia of a System of ParticlesThe moment of inertia of a system of particles about an axis is given by,

I = m ri ii

2 ...(ii)

m1

m2

m3r3

r2

r1

where ri is the perpendicular distance from the axis to the ith particle, which has a mass mi.Ex.1 Two heavy particles having masses m1 & m2 are situated in a plane perpendicular to

line AB at a distance of r1 and r2 respectively.

r1 r2

A

B

m1 m2

C

D

E F

(i) What is the moment of inertia of the system about axis AB?(ii) What is the moment of inertia of the system about an axis passing through m1

and perpendicular to the line joining m1 and m2 ?(iii) What is the moment of inertia of the system about an axis passing through m1

and m2?



Sol. (i) Moment of inertia of particle on left is I1 = m1r12.

Moment of Inertia of particle on right is I2 = m2r22.

Moment of Inertia of the system about AB isI = I1+ I2 = m1r1

2 + m2r22

(ii) Moment of inertia of particle on left is I1 = 0Moment of Inertia of the system about CD isI = I1 + I2 = 0 + m2(r1 + r2)2

(iii) Moment of inertia of particle on left is I1 = 0Moment of inertia of particle on right is I2 = 0Moment of Inertia of the system about EF isI = I1 + I2 = 0 + 0

Ex.2 Three light rods, each of length 2 , are joined together to form a triangle. Three particlesA, B, C of masses m, 2m, 3m are fixed to the vertices of the triangle. Find the momentof inertia of the resulting body about(a) an axis through A perpendicular to the plane ABC,(b) an axis passing through A and the midpoint of BC.

Sol. (a) B is at a distant 2 from the axis XY so the moment ofinertia of B (IB) about XY is 2 m (2 )2

Similarly Ic about XY is 3m (2 )2 and IA about XY is m(0)2

B

AX

Y

2l

C3m2m

m

2l

Therefore the moment of inertia of the body about XY is2m (2 )2 + 3 m(2 )2 + m(0)2 = 20 m 2

(b) IA about X' Y' = m(0)2

IB about X' Y' = 2m ( )2

IC about X' Y' = 3m ( )2

Therefore the moment of inertia of the body about X' Y' ism(0)2 + 2m( )2 + 3m( )2 = 5 m 2

C3m2m

A m

X'

B

Y'Ex.3 Four particles each of mass m are kept at the four corners of a square of edge a. Find

the moment of inertia of the system about a line perpendicular to the plane of thesquare and passing through the centre of the square.

Sol. The perpendicular distance of every particle fromthe given line is a / 2 . The moment of inertia of

one particle is, therefore, m a( / )2 2 =12

2ma . The

moment of inertia of the system is,

therefore, 412

2ma = 2 ma2.

m m

mm



2.3 Moment of Inertia of Rigid BodiesFor a continuous mass distribution such as foundin a rigid body, we replace the summation of

I m ri ii

2 by an integral. If the system is divided

into infinitesimal element of mass dm and if r isthe distance from a mass element to the axis ofrotation, the moment of inertia is,

I = r dm2

r

where the integral is taken over the system.

(A) Uniform rod about a perpendicular bisectorConsider a uniform rod of mass M and length l figure and suppose the moment of inertia is tobe calculated about the bisector AB. Take the origin at the middle point O of the rod. Considerthe element of the rod between a distance x and x + dx from the origin. As the rod is uniform,Mass per unit length of the rod = M/ lso that the mass of the element = (M/l)dx. x dx

B

A

0The perpendicular distance of the element fromthe line AB is x. The moment of inertia of thiselement about AB is

d M dx xIl

2 .

When x = – l/2, the element is at the left end of the rod. As x is changed from – l/2 to l/2, theelements cover the whole rod.Thus, the moment of inertia of the entire rod about AB is

12M

3xMdxxM 22/

2/

32/

2/

2 lll

l

l–

l

l

I

(B) Moment of inertia of a rectangular plate about a line parallel to an edge and passingthrough the centreThe situation is shown in figure. Draw a line parallel to AB at a distance x from it and anotherat a distance x + dx. We can take the strip enclosed between the two lines as the smallelement.

x

dx

b

B

A

lIt is “small” because the perpendiculars from different points of the strip to AB differ by notmore than dx. As the plate is uniform,

its mass per unit area = Mbl

Mass of the strip = Mb

bdx M dxl l

.



The perpendicular distance of the strip from AB = x.

The moment of inertia of the strip about AB = dI = M dx xl

2 . The moment of inertia of the givenplate is, therefore,

12MdxxM 22/

2/

2 ll

l

l

I

The moment of inertia of the plate about the line parallel to the other edge and passingthrough the centre may be obtained from the above formula by replacing l by b and thus,

I Mb2

12.

(C) Moment of inertia of a circular ring about its axis (the line perpendicular to the plane ofthe ring through its centre)Suppose the radius of the ring is R and its mass is M. As all the elements of the ring are at thesame perpendicular distance R from the axis, the moment of inertia of the ring is

I r dm R dm R dm MR2 2 2 2 .

(D) Moment of inertia of a uniform circular plate about its axisLet the mass of the plate be M and its radius R. The centre is at O and the axis OX isperpendicular to the plane of the plate.

R

x0

X

dx

Draw two concentric circles of radii x and x + dx, both centred at O and consider the area ofthe plate in between the two circles.This part of the plate may be considered to be a circular ring of radius x. As the periphery ofthe ring is 2 x and its width is dx, the area of this elementary ring is 2 xdx. The area of theplate is R2. As the plate is uniform,

Its mass per unit area = MR2

Mass of the ring MR

xdx MxdxR2 22 2

Using the result obtained above for a circular ring, the moment of inertia of the elementaryring about OX is

d MxdxR

xI 22

2.

The moment of inertia of the plate about OX is

2MR

dxxRM2 2

3R

02I .

(E) Moment of inertia of a hollow cylinder about its axisSuppose the radius of the cylinder is R and its mass is M. As every element of this cylinder isat the same perpendicular distance R from the axis, the moment of inertia of the hollowcylinder about its axis is

222 MRdmRdmrI



(F) Moment of inertia of a uniform solid cylinder about its axisLet the mass of the cylinder be M and its radius R. Draw two cylindrical surface of radii x andx + dx coaxial with the given cylinder. Consider the part of the cylinder in between the twosurface. This part of the cylinder may be considered to be a hollow cylinder of radius x. Thearea of cross-section of the wall of this hollow cylinder is 2 x dx. If the length of the cylinderis l, the volume of the material of this elementary hollow cylinder is 2 x dxl.The volume of the solid cylinder is R2 l and it is uniform, hence its mass per unit volume is

MR2 l

The mass of the hollow cylinder considered is

MR

xdxM

Rxdx2 22

2l

l .dx

xAs its radius is x, its moment of inertia about the given axis is

dM

Rxdx xI

22

2.

The moment of inertia of the solid cylinder is, therefore,

I 222

3

0

2MR

x dx MRR

.

Note that the formula does not depend on the length of the cylinder.(G) Moment of inertia of a uniform hollow sphere about a diameter

Let M and R be the mass and the radius of the sphere, O its centre and OX the given axis(figure). The mass is spread over the surface of the sphere and the inside is hollow.Let us consider a radius OA of the sphere at an angle with the axis OX and rotate this radiusabout OX. The point A traces a circle on the sphere. Now change to + d and get anothercircle of somewhat larger radius on the sphere. The part of the sphere between these twocircles, shown in the figure, forms a ring of radius R sin . The width of this ring is Rd and itsperiphery is 2 R sin . Hence,the area of the ring = (2 R sin ) (Rd ).

Mass per unit area of the sphereMR4 2 .

The mass of the ring MR

R Rd M d4

222 ( sin )( ) sin .

R sinA

Rd

Rd

x

0

The moment of inertia of this elemental ring about OX is

d M d RI2

2sin . ( sin ) MR d2

2 3sin

As increases from 0 to , the elemental rings cover the whole spherical surface. Themoment of inertia of the hollow sphere is, therefore,

I MR d MR d2 2

12 3

0

22

0

sin ( cos ) sin MR d2

2

02

1( cos ) (cos )

MR MR2 3

0

2

2 323

cos cos



(H) Moment of inertia of a uniform solid sphere about a diameterLet M and R be the mass and radius of the given solid sphere.Let O be centre and OX the given axis. Draw two spheres ofradii x and x + dx concentric with the given solid sphere. Thethin spherical shell trapped between these spheres may be treatedas a hollow sphere of radius x.The mass per unit volume of the solid sphere

x0

x

= M

R

MR4

3

343 3

The thin hollow sphere considered above has a surface area 4 x2 and thickness dx. Its volumei s4 x2 dx and hence its mass is

= 3

44

32M

Rx dx( ) =

33

2MR

x dx

Its moment of inertia about the diameter OX is, therefore,

dl = 23

33

2MR

x dx x2 = 2

34M

Rx dx

If x = 0, the shell is formed at the centre of the solid sphere. As x increases from 0 to R, theshells cover the whole solid sphere.The moment of inertia of the solid sphere about OX is, therefore,

I = 23

4

0

MR

x dxR

= 25

2MR .

Ex.4 Find the moment of Inertia of a cuboid along the axis as shown in the figure.

Ib

ca

Sol. After compressing the cuboid parallel to the axis I = 12

)ba(M 22

3. THEOREMS OF MOMENT OF INERTIAThere are two important theorems on moment of inertia, which, in some cases enable themoment of inertia of a body to be determined about an axis, if its moment of inertia aboutsome other axis is known. Let us now discuss both of them.



3.1 Theorem of parallel axesA very useful theorem, called the parallel axes theorem relatesthe moment of inertia of a rigid body about two parallel axes,one of which passesthrough the centre of mass. COM

rTwo such axes are shown in figure for a body of mass M. If r isthe distance between the axes and ICOM and I are the respectivemoments of inertia about them, these moments are related by,

I = ICOM + Mr2

* Theorem of parallel axis is applicable for any type of rigid body whether it is a two dimensionalor three dimensional

Ex 5. Three rods each of mass m and length l are joinedtogether to form an equilateral triangle as shown infigure. Find the moment of inertia of the systemabout an axis passing through its centre of mass andperpendicular to the planeof triangle.

A

B C

COM

Sol. Moment of inertia of rod BC about an axis perpendicularto plane of triangle ABC and passing through the mid-point of rod BC (i.e., D) is

I1 = ml2

12From theorem of parallel axes, moment of inertia of thisrod about the asked axis is

I2 = I1 + mr2 = m m ml l

2 3l2 2 2

12 6

A

B C30°

r

COM

D Moment of inertia of all the three rod is

I I m m3 36 22

2 2l l

Ex.6. Find the moment of inertia of a solid sphere of mass M and radius R about an axis XXshown in figure.

x

x

Sol. From theorem of parallel axis,IXX = ICOM + Mr2

= 25

2 2MR MR

x

x

COM

r=R=

75

2MR



Ex.7. Consider a uniform rod of mass m and length 2l with two particles of mass m each atits ends. Let AB be a line perpendicular to the length of the rod passing through itscentre. Find the moment of inertia of the system about AB.

Sol. IAB = Irod + Iboth particles

m m( ) ( )212

22

2l l

A

I Im m

B73

2ml Ans.

3.2 Theorem of perpendicular axesThe theorem states that the moment of inertia of a plane lamina about an axis perpendicularto the plane of the lamina is equal to the sum of the moments of inertia of the lamina abouttwo axes perpendicular to each other, in its own plane and intersecting each other, at thepoint where the perpendicular axis passes through it.Let x and y axes be chosen in the plane of the body and z-axis perpendicular, to this plane,three axes being mutually perpendicular, then the theorem states that.

zy

xO

xi

yiri

P

Iz = Ix + Iy

Important point in perpendicular axis theorem(i) This theorem is applicable only for the plane bodies (two dimensional).(ii) In theorem of perpendicular axes, all the three axes (x, y and z) intersect each other and this

point may be any point on the plane of the body (it may even lie outside the body).(iii) Intersection point may or may not be the centre of mass of the body.

Ex.8 Find the moment of inertia of uniform ring of mass M and radius R about a diameter.

Z

0

A

DC

B

Sol. Let AB and CD be two mutually perpendicular diameters of the ring. Take them ax X and Y-axes and the line perpendicular to the plane of the ring through the centre as the Z-axis. Themoment of inertia of the ring about the Z-axis is I = MR2. As the ring is uniform, all of itsdiameter equivalent and so Ix = Iy, From perpendicular axes theorem,

Iz = Ix + Iy Hence Ix = Iz

2 = MR2

2Similarly, the moment of inertia of a uniform disc about a diameter is MR2/4



Ex.9 Two uniform identical rods each of mass M and length are joined to form a cross asshown in figure. Find the moment of inertia of the cross about a bisector as showndotted in the figure.

Sol. Consider the line perpendicular to the plane of the figure through the centre of the cross. The

moment of inertia of each rod about this line is M2

12 and hence the moment of inertia of the

cross is M 2

6. The moment of inertia of the cross about the two bisector are equal by

symmetry and according to the theorem of perpendicular axes, the moment of inertia of the

cross about the bisector is M 2

12.

Ex.10 In the figure shown find moment of inertia of a plate having mass M, length andwidth b about axis 1,2,3 and 4. Assume that C is centre and mass is uniformly distributed

b3

14 2

C

Sol. Moment of inertia of the plate about axis 1 (by taking rods perpendicular to axis 1)l1 = Mb2/3

Moment of inertia of the plate about axis 2 (by taking rods perpendicular to axis 2)I2 = M 2/12

Moment of inertia of the plate about axis 3 (by taking rods perpendicular to axis 3)

IMb

3

2

12Moment of inertia of the plate about axis 4(by taking rods perpendicular to axis 4)

I4 = M 2/3

3.3 Moment of Inertia of Compound BodiesConsider two bodies A and B, rigidly joined together. The moment of inertia of this compoundbody, about an axis XY, is required. If IA is the moment of inertia of body A about XY. IB is themoment of inertia of body B about XY.Then, moment of Inertia of compound body I = IA + IBExtending this argument to cover any number of bodies rigidly joined together, we see thatthe moment of inertia of the compound body, about a specified axis, is the sum of themoments of inertia of the separate parts of the body about the same axis.

Y

B

AX



Ex.11 Two rods each having length l and mass m joined together at point B as shown infigure.Then findout moment of inertia about axis passing thorugh A and perpendicularto the plane of page as shown in figure.

× AB

CSol. We find the resultant moment of inertia I by dividing in two parts such as

I = M.I of rod AB about A + M.I of rod BC about AI = I1 + I2 ... (1)

first calculate I1 :B A

×

I1 = m 2

3...(2)

Calculation of I2 :use parallel axis theorem

I2 = ICM + md2

= 2

22

4m

12m

= m4

512

m 22...(3)

×

×

/ 2

COMd

Put value from eq. (2) & (3) into (1)

I = 4

m512

m3

m 222

I = )1514(12

m 2 I = 5

3

2m

4. CAVITY PROBLEMS :

Ex.12 A uniform disc having radius 2R and mass density as shown in figure. If a small discof radius R is cut from the disc as shown. Then find out the moment of inertia ofremaining disc around the axis that passes through O and is perpendicular to the planeof the page.

2R O R

Sol. We assume that in remaning part a disc of radius R and mass density ± is placed. Then

2R O R

2R × I1

M R122( )

when is taken

+ × RI2

M R22–

–when is takes

Total Moment of Inertia I = I1 + I2

I1 = M R122

2( )



I1 = 4 4

2

2 2R R. = 8 R4

To calculate I2 we use parallel axis theorem.I2 = ICM + M2R2

I2 = M R2

2

2 + M2R2

I2 = 32 2

2M R = 32

2 2(– )R R I2 = –32

4R

Now I = I1 + I2

I = 8 32

4 4R R– I = 132

4R

Ex.13 A uniform disc of radius R has a round disc of radius R/3 cut as shown in Fig. The massof the remaining (shaded) portion of the disc equals M. Find the moment of inertia ofsuch a disc relative to the axis passing through geometrical centre of original disc andperpendicular to the plane of the disc.

OR

Sol. Let the mass per unit area of the material of discbe . Now the empty space can be considered ashaving density – and .

Now I0 = I + I–

( R2)R2/2 = M.I of about O

I– = – ( / ) ( / ) [– ( / ) ]( / )R R R R3 32

3 2 32 2

2 2

= M.I of – about 0

I0 = 49

4R Ans.

5. TORQUE :Torque represents the capability of a force to produce change in the rotational motion of thebody

FP

Qr sin

Line of action of force

r

5.1 Torque about point :

Torque of force F about a point r Fwhere F = force applied

P = point of application of forceQ = point about which we want to calculate the torque.



r = position vector of the point of application of force from the point about which wewant to determine the torque.

rFsin = r F = rFwhere = angle between the direction of force and the position vector of P wrt. Q.

r = perpendicular distance of line of action of force from point Q.

F = force armSI unit to torque is Nm

Torque is a vector quantity and its direction is determined using right hand thumb rule.

Ex.14 A particle of mass M is released in vertical plane from a point P at x = x0 on the x-axisit falls vertically along the y-axis. Find the torque acting on the particle at a time tabout origin?

Sol.

P

mg

x0O x

r

Torque is produced by the force of gravity

rF ksin

or r F x mg0

Ex.15 Calculate the total torque acting on the body shown in figure about the point O

15N37°

5N150°

20N

90°

10N

O

30°

Sol.

15N

37°

5N

150°

90°

10N

O

30°

15sin37°

20N

20sin30°

4cm

0 = 15sin37 × 6 + 20 sin 30° × 4 – 10 × 4= 54 + 40 – 40 = 54 N-cm

0 = 0.54 N-m



Ex.16 A particle having mass m is projected with a velocityv0 from a point P on a horizontal ground making anangle with horizontal. Find out the torque aboutthe point of projection acting on the particle when(a) it is at its maximum height ?

V0

P Q(b) It is just about to hit the ground back ?

Sol.

(a) Particle is at maximum height then about point P is p r F

F = mg ; rR2

r

v0

mg

P P =

R mg2

= mgv

g02 2

2sin

p = mv0

2 22sin

(b) when particle is at point Q then about point P is p r F'

r R ; F = mgP

Q

mgp mgR' = mg g

2sinv20

Ex.17 In the previous question, during the motion of particle from P to Q. Torque ofgravitational force about P is :(A) increasing (B) decreasing(C) remains constant (D) first increasing then decreasing

Sol. Torque of gravitational force about P is increasing because r is increasing from O to R.(Range)

5.2 Torque about axis :

r F

where = torque acting on the body about the axis of rotation

r = position vector of the point of application of force about the axis of rotation.

F = force applied on the body..

net 1 2 3 .....To understand the concept of torque about axis wetake a general example which comes out in daily life.Figure shows a door ABCD. Which can rotate aboutaxis AB. Now if we apply force. F at point. r ×

A D

y

x

B C

F

in inward direction then AB = r F and direction of this

AB is along y axis from right hand thumb rule. Whichis parallel to AB so gives the resultant torque.Now we apply force at point C in the direction as shown

figure. At this time F&r are perpendicular to each otherwhich gives

rFAB



But door can’t move when force is applied in this direction because the direction of AB isperpendicular to AB according to right hand thumb rule.

So there is no component of along AB which gives 0res

Now conclude Torque about axis is the component of Fr parallel to axis of rotation.

Note : The direction of torque is calculated using right hand thumb rule and it is alwaysperpendicular to the plane of rotation of the body.

×F3 r3 r1F1

F2

r2

If F1 or F2 is applied to body, body revolves in anti-clockwise direction and F3 makes bodyrevolve in clockwise direction. If all three are applied.

resul t F r F r F rtan –1 1 2 2 3 3 (in anti-clockwise direction)

6. BODY IS IN EQUILIBRIUM : -We can say rigid body is in equillibrium when it is in(a) Translational equilibrium

i.e. Fnet 0

Fnet x = 0 and Fnet y = 0 and(b) Rotational equillibrium

net 0

i.e., torque about any point is zero

Note :(i) If net force on the body is zero then net torque of the forces may or may not be zero.example.

A pair of forces each of same magnitude and acting in opposite direction on the rod.

A BFC

F 2

A F2(2) If net force on the body is zero then torque of the forces about each and every point is same

about B B F +F

B F2

about C C F2



Ex.18 Determine the point of application of third force for which body is in equillibrium whenforces of 20 N & 30 N are acting on the rod as shown in figure

A

20N

10cm C 20cm30N

B

Sol. Let the magnitude of third force is F, is applied in upward direction then the body is in theequilibrium when

(i) Fnet 0 (Translational Equillibrium)

20 + F = 30 F = 10 NSo the body is in translational equilibrium when 10 N force act on it in upward direction.

(ii) Let us assume that this 10 N force act.Then keep the body in rotational equilibriumSo Torque about C = 0i.e. c = 0 A C B

30N

10N20N

x

20cm 30 × 20 = 10 x

x = 60 cmso 10 N force is applied at 70 cm from point A to keep the body in equilibrium.

Ex.19 Determine the point of application of force, when forces are acting on the rod as shown infigure.

3N

5N5cm 5cm

10N

Sol. Since the body is in equillibrium so we conclude F net 0 and torque about any point is zero

i.e., net 0

3N

5N

AF

F1

F2 x 37°

10N6

8N

Let us assume that we apply F force downward at A angle from the horizontal, at x distancefrom B

From F net 0

Fnet x = 0 which givesF2 = 8 N

From Fnet y = 0 5 + 6 = F1 + 3F1 = 8 N

If body is in equillibrium then torque about point B is zero.3 × 5 + F1. x – 5 × 10 = 015 + 8x – 50 = 0

x = 359

x = 4.375 cm



Ex.20 A uniform rod length , mass m is hung from two strings of equal length from a ceilingas shown in figure. Determine the tensions in the strings ?

/ 4

A BSol. Let us assume that tension in left and right string is TA and TB respectively. Then

Rod is in equilibrium then 0Fnet & 0net

From 0Fnet

mg = TA + TB ...(1)

/ 4

A B

TBTA2/

mg

From net = 0 about A

0T43

2mg B

TB = 3mg2

from eq. (1) 3mg2TA = mg TA =

3mg

Ladder Problems :Ex.21 A stationary uniform rod of mass ‘m’, length ‘ ’ leans against a smooth vertical wall

making an angle with rough horizontal floor. Find the normal force & frictional forcethat is exerted by the floor on the rod?

smooth

roughSol. As the rod is stationary so the linear acceleration and angular acceleration of rod is zero.

i.e., acm = 0 ; = 0.

N = fN = mg

2

1a =0cm

Torque about any point of the rod should also be zero

Free Body Diagram

mg

N2

A

N1

Bf

= 0

A = 0 mg cos 2

+ f sin = N1 cos .

N1 cos = sin f + mgcos

2

f = mgcos

sin2 =

mgcot2



Ex.22 The ladder shown in figure has negligible mass and rests on a frictionless floor. Thecrossbar connects the two legs of the ladder at the middle. The angle between the twolegs is 60°. The fat person sitting on the ladder has a mass of 80 kg. Find the contanctforce exerted by the floor on each leg and the tension in the crossbar.

60°

N1m

1m

N

W

T

Sol. The forces acting on different parts are shown in figure. Consider the vertical equilibrium of“the ladder plus the person” system. The forces acting on this system are its weight (80 kg)g and the contact force N + N = 2 N due to the floor. Thus2 N = (80 kg) g or N = (40 kg) (9.8 m/s2) = 392 NNext consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting onit about the upper end,

N (2m) tan 30° = T (1m) or T = N 23 = (392 N) ×

23 = 450 N

Ex.23 A thin plank of mass m and length is pivoted at one end and it is held stationary inhorizontal position by means of a light thread as shown in the figure then find out theforce on the pivot.

Sol. Free body diagram of the plank is shown in figure. Plank is in equilibrium condition

N1

N2

O Amg

T

So Fnet & net on the plank is zero(i) from Fnet = 0

Fnet x = 0N1 = 0

Now 0Fynet

N2 + T = mg ...(i)from net = 0

net about point A is zeroso N2 . = mg . /2

N mg2 2



Ex.24 A square plate is hinged as shown in figure and it is held stationary by means of a lightthread as shown in figure. Then find out force exerted by the hinge.

square plate

Sol. F.B.D.

Body is in equilibrium and

mg

T

NT and mg force passing through one line so

from net = 0, N = 0



(A) MOMENT OF INERTIA1. The moment of inertia of a body dependsupon -(A) mass only(B) angular velocity only(C) distribution of particles only(D) mass and distribution of mass about the axisSol.

2. Two spheres of same mass and radius arein contact with each other. If the moment ofinertia of a sphere about its diameter is I, thenthe moment of inertia of both the spheresabout the tangent at their common point wouldbe -(A) 3I (B) 7I (C) 4I (D) 5ISol.

3. A disc of metal is melted to recast in theform of a solid sphere. The moment of inertiasabout a vertical axis passing through the centrewould -(A) decrease (B) increase(C) remains same (D) nothing can be said

Sol.

4. The M.I. of a disc about its diameter is 2units. Its M.I. about axis through a point onits rim and in the plane of the disc is(A) 4 units. (B) 6 units(C) 8 units (D) 10 unitsSol.

5. A solid sphere and a hollow sphere of thesame mass have the same moments of inertiaabout their respective diameters, the ratio oftheir radii is(A) (5)1/2 : (3)1/2 (B) (3)1/2 : (5)1/2

(C) 3 : 2 (D) 2 : 3Sol.

Exercise - I OBJECTIVE PROBLEMS (JEE MAIN)



6. A stone of mass 4kg is whirled in a horizontalcircle of radius 1m and makes 2 rev/sec. Themoment of inertia of the stone about the axisof rotation is(A) 64 kg × m2 (B) 4 kg × m2

(C) 16 kg × m2 (D) 1 kg × m2

Sol.

7. Three rings, each of mass P and radius Qare arranged as shown in the figure. Themoment of inertia of the arrangement aboutYY’ axis will be

(A) 72

PQ2 (B) 27

PQ2

(C) 25

PQ2 (D) 52

PQ2

Sol.

8. A circular disc A of radius r is made froman iron plate of thickness t and another circulardisc B of radius 4r is made from an iron plateof thickness t/4. The relation between themoments of inertia IA and IB is(A) IA > IB(B) IA = IB(C) IA < IB(D) depends on the actual values of t and r.Sol.

9. The moment of inertia of a uniform semicircularwire of mass M and radius r about a lineperpendicular to the plane of the wire throughthe centre is

(A) Mr2 (B) 12

2Mr (C) 14

2Mr (D) 25

2Mr

Sol.

10. Let IA and IB be moments of inertia of a bodyabout two axes A and B respectively. The axis Apasses through the centre of mass of the bodybut B does not.(A) IA < IB(B) If IA < IB, the axes are parallel.(C) If the axes are parallel, IA < IB(D) If the axes are not parallel, IA IB



Sol.

11. For the same total mass which of the followingwill have the largest moment of inertia about anaxis passing through its centre of mass andperpendicular to the plane of the body(A) a disc of radius a(B) a ring of radius a(C) a square lamina of side 2a(D) four rods forming a square of side 2aSol.

12. Moment of inertia of a thin semicircular disc(mass = M & radius = R) about an axis throughpoint O and perpendicular to plane of disc, isgiven by :

R

O

(A) 14

2MR (B) 12

2MR

(C) 18

2MR (D) MR2

Sol.

13. A rigid body can be hinged about any pointon the x-axis. When it is hinged such that thehinge is at x, the moment of inertia is given byI = 2x2 – 12x + 27 The x-coordinate of centre ofmass is(A) x = 2 (B) x = 0(C) x = 1 (D) x = 3Sol.

14. Consider the following statementsAssertion (A) : The moment of inertia of a rigidbody reduces to its minimum value as comparedto any other parallel axis when the axis of rotationpasses through its centre of mass.Reason (R) : The weight of a rigid body alwaysacts through its centre of mass in uniformgravitational field. Of these statements :(A) both A and R are true and R is the correctexplanation of A(B) both A and R are true but R is not a correctexplanation of A(C) A is true but R is false(D) A is false but R is trueSol.



15. A body is rotating uniformly about a verticalaxis fixed in an inertial frame. The resultant forceon a particle of the body not on the axis is(A) vertical(B) horizontal and skew with the axis(C) horizontal and intersecting the axis(D) none of theseSol.

16. One end of a uniform rod of mass m andlength I is clamped. The rod lies on a smoothhorizontal surface and rotates on it about theclamped end at a uniform angular velocity . Theforce exerted by the clamp on the rod has ahorizontal component(A) m 2l (B) zero

(C) mg (D) 2m21

Sol.

17. A rod of length 'L' is hinged from one end. Itis brought to a horizontal position and released.The angular velocity of the rod when it is in verticalposition is

(A) 2gL

(B) 3gL

(C) gL2

(D) gL

Sol.

(B) TORQUE AND PUREROTATIONAL MOTION

18. A disc of radius 2m and mass 200kg isacted upon by a torque 100N-m. Its angularacceleration would be(A) 1 rad/sec2 (B) 0.25 rad/sec2

(C) 0.5 rad/sec2. (D) 2 rad/sec2.Sol.

19. On applying a constant torque on a body-(A) linear velocity increases(B) angular velocity increases(C) it will rotate with constant angular velocity(D) it wil l move with constant velocitySol.

20. A wheel starting with angular velocity of10 radian/sec acquires angular velocity of 100radian/sec in 15 seconds. If moment of inertiais 10kg-m2, then applied torque (in newton-metre) is(A) 900 (B) 100 (C) 90 (D) 60



Sol.

21. An automobile engine develops 100H.P.when rotating at a speed of 1800 rad/min. Thetorque it delivers is(A) 3.33 W-s (B) 200W-s(C) 248.7 W-s (D) 2487 W-sSol.

22. The moment of inertia and rotational kineticenergy of a fly wheel are 20kg-m2 and 1000joule respectively. Its angular frequency perminute would be -

(A) 600

(B) 25

2 (C) 5

(D) 300

Sol.

23. The angular velocity of a body is

= 2 i + 3 j + 4 k and a torque

= i + 2 j + 3 k acts on it. The rotationalpower will be(A) 20 watt (B) 15 watt(C) 17 watt (D) 14 watt

Sol.

24. A torque of 2 newton-m produces anangular acceleration of 2 rad/sec2 a body. Ifits radius of gyration is 2m, its mass will be:(A) 2kg (B) 4 kg (C) 1/2 kg (D) 1/4 kgSol.

25. A particle is at a distance r from the axisof rotation. A given torque produces someangular acceleration in it. If the mass of theparticle is doubled and its distance from theaxis is halved, the value of torque to producethe same angular acceleration is(A) /2 (B) (C) 2 (D) 4



Sol.

26. A weightless rod is acted on by upward parallelforces of 2N and 4N ends A and B respectively.The total length of the rod AB = 3m. To keep therod in equilibrium a force of 6N should act in thefollowing manner :(A) Downwards at any point between A and B.(B) Downwards at mid point of AB.(C) Downwards at a point C such that AC = 1m.(D) Downwards at a point D such that BD = 1m.Sol.

27. A right triangular plate ABC of mass m is freeto rotate in the vertical plane about a fixedhorizontal axis through A. It is supported by astring such that the side AB is horizontal. Thereaction at the support A is :

A l

l

C

B

(A) mg3

(B) 2

3mg

(C) mg2

(D) mg

Sol.

28. In an experiment with a beam balance onunknown mass m is balanced by two known massm is balanced by two known masses of 16 kg and4 kg as shown in figure.

m16kg

l1 l2

m4kg

l1 l2

The value of the unknown mass m is(A) 10 kg (B) 6 kg (C) 8 kg (D) 12 kgSol.

29. A homogeneous cubical brick lies motionlesson a rough inclined surface. The half of the brickwhich applies greater pressure on the plane is :

(A) left half(B) right half(C) both applies equal pressure(D) the answer depend upon coefficient of frictionSol.



30. Consider the following statementsAssertion (A) : A cyclist always bends inwardswhile negotiating a curveReason (R) : By bending he lowers his centre ofgravity Of these statements,(A) both A and R are true and R is the correctexplanation of A(B) both A and R are true but R is not the correctexplanation of A(C) A is true but R is false(D) A is false but R is trueSol.

31. A rod is hinged at its centre and rotated byapplying a constant torque starting from rest.The power developed by the external torque as afunction of time is :

(A)

Pex t

time

(B)

Pex t

time

(C)

Pex t

time

(D)

Pex t

time

Sol.

32. A pulley is hinged at the centre and a masslessthread is wrapped around it. The thread is pulledwith a constant force F starting from rest. Asthe time increases,

F

(A) its angular velocity increases, but force onhinge remains constant(B) its angular velocity remains same, but forceon hinge increases(C) its angular velocity increases and force onhinge increases(D) its angular velocity remains same and forceon hinge is constant.

33. The angular momentum of a flywheel havinga moment of inertia of 0.4 kg m2 decreases from30 to 20 kg m2/s in a period of 2 second. Theaverage torque acting on the flywheel during thisperiod is :(A) 10 N.m (B) 2.5 N.m(C) 5 N.m (D) 1.5 N.mSol.



34. A particle starts from the point (0m, 8m) and

moves with uniform velocity of 3 i m/s. After 5seconds, the angular velocity of the particle aboutthe origin will be :

3m/s

8m

y

xO

(A) 8

289rad s/ (B)

38

rad s/

(C) 24289

rad s/ (D) 8

17rad s/

Sol.

(C) ANGULAR MOMENTUM35. The rate of change of angular momentumis called(A) angular velocity (B)angular acceleration(C) force (D) torqueSol.

36. The rotational kinetic energy of a rigidbody of moment of inertia 5 kg-m2 is 10 joules.The angular momentum about the axis ofrotation would be -(A) 100 joule-sec (B) 50 joule-sec(C) 10 joule-sec (D) 2 joule -sec

Sol.

37. The angular velocity of a body changesfrom one revolution per 9second to 1 revolutionper second without applying any torque. The ratioof its radius of gyration in the two cases is(A) 1 : 9 (B) 3 : 1(C) 9 : 1 (D) 1 : 3Sol.

38. A dog of mass m is walking on a pivoteddisc of radius R and mass M in a circle ofradius R/2 with an angular frequency n: thedisc will revolve in opposite direction withfrequency -

RR/2

(A) mnM

(B) mnM2

(C) 2mnM

(D) 2MnM

Sol.



39. A particle moves with a constant velocityparallel to the X-axis. Its angular momentum withrespect to the origin.(A) is zero(B) remains constant(C) goes on increasing(D) goes on decreasing.Sol.

40. A person sitting firmly over a rotating stoolhas his arms streatched. If he folds his arms, hisangular momentum about the axis of rotation(A) increases (B) decreases(C) remains unchanged (D) doubles.Sol.

41. A man, sitting firmly over a rotating stool hashis arms streched. If he folds his arms, the workdone by the man is(A) zero(B) positive(C) negative(D) may be positive or negative.

Sol.

42. A particle of mass 2 kg located at the position

( )i j m has a velocity 2( – )i j k m/s. Itsangular momentum about z-axis in kg-m2 /s is :(A) zero (B) +8(C) 12 (D) – 8Sol.

43. A ball of mass m moving with velocity v, collidewith the wall elastically as shown in the figure.After impact the change in angular momentumabout P is : P

d

(A) 2 mvd (B) 2 mvd cos(C) 2 mvd sin (D) zeroSol.



44. A uniform rod of mass M has an impulse appliedat right angles to one end. If the other end beginsto move with speed V, the magnitude of theimpulse is

(A) MV (B) MV2 (C) 2MV (D)

23MV

Sol.

(D) COMBINED TRANSLATIONAL+ ROTATIONAL MOTION

45. A circular ring of wire of mass M andradius R is making n revolutions/sec about anaxis passing through a point on its rim andperpendicular to its plane. The kinetic energyof rotation of the ring is given by-(A) 4 2MR2n2 (B) 2 2MR2n2

(C) 12

2MR2n2 (D) 8 2MR2n2

Sol.

46. Rotational kinetic energy of a disc ofconstant moment of inertia is -(A) directly proportional to angular velocity(B) inversely proportional to angular velocity(C) inversely proportional to square of angularvelocity(D) directly proportional to square of angularvelocity

Sol.

47. A circular disc has a mass of 1kg andradius 40 cm. It is rotating about an axispassing through its centre and perpendicularto its plane with a speed of 10rev/s. The workdone in joules in stopping it would be-(A) 4 (B) 47.5 (C) 79 (D) 158Sol.

48. A disc rolls on a table. The ratio of its K.E.of rotation to the total K.E. is -(A) 2/5 (B) 1/3 (C) 5/6 (D) 2/3Sol.



49. A disk and a ring of the same mass arerolling to have the same kinetic energy. Whatis ratio of their velocities of centre of mass(A) (4:3)1/2 (B) (3 : 4)1/2

(C) (2)1/2 : (3) 1/2 (D) (3)1/2 : (2)1/2

Sol.

50. If the applied torque is directly proportionalto the angular displacement , then the workdone in rotating the body through an angle would be - (C is constant of proportionality)

(A) C (B) 12 C (C)

12 C 2 (D) C 2

Sol.

51. The centre of a wheel rolling without slippingin a plane surface moves with speed v0. A particleon the rim of the wheel at the same level as thecentre will be moving at speed.(A) zero (B) v0 (C) 2v0 (D) 2v0

Sol.

52. A solid sphere, a hollow sphere and a disc, allhaving smooth incline and released. Least timewill be taken in reaching the bottom by(A) the solid sphere (B) the hollow sphere(C) the disc (D) all will take same time.Sol.

53. A wheel of radius r rolling on a straight line,the velocity of its centre being v. At a certaininstant the point of contact of the wheel withthe grounds is M and N is the highest point onthe wheel (diametrically opposite to M). Theincorrect statement is :(A) The velocity of any point P of the wheel isproportional to MP.(B) Points of the wheel moving with velocitygreater than v form a larger area of the wheelthan points moving with velocity less than v.(C) The point of contact M is instantaneously atrest.(D) The velocities of any two parts of the wheelwhich are equidistant from centre are equal.



Sol.

54. There is rod of length l. The velocities of itstwo ends are v1 and v2 in opposite directionsnormal to the rod. The di stance of theinstantaneous axis of rotation from v1 is :

(A) zero (B) v

v v2

1 2l (C)

21

1

vvv l

(D) l/2

Sol.

55. A ring of radius R rolls without sliding with aconstant velocity. The radius of curvature of thepath followed by any particle of the ring at thehighest point of its path will be(A) R (B) 2R (C) 4R (D) none

Sol.

56. The linear speed of a uniform spherical shellafter rolling down an inclined plane of vertical heighth from rest, is :

(A) 10

7gh

(B) 45gh

(C) 65gh

(D) 2gh

Sol.

57. A body kept on a smooth horizontal surfaceis pulled by a constant horizontal force applied atthe top point of the body. If the body rolls purelyon the surface, its shape can be :(A) thin pipe (B) uniform cylinder(C) uniform sphere (D) thin spherical shell



Sol.

58. A solid sphere with a velocity (of centre ofmass) v and angular velocity is gently placedon a rough horizontal surface. The frictional forceon the sphere :(A) must be forward (in direction of v)(B) must be backward (opposite to v)(C) cannot be zero(D) none of the aboveSol.

59. A cylinder is pure rolling up an incline plane.It stops momentarily and then rolls back. Theforce of friction.(A) on the cylinder is zero throughout the journey(B) is directed opposite to the velocity of thecentre of mass throughout the journey(C) is directed up the plane throughout the journey(D) is directed down the plane throughout thejourneySol.

60. A uniform circular disc placed on a roughhorizontal surface has initially a velocity v0 andan angular velocity 0 as shown in the figure.The disc comes to rest after moving some distance

in the direction of motion. Then vr

0

0is

v0

0

(A) 12

(B) 1 (C) 32

(D) 2

Sol.

61. A Cubical bloc of mass M and edge a slidesdown a rough inclined plane of inclination witha uniform velocity. The torque of the normal forceon the block about its centre has a magnitude.(A) zero (B) Mga

(C) Mga sin (D) 12

Mgasin

Sol.



(A) MOMENT OF INERTIA1. Three bodies have equal masses m. Body A issolid cylinder of radius R, body B is a square laminaof side R, and body C is a solid sphere of radius R.Which body has the smallest moment of inertiaabout an axis passing through their centre of massand perpendicular to the plane (in case of lamina)(A) A (B) B(C) C (D) A and C bothSol.

2. Two rods of equal mass m and length l lie alongthe x axis and y axis with their centres origin.What is the moment of inertia of both about theline x = y :

(A) ml2

3(B)

ml2

4

(C) ml2

12(D)

ml2

6Sol.

3. Moment of inertia of a rectangular plate aboutan axis passing through P and perpendicular tothe plate is I. Then moment of PQR about an axisperpendicular to the plane of the plate :

P Q

S R(A) about P = I/2 (B) about R = I/2(C) about P > I/2 (D) about R > I/2

Sol.

4. A thin uniform rod of mass M and length L hasits moment of inertia I1 about its perpendicularbisector. The rod is bend in the form of asemicircular arc. Now its moment of inertiathrough the centre of the semi circular arc andperpendicular to its plane is I2. The ratio of I1 : I2will be _________________(A) < 1 (B) > 1 (C) = 1 (D) can’t be saidSol.

5. A square plate of mass M and edge L is shown infigure. The moment of inertia of the plate aboutthe axis in the plane of plate passing through oneof its vertex making an angle 15° from horizontal is.

15°

axis

L

L

(A) ML2

12 (B)

1124

2ML (C)

712

2ML (D) none

Sol.

Exercise - II



Question No. 6 to 9 (4 questions)The figure shows an isosceles triangular plate ofmass M and base L. The angle at the apex is 90°.The apex lies at the origin and the base is parallelto X - axis.

Y

M

X6. The moment of inertia of the plate about thez-axis is

(A) ML2

12 (B)

ML2

24(C)

ML2

6 (D) none of these

Sol.

7. The moment of inertia of the plate about thex-axis is

(A) ML2

8 (B)

ML2

32 (C)

ML2

24(D)

ML2

6Sol.

8. The moment of inertia of the plate about itsbase parallel to the x-axis is

(A) ML2

18 (B)

ML2

36 (C)

ML2

24 (D) none of these

Sol.

9. The moment of inertia of the plate about they-axis is

(A) ML2

6(B) ML2

8

(C) ML2

24(D) none of these

Sol.

10. ABCD is a square plate with centre O. Themoments of inertia of the plate about the per-pendicular axis through O is I and about the axes1, 2, 3 & 4 are I1, I2, I3 & I4 respectively. It followsthat :

O

D4

C

3

B2

1

A

(A) I2 = I3 (B) I = I1 + I4(C) I = I2 + I4 (D) I1 = I3Sol.



(B) TORQUE & PUREROTATIONAL MOTION

11. A horizontal force F = mg/3 is applied on theupper surface of a uniform cube of mass ‘m’ andside ‘a’ which is resting on a rough horizontalsurface having s = 1/2. The distance betweenlines of action of ‘mg’ and normal reaction ‘N’ is :(A) a/2 (B) a/3 (C) a/4 (D) NoneSol.

12. A man can move on a horizontal planksupported symmetrically as shown. The variationof normal reaction on support A with distance xof the man from the end of the plank is bestrepresented by :

A B

1m 4m 1m

x=0

(A)

N

x(B)

N

x

(C)

N

x(D)

N

xSol.

13. A uniform cube of side ‘b’ and mass M rest ona rough horizontal table. A horizontal force F isapplied normal to one of the face at a point, at aheight 3b/4 above the base. What should be thecoefficient of friction ( ) between cube and tableso that is will tip about an edge before it startsslipping?

b3b/4

F

(A) 23

(B) 13

(C) 32

(D) none

Sol.

14. A solid cone hangs from a frictionless pivot

at the origin O, as shown. If i , j and k are unitvectors, and a, b, and c are positive constants,which of the following forces F applied to the rimof the cone at a point P results in a torque onthe cone with a negative component Z ?

b

z

x

yo

c

ki j

(A) F = ak , P is (0, b, –c)

(B) F = –ak , P is (0, –b, –c)

(C) F = a j , P is (–b, 0, –c)(D) NoneSol.



15. A block of mass m is attached to a pulleydisc of equal mass m, radius r by means of aslack string as shown. The pulley is hinged aboutits centre on a horizontal table and the block isprojected with an initial velocity of 5 m/s. Itsvelocity when the string becomes taut will be

(A) 3 m/s (B) 2.5 m/s(C) 5/3 m/s (D) 10/3 m/sSol.

16. A rod of weight w is supported by two paral-lel knife edges A and B and is in equilibrium in ahorizontal position. The knives are at a distanced from each other. The centre of mass of the rodis at a distance x from A.

(A) the normal reaction at A is wxd

(B) the normal reaction at A is w d x

d( )

(C) the normal reaction at B is wxd

(D) the normal reaction at B is w d x

d( )

Sol.

17. A block with a square base measuring axaand height h, is placed on an inclined plane. Thecoefficient of friction is . The angle of inclina-tion ( ) of the plane is gradually increased. Theblock will

(A) topple before sliding if ah

(B) topple before sliding if ah

(C) slide before toppling if ah

(D) slide before toppling if ah

Sol.

18. A body is in equilibrium under the influence ofa number of forces. Each force has a differentline of action. The minimum number of forces re-quired is(A) 2, if their lines of action pass through thecentre of mass of the body(B) 3, if their lines of action are not parallel(C) 3, if their lines of action are parallel(D) 4, if their lines of action are parallel and allthe forces have the same magnitudeSol.



19. A block of mass m moves on a horizontalrough surface with initial velocity v. The heightof the centre of mass of the block is h from thesurface. Consider a point A on the surface(A) angular momentum about A is mvh initially(B) the velocity of the block decreases at timepasses(C) torque of the forces acting on block is zeroabout A(D) angular momentum is not conserved about ASol.

20. Four point masses are fastened to the cor-ners of a frame of negligible mass lying in the xyplane. Let w be the angular speed of rotation.Then

y-axism

Mm

M x-axis

z-axis a

b

(A) rotational kinetic energy associated with agiven angular speed depends on the axis of rota-tion.(B) rotational kinetic energy about y-axis is inde-pendent of m and its value is Ma2 2

(C) rotational kinetic energy about z-axis dependson m and its value is (Ma2 + mb2) 2

(D) rotational kinetic energy about z-axis is inde-pendent of m and its value is Mb2 2

Sol.

21. A particle falls freely near the surface of theearth. Consider a fixed point O (not verticallybelow the particle) on the ground.(A) Angular momentum of the particle about O isincreasing(B) Torque of the gravitational force on the par-ticle about O is decreasing(C) The moment of inertia of the particle about Ois decreasing(D) The angular velocity of the particle about Ois increasingSol.



22. A rod hinged at one end is released from thehorizontal position as shown in the figure. Whenit becomes vertical its lower half separates withoutexerting any reaction at the breaking point. Thenthe maximum angle ‘ ’ made by the hinged upperhalf with the vertical is :

C B A

B

C

B

(A) 30° (B) 45° (C) 60° (D) 90°Sol.

(C) ANGULAR MOMENTUM23. If a person sitting on a rotating stool with hishands outstretched, suddenly lowers his hands,then his(A) Kinetic energy will decrease(B) Moment of inertia will decrease(C) Angular momentum will increase(D) Angular velocity will remain constantSol.

24. A man spinning in free space changes theshape of his body, eg. by spreading his arms orcurling up. By doing this, he can change his(A) moment of inertia(B) angular momentum(C) angular velocity(D) rotational kinetic energy

Sol.

25. A thin circular ring of mass 'M' and radius 'R'is rotating about its axis with a constant angularvelocity . Two objects each of mass m, areattached gently to the opposite ends of a diameterof the ring. The ring now rotates with an angularvelcoity.

(A) M

M m( ) (B) M

M m( )2

(C) M

M m( – )2 (D) ( )M m

M3

Sol.

26. A small bead of mass m moving with velocityv gets threaded on a stationary semicircular ringof mass m and radius R kept on a horizontal table.The ring can freely rotate about its centre. Thebead comes to rest relative to the ring. What willbe the final angular velocity of the system?

mv

RO

(A) v/R (B) 2v/R (C) v/2R (D) 3v/R



Sol.

27. A thin uniform straight rod of mass 2 kg andlength 1 m is free to rotate about its upper endwhen at rest. It receives an impulsive blow of 10Ns at its lowest point, normal to its length asshown in figure. The kinetic energy of rod just afterimpact is

10 NS

(A) 75 J (B) 100 J(C) 200 J (D) noneSol.

28. A child with mass m is standing at the edgeof a disc with moment of inertia I, radius R, andinitial angular velocity . See figure given below.The child jumps off the edge of the disc withtangential velocity v with respect to the ground.The new angular velocity of the disc is

v

(A) I

I

2 2– mv(B)

(I + mR )I

2 2 – mv2

(C) I – mvR

I(D) ( )I + mR

I

2 mvR

Sol.

Question No. 29& 30 (2 questions)A uniform rod is fixed to a rotating turntable sothat its lower end is on the axis of the turntableand it makes an angle of 20° to the vertical.(The rod is thus rotating with uniform angularvelocity about a vertical axis passing through oneend.) If the turntable is rotating clockwise asseen from above.

20°

29. What is the direction of the rod’s angularmomentum vector (calculated about its lower end)(A) vertically downwards(B) down at 20° to the horizontal(C) up at 20° to the horizontal(D) vertically upwardsSol.

30. Is there a torque acting on it, and if so inwhat direction?(A) yes, vertically (B) yes, horizontally(C) yes at 20° to the horizontal(D) noSol.



31. One ice skater of mass m moves with speed2v to the right, while another of the same massm moves with speed v toward the left, as shownin figure I. Their paths are separated by a distanceb. At t = 0, when they are both at x = 0, theygrasp a pole of length b and negligible mass. Fort > 0, consider the system as a rigid body of twomasses m separated by distance b, as shown infigure II. Which of the following is the correctformula for the motion after t = 0 of the skaterinitially at y = b/2 ?

2vm

b

(t<0)

y

x

vm

Figure 1

x

y

b/2

t=0

Figure II(A) x = 2vt, y = b/2(B) x = vt + 0.5b sin (3vt/b), y = 0.5b cos(3vt/b)(C) x = 0.5vt + 0.5b sin (3vt/b), y = 0.5b cos(3vt/b)(D) x = 0.5vt + 0.5b sin (6vt/b), y = 0.5b cos(6vt/b)Sol.

32. A uniform rod AB of length L and mass M islying on a smooth table. A small particle of massm strike the rod with a velocity v0 at point C at adistance x from the centre O. The particle comesto rest after collision. The value of x, so thatpoint A of the rod remains ststionary just aftercollision is :

C

Ox

B

A

m v0

(A) L/3 (B) L/6 (C) L/4 (D) L/12

Sol.

33. A uniform rod AB of mass m and length l is atrest on a smooth horizontal surface. An impulse Jis applied to the end B, perpendicular to the rodin the horizontal direction. Speed of particle P at

a distance l6

from the centre towards A of the

rod after time t mJl

12 is

(A) 2 Jm

(B) Jm2 (C)

Jm (D) 2

Jm

Sol.

34. A uniform rod of mass M is hinged at its upperend. A particle of mass m moving horizontallystrikes the rod at its mid point elastically. If theparticle comes to rest after collision find the valueof M/m = ?

M

vm

(A) 3/4 (B) 4/3(C) 2/3 (D) none



Sol.

35. Two equal masses each of mass M are joinedby a massless rod of length L. Now an impulse MVis given to the mass M making an angle of 30ºwith the length of the rod. The angular velocityof the rod just after imparting the impulse is

30°

MV

MM

(A) vL

(B) 2vL

(C) vL2 (D) none of these

Sol.

36. Two particles of equal mass m at A and B areconnected by a rigid light rod AB lying on a smoothhorizontal table. An impulse J is applied at A in theplane of the table and perpendicular at AB. Thenthe velocity of particle at A is :

(A) Jm2 (B)

Jm (C)

2Jm (D) zero

Sol.

(D) COMBINED TRANSLATIONAL+ ROTATIONAL MOTION

37. A ring rolls without slipping on the ground.Its centre C moves with a constant speed u. P isany point on the ring. The speed of P with re-spect to the ground is v.(A) 0 v 2u(B) v = u, if CP is horizontal(C) v = u, if CP makes an angle of 30º with thehorizontal and P is below the horizontal level of C

(D) v u2 , if CP is horizontal

Sol.

38. A yo-yo is resting on a perfectly rough hori-zontal table. Forces F1, F2 and F3 are appliedseparately as shown.The correct statement is

F1

F3F2

(A) when F3 is applied the centre of mass willmove to the right(B) when F2 is applied the centre of mass willmove to the left(C) when F1 is applied the centre of mass willmove to the right(D) when F2 is applied the centre of mass willmove to the right



Sol.

39. A disc of circumference s is at rest at a pointA on a horizontal surface when a constant hori-zontal force begins to act on its centre. BetweenA and B there is sufficient friction to preventslipping, and the surface is smooth to the right ofB. AB = s. The disc moves from A to B in time T.To the right of B,

Force

A B(A) the angular acceleration of the disc will dis-appear, linear acceleration will remain unchanged(B) linear acceleration of the disc will increase(C) the disc will make one rotation in time T/2(D) the disc will cover a distance greater than sin further time T.Sol.

40. A plank with a uniform sphere placed on it,rests on a smooth horizontal plane. Plank is pulledto right by a constant force F. If the sphere doesnot slip over the plank.

F

(A) acceleration of centre of sphere is less thanthat of the plank(B) acceleration of centre of sphere is greaterthan the plank because friction acts rightwardon the sphere(C) acceleration of the centre of sphere may betowards left(D) acceleration of the centre of sphere relativeto plank may be greater than that of the plankrelative to floorSol.

41. A hollow sphere of radius R and mass m isfully filled with water of mass m. It is rolled downa horizontal plane such that its centre of massmoves with a velocity v. If it purely rolls

(A) Kinetic energy of the sphere is 56

2mv

(B) Kinetic energy of the sphere is 45

2mv

(C) Angular momentum of the sphere about a

fixed point on ground is 83

mvR

(D) Angular momentum of the sphere about a

fixed point on ground is 145

mvR

Sol.



42. In the figure shown, the plank is being pulledto the right with a constant speed v. If the cylin-der does not slip then :

R

v

(A) the speed of the centre of mass of the cylin-der is 2v(B) the speed of the centre of mass of the cylin-der is zero(C) the angular velocity of the cylinder is v/R(D) the angular velocity of the cylinder is zeroSol.

,43. If a cylinder is rolling down the incline withsliding(A) after some time it may start pure rolling(B) after sometime it will start pure rolling(C) it may be possible that it will never start purerolling(D) none of theseSol.

44. Which of the following statements are cor-rect(A) friction acting on a cylinder without sliding onan inclined surface is always upward along theincline irrespective of any external force actingon it.(B) friction acting on a cylinder without sliding onan inclined surface is may be upward may bedownwards depending on the external force act-ing on it.(C) friction acting on a cylinder rolling withoutsliding may be zero depending on the externalforce acting on it.(D) nothing can be said exactly about it as itdepends on the friction coefficient on inclinedplaneSol.

Question No. 45 to 47 (3 Questions)A cylinder and a ring of same mass M and radiusR are placed on the top of a rough inclined planeof inclination . Both are released simultaneouslyfrom the same height h.

45. Choose the correct statement(s) related tothe motion of each body(A) The friction force acting on each bodyopposes the motion of its centre of mass(B) The friction force provides the necessary torqueto rotate the body about its centre of mass(C) without friction none of the two bodies can roll(D) The friction force ensures that the point ofcontact must remain stationarySol.



46. Identify the correct statement(s)(A) The friction force acting on the cylinder maybe more than that acting on the ring(B) The friction force acting on the ring may bemore than that acting on the cylinder(C) If the friction is sufficient to roll the cylinderthen the ring will also roll(D) If the friction is sufficient to roll the ring thenthe cylinder will also rollSol.

47. When these bodies roll down to the foot ofthe inclined plane, then(A) the mechanical energy of each body is con-served(B) the velocity of centre of mass of the cylinder

is 23gh

(C) the velocity of centre of mass of the ring isgh

(D) the velocity of centre of mass of each bodyis 2ghSol.

Question No. 48 to 51 (4 Questions)A ring of mass M and radius R sliding with a ve-locity v0 suddenly enters into rough surface wherethe coefficient of friction is , as shown in figure.

v0

Rough ( )48. Choose the correct statement(s)(A) As the ring enters on the rough surface, thelimiting friction force acts on it(B) The direction of friction is opposite to thedirection of motion(C) The friction force accelerates the ring in theclockwise sense about its centre of mass(D) As the ring enters on the rough surface itstarts rollingSol.

49. Choose the correct statement(s)(A) The momentum of the ring is conserved(B) The angular momentum of the ring is con-served about its centre of mass(C) The angular momentum of the ring conservedabout any point on the horizontal surface(D) The mechanical energy of the ring is con-servedSol.



50. Choose the correct statement(s)(A) The ring starts its rolling motion when thecentre of mass stationary(B) The ring starts rolling motion when the pointof contact becomes stationary(C) The time after which the ring starts rolling is

vg

0

2

(D) The rolling velocity is v0

2Sol.

51. Choose the correct alternative(s)(A) The linear distance moved by the centre of

mass before the ring starts rolling is 38

02vg

(B) The net work done by friction force is 38 0

2mv

(C) The loss is kinetic energy of the ring is mv0

2

4

(D) The gain in rotational kinetic energy is mv0

2

8Sol.

52. Consider a sphere of mass ‘m’ radius ‘R’ doingpure rolling motion on a rough surface havingvelocity v0 as shown in the Figure. It makes anelastic impact with the smooth wall and movesback and starts pure rolling after some time again.

v0

O(A) Change in angular momentum about ‘O’ in theentire motion equals 2mv0 R in magnitude.(B) Moment of impulse provided by wall duringimpact about O equals 2mv0R in magnitude

(C) Final velocity of ball will be 37 0v

(D) Final velocity of ball will be 37 0v

Sol.

53. A solid sphere, a hollow sphere and a disc, allhaving same mass and radius, are placed at thetop of an incline and released. The frictioncoefficients between the objects and the inclineare same and not sufficient to allow pure rolling.The smallest kinetic energy at the bottom of theincline will be achieved by(A) the solid sphere (B) the hollow sphere(C) the disc(D) all will achieve same kinetic energy.Sol.



54. Fig. shows a smooth inclined plane fixed in acar accelerating on a horizontal road. The angleof incline is related to the acceleration a of thecar as a = g tan . If the sphere is set in purerotation on the incline.

a

(A) it will continue pure rolling(B) it will slip down the plane(C) its linear velocity will increase(D) its linear velocity will decrease.Sol.

55. A straight rod of length L is released on africtionless horizontal floor in a vertical position.As it falls + slips, the distance of a point on therod from the lower end, which follows a quartercircular locus is(A) L/2 (B) L/4(C) L/8 (D) NoneSol.

56. A ladder of length L is slipping with its endsagainst a vertical wall and a horizontal floor. At acertain moment, the speed of the end in contactwith the horizontal floor is v and the ladder makesan angle = 30º with the horizontal. Then thespeed of the ladder’s center must be

(A) 2 3v / (B) v/2(C) v (D) noneSol.

57. In the previous question, if dv/dt = 0, thenthe angular acceleration of the ladder when =45º is(A) 2v2/L2 (B) v2/2L2

(C) 2 2 2[ / ]v L (D) NoneSol.

58. A time varying force F = 2t is applied on aspool rolling as shown in figure. The angularmomentum of the spool at time t about bottommost point is :

R

rF=2t

(A) r tR

2 2(B)

( )R rr

t2

2

(C) (R + r)t2 (D) data is insufficient



Sol.

59. A ball rolls down an inclined plane, figure.The ball is first released from rest from P andthen later from Q. Which of the followingstatement is/ are correct ?

Q

2hh

P

O(i) The ball takes twice as much time to roll fromQ to O as it does to roll from P to O.(ii) The acceleration of the ball at Q is twice aslarge as the acceleration at P.(iii) The ball has twice as much K.E. at O whenrolling from Q as it does when rolling from P.(A) i, ii only (B) ii, iii only(C) i only (D) iii onlySol.

60. Starting from the rest, at the same time, aring, a coin and a solid ball of same mass rolldown an incline without slipping. The ratio of theirtranslational kinetic energies at the bottom willbe(A) 1 : 1 : 1 (B) 10 : 5 : 4(C) 21 : 28 : 30 (D) none

Sol.

61. In the figure shown a ring A is initially rollingwithout sliding with a velocity v on the horizontalsurface of the body B (of same mass as A). Allsurfaces are smooth. B has no initial velocity.What will be the maximum height reached by Aon B.

A v

B Smooth

(A) 34

2vg (B)

vg

2

4

(C) vg

2

2 (D) vg

2

3Sol.



62. Inner and outer radii of a spool are r and Rrespectively. A thread is wound over its innersurface and placed over a rough horizontalsurface. Thread is pulled by a force F as shownin fig. then in case of pure rolling

F

(A) Thread unwinds, spool rotates anticlockwiseand friction act leftwards(B) Thread winds, spool rotates clockwise andfriction acts leftwards(C) Thread winds, spool moves to the right anffriction act rightwards(D) Thread winds, spool moves to the right andfriction does not come into existence.Sol.

63. Portion AB of the wedge shown in figure isrough and BC is smooth. A solid cylinder rollswithout slipping from A to B. The ratio oftranslational kinetic energy to rotational kineticenergy, when the cylinder reaches point C is :

A

B

AB=BCD C

(A) 3/4 (B) 5 (C) 7/5 (D) 8/3

Sol.

64. A plank of mass M is placed over smoothinclined plane and a sphere is also placed overthe plank. Friction is sufficient between sphereand plank. If plank and sphere are released fromrest, the frictional force on sphere is :

(A) up the plane (B) down the plane(C) horizontal (D) zeroSol.



65. A plank with a uniform sphere placed on itrests on a smooth horizontal plane. Plank is pulledto right by a constant force F. If sphere does notslip over the plank. Which of the following isincorrect.

F

(A) Acceleration of the centre of sphere is lessthan that of the plank(B) Work done by friction acting on the sphere isequal to its total kinetic energy.(C) Total kinetic energy of the system is equal towork done by the force F(D) None of the aboveSol.

66. A ring of mass m and radius R has threeparticles attached to the ring as shown in thefigure. The centre of the ring has speed v0. Thekinetic energy of the system is (Slipping is absent)

2mm

m

(A) 6mv02 (B) 12 mv0

2

(C) 4 mv02 (D) 8 mv0

2

Sol.

67. A uniform sphere of radius R is placed on arough horizontal surface and given a linear velocityv0 angular velocity 0 as shown. The sphere comesto rest after moving some distance to the right.It follows that :

0

v0

(A) v0 = 0R (B) 2v0 = 5 0R(C) 5v0 = 2 0R (D) 2v0 = 0RSol.



(A) MOMENT OF INERTIA1. Find the moment of inertia of a uniform half-disc about an axis perpendicular to the plane andpassing through its centre of mass. Mass of thisdisc is M and radius is R.Sol.

2. Find the moment of inertia of a pair of solidspheres, each having a mass m and radius r, keptin contact about the tangent passing throughthe point of contact.Sol.

3. Find the radius of gyration of a circular ring ofradius r about a line perpendicular to the plane ofthis ring and tangent to the ring.

Sol.

4. Moment of inertial of a triangle plane of massM shown in figure about vertical axis AB is :

45°

l

l m

A

B

Sol.

Exercise - III (JEE ADVANCED)



5. A uniform rod of mass m is bent into the formof a semicircle of radius R. The moment of inertiaof the rod about an axis passing through A andperpendicular to the plane of the paper is

AR

Sol.

(B) TORQUE & PURE ROTATIONALMOTION

6. A simple pendulum of length is pulled aside tomade an angle with the vertical. Find themagnitude of the torque of the weight w of thebob about the point of suspension. When is thetorque zero ?Sol.

7. Two forces F i j k1 2 5 6– – and F i j k2 2– –are acting on a body at the points (1, 1, 0) and(0, 1, 2). Find torque acting on the body aboutpoint (–1, 0, 1).Sol.

8. Assuming frictionless contacts, determine themagnitude of external horizontal force P appliedat the lower end for equilibrium of the rod. Therod is uniform and its mass is 'm'.

Wall

P

Sol.



9. A rod of mass m and length L, lying horizontally,is free to rotate about a vertical axis through itscentre. A horizontal force of constant magnitudeF acts on the rod at a distance of L/4 from thecentre. The force is always perpendicular to therod. Find the angle rotated by the rod during thetime t after the motion starts.Sol.

10. The uniform rod AB of mass m is releasedfrom rest when = 60°. Assuming that the frictionforce between end A and the surface is largeenough to prevent sliding, determine (for theinstant just after release)

L

A

B

(a) The angular acceleration of the rod(b) The normal reaction and the friction force at A.(c) The minimum value of , compatible with thedescribed motion.Sol.

11. Figure shows two blocks of mass m and mconnected by a string passing over a pulley. Thehorizontal table over which the mass m slides issmooth. The pulley (uniform disc) has mass mand it can freely rotate about this axis. Find theacceleration of the mass m assuming that thestring does not slip on the pulley.

mm

m

Sol.

12. A solid cylinder of mass M = 1kg & radius R =0.5m is pivoted at its centre & has three particlesof mass m = 0.1 kg mounted at its perimeter asshown in the figure. The system is originally atrest. Find the angular speed of the cylinder, whenit has swung through 90° in anticlockwisedirection.

Sol.



13. A cube is in limiting equilibrium on an inclinedplane forming an angle of 30° with the horizontal.The line of action of the normal reaction of theplane on the cube isSol.

14. A body weighs 6 gms when placed in one panand 24 gms when placed on the other pan of afalse balance. If the beam is horizontal when boththe pans are ampty, the true weight of the bodyis :Sol.

15. An inverted “V” is made up of two uniformboards each weighing 200 N. Each side has thesame length and makes an angle 30° with thevertical as shown in figure. The magnitude of thestatic frictional force that acts on each of thelower end of the V is

l

P

30°30°

Sol.

16. A uniform sphere of weight W and radius 5cm is being held by a string as shown in thefigure. The wall is smooth. The tension in thestring will be

8cm

Sol.



17. A light string is wrapped around a cylinder ofmass ‘m and radius ‘R’. The string is pulled verticallyupward to prevent the centre of mass from fallingas the cylinder unwinds the string. Then lengthof the string unwound when the cylinder hasreached a speed will be :Sol.

18. The moment of inertia of the pulley systemas shown in the figure is 4 kgm2. The radii ofbigger and smaller pulleys 2m and 1m respectively.The angular acceleration of the pulley system is

1m2m

5kg4kg

Sol.

19. The two small spheres each have a mass of 3kg and are attached to the rod of negligible mass.A torque M = 8t Nm, where t is in seconds isapplied to the rod. Find the value of time wheneach sphere attains a speed of 3 m/s startingfrom rest.

1m 1m3kg 3kg

MSol.

20. A rectangular plate of mass 20 kg is suspendedfrom points A and B as shown. If pin B is removeddetermine the initial angular acceleration (in rad/s2) of plate. (g = 10m/s2)

0.2m

A B0.15m

Sol.

21. A solid homogeneous cylinder of height h andbase radius r is kept vertically on a conveyer beltmoving horizontally with an increasing velocityv = a + bt2. If the cylinder is not allowed to slipfind the time when the cylinder is about to topple.Sol.



22. A square frame made up of a wire of mass m& length l is held in horizontal plane. It is free torotate about AD. If the frame is released, thework done by gravity during the time framerotates through an angle of 90° is equal to

A

B

C

DP

Sol.

23. Three equal masses m are rigidly connectedto each other by massless rods of length l formingan equilateral triangle, as shown above. Theassembly is to be given an angular velocity about an axis perpendicular to the triangle. Forfixed , the ratio of the kinetic energy of theassembly for an axis through B compared withthat for an axis through A is equal to

ll

l

m

mB mA

Sol.

24. In the figure A & B are two blocks of mass 4kg & 2 kg respectively attached to the two endsof a light string passing over a disc C of mass 40kg and radius 0.1m. The disc is free to rotateabout a fixed horizontal axes, coinciding with itsown axis. The system is released from rest andthe string does not slip over the disc. Find :

A

B(i) the linear acceleration of mass B.(ii) the number of revolutions made by the discat the end of 10 sec. from the start.(iii) the tension in the string segment supportingthe block A.Sol.



25. A mass m is attached to a pulley through acord as shown in the fig. The pulley is a solid diskwith radius R. The cord does not slip on the disk.The mass is released from rest at a height h fromthe ground and at the instant the mass reachesthe ground, the disk is rotating with angularvelocity . Find the mass of the disk.

m

R

h

Sol.

(C) ANGULAR MOMENTUM26. A particle having mass 2 kg is moving alongstraight line 3x+ 4 y = 5 with speed 8m/s. Findangular momentum of the particle about origin, xand y are in meters.Sol.

27. A particle having mass 2 kg is moving with

velcoity ( ) /2 3i j m s . Find angular momentum of

the particle about origin when it is at (1, 1, 0).Sol.

28. A uniform square plate of mass 2.0 kg andedge 10 cm rotates about one of its diagonalsunder the action of a constant torque of 0.10N.m. Calculate the angular momentum and thekinetic energy of the plate at the end of the fifthsecond after the start.Sol.



29. A wheel of moment of inertia 0.500 kg-m2

and radius 20.0 cm is rotating about its axis atan angular speed of 20.0 rad/s. It picks up astationary particle of mass 200 g at its edge.Find new angular speed of the wheel.Sol.

30. A uniform circular disc can rotate freely abouta rigid vertical axis through its centre O. A manstands at rest at A on the edge due east of O.The mass of the disc is 22 times the mass of theman. The man starts walking anticlockwise. Whenhe reaches the point A after completing onerotation relative to the disc he will be :Sol.

31. Two identical disks are positioned on a verticalaxis. The bottom disk is rotating at angularvelocity 0 and has rotational kinetic energy KE0.The top disk is initially at rest. It is allowed tofall, and sticks to the bottom disk. What is therotational kinetic energy of the system after thecollision?

0

Sol.

32. A uniform ring is rotating about vertical axiswith angular velocity initially. A point insect (S)having the same mass as that of the ring startswalking from the lowest point P1 and finallyreaches the point P2 (as shown in figure). Thefinal angular velocity of the ring will be equal to

90°P2

O

axis of rotation

SP1

Sol.



33. A particle of mass 10 kg is moving with auniform speed of 6m/sec. in x-y plane along theline 3y = 4x+ 10 the magnitude of its angularmomentum about the origin in kg –m2/s isSol.

(D) COMBINED TRANSLATIONAL +ROTATIONAL MOTION

34. A sphere of mass m rolls on a plane surface.Find its kinetic energy at an instant when itscentre moves with speed v.Sol.

35. A cylinder rolls on a horizontal plane surface.If the speed of the centre is 25 m/s, what is thespeed of the highest point ?Sol.

36. A small spherical ball is released from a pointat a height h on a rough track shown in figure.Assuming that it does not slip anywhere, find itslinear speed when it rolls on the horizontal partof the track.

h

Sol.



37. A sphere starts rolling down an incline ofinclination . Find the speed of its centre when ithas covered a distance .Sol.

38. A solid uniform sphere of mass m is releasedfrom rest from the rim of a hemispherical cup sothat it rolls without sliding along the surface. Ifthe rim of the hemisphere is kept horizotnal, findthe normal force exerted by the cup on the ballwhen the ball reaches the bottom of the cup.Sol.

39. Two small spheres A & B respectively of massm & 2m are connected by a rigid rod of length &negligible mass. The two spheres are resting on ahorizontal, frictionless surface. When A is suddenlygiven the velocity v0 as shown. Find velocities ofA & B after the rod has rotated through 180°.

v0A

BSol.

40. A uniform rod of mass m and length is struckat an end by a force F perpendicular to the rodfor a short time interval t. Calculate(a) the speed of the centre of mass,(b) the angular speed of the rod about the centreof mass,(c) the kinetic energy of the rod and(d) the angular momentum of the rod about thecentre of mass after the force has stopped toact. Assume that t is so small that the rod doesnot appreciably change its direction while the forceacts.Sol.



41. A hollow cylinder with inner radius R, outerradius 2R mass M is rolling with speed of its axis v.Its kinetic energy is

R

Sol.

42. The cylinder shown, with mass M and radiusR, has a radially dependent density. Thecylinder starts from rest and rolls withoutslipping down an inclined plane of height H. Atthe bottom of the plane of height H. At thebottom of the plane its translational speed is(8gH/7)1/2. Which of the following is therotational inertia of the cylinder?

MR

H

Sol.



1. A thin uniform rod of mass M and length L ishinged at its upper end, and released from rest ina horizontal position. The tension at a pointlocated at a distance L/3 from the hinge point,when the rod becomes vertical, will be

2. A rigid horizontal smooth rod AB of mass 0.75kg and length 40 cm can rotate freely about afixed vertical axis through its mid point O. Tworings each of mass 1 kg are initially at rest adistance of 10 cm from O on either side of therod. The rod is set in rotation with an angularvelocity of 30 radians per second. The velocityof each ring along the length of the rod in m/sthen they reach the ends of the rod is

C DBA

O

3. A straight rod AB of mass M and length L isplaced on a frictionless horizontal surface. Ahorizontal force having constant magnitude F anda fixed direction starts acting at the end A. Therod is initially perpendicular to the force. The initialacceleration of end B is

4. A wheel is made to roll without slipping, towardsright, by pulling a string wrapped around a coaxialspool as shown in figure. With what velocity thestring should be pulled so that the centre of thewheel moves with a velocity of 3 m/s?

C0.3m

0.1m

B StringA

5. A solid uniform disk of mass m rolls withoutslipping down a fixed inclined plane with anacceleration a. The frictional force on the diskdue to surface of the plane is :

6. A carpet of mass ‘M’ made of inextensiblematerial is rolled along its length in the form of acylinder of radius ‘R’ and is kept on a rough floor.The carpet starts unrolling without sliding on thefloor when a negligibly small push is given to it.The horizontal velocity of the axis of the cylindricalpart of the carpet when its radius reduces to R/2will be :

R/2

7. A slightly loosely fit window is balanced bytwo strings which are connected to weights w/2each. The strings pass over the frictionless pulleysas shown in the figure. The strings are tied almostat the corner of the window. The string on theright is cut and then the window acceleratesdownwards. If the coefficients of friction betweenthe window and the side supports is thencalculate the acceleration of the window in termsof , a, b and g, where a is width and b is thelength of the window.

ww/2b

a

w/2

fixed window support

8. A uniform wood door has mass m, height h,and width w. It is hanging from two hingesattached to one side; the hinges are located h/3and 2h/3 from the bottom of the door. Supposethat m = 20.0 kg, h = 2.20 m, and w = 1.00 mand the bottom smooth hinge is not screwed intothe door frame. Find the forces acting on thedoor.

h

W

Hingescom

9. A hole of radius R/2 is cut from a solid sphereof radius R. If the mass of the remaining plate isM, then moment of inertia of the body about anaxis through O perpendicular to plane is_________.

RO R/2



10. A uniform beam of length L and mass m issupported as shown. If the cable suddenly breaks,determine ;

L

(1/4)L

A B

(a) the acceleration of end B.(b) the reaction at the pin support.

11. A thin rod AB of length a has variable mass

per unit length 0 1 xa where x is the distance

measured from A and 0 is a constant.(a) Find the mass M of the rod.(b) Find the position of centre of mass of therod.(c) Find moment of inertia of the rod about anaxis passing through A and perpendicular to AB.Rod is freely pivoted at A and is hanging inequilibrium when it is struck by a horizontal impulseof magnitude P at the point B.(d) Find the angular velocity with which the rodbegins to rotate.(e) Find minimum value of impulse P if B passesthrough a point vertically above A.

12. Two separate cylinders of masses m (= 1kg)and 4m and radii R(=10cm) and 2R rotating inclockwise direction with 1 = 100 rad/sec and 2= 200 rad/sec. Now they are held in contact witheach other as in fig. Determine their angularvelocities after the slipping between the cylindersstops.

13. A circular disc of mass 300 gm and radius 20cm can rotate freely about a vertical axis passingthrough its centre of O. A small insect of mass100 gm is initially at a point A on the disc (whichis initially stationary) the insect starts walkingfrom rest along the rim of the disc with such atime varying relative velocity that the disc rotatesin the opposite direction with a constant angularacceleration = 2 rad/s2. After some time T, theinsect is back at the point A. By what angle hasthe disc rotated till now ; as seen by a stationaryearth observer ? Also find the time T.

14. A spool of inner radius R and outer radius 3Rhas a moment of inertia = MR2 about an axispassing through its geometric centre, where M isthe mass of the spool. A thread woudn on theinner surface of the spool is pulled horizontallywith a constant force = Mg. Find the accelerationof the point on the thread which is being pulledassuming that the spool rolls purely on the floor.

15. A sphere of mass m and radius r is pushedonto the fixed horizontal surface such that it rollswithout slipping from the beginning. Determinethe minimum speed v of its mass centre at thebottom so that it rolls completely around the loopof radius (R + r) without leaving the track inbetween.

(R+r)Sphere

r V

16. Two uniform cylinders, each of mass m = 10kg and radius r = 150 mm, are connected by arough belt as shown. If the system is releasedfrom rest, determine

r

r

(a) the velocity of the centre of cylinder A afterit has moved through 1.2 m &(b) the tension in the portion of the beltconnecting the two cylinders.

17. A uniform rod of mass m and length l is restingon a smooth horizontal surface. A particle of massm/2 travelling with a speed v0 hits the rod normallyand elastically. Find final velocity of particle andthe angular velocity of the rod.

v0

m/2Top view

(m, )lRodCl/4



18. One side of a spring of initial, unstretchedlength l0 = 1m, lying on a frictionless table, isfixed, the other one is fastened to a small puck ofmass m = 0.1kg. The puck is given velocity in adirection perpendicular to the spring, at an initialspeed v0 = 11 m/s. In the course of the motion,the maximum elongation of the spring is l = l0/10.What is the force constant of the spring (in SIunits) ?

ml0

v0

19. A block X of mass 0.5 kg is held by a longmassless string on a frictionless inclined plane ofinclination 30º to the horizontal. The string iswound on a uniform solid cylindrical drum Y ofmass 2kg and of radius 0.2 m as shown in thefigure. The drum is given an initial angular velocitysuch that the block X starts moving up the plane.

X

Y

(i) Find the tension in the string during the motion(ii) At a certain instant of time the magnitude ofthe angular velocity of Y is 10 rad/sec. Calculatethe distance travelled by X from that instant oftime until it comes to rest.



Exercise - IV PREVIOUS YEAR QUESTIONS

LEVEL - I JEE MAIN

1. A solid sphere, a hollow sphere and a ring arereleased from top of an inclined plane (frictionless)so that they slide down the plane. Then maximumacceleration down the plane is for (no rolling)

[AIEEE 2002](A) solid sphere (B) hollow sphere(C) ring (D) All sameSol.

2. Moment of inertia of a circular wire of mass Mand radius R about its diameter is [AIEEE 2002]

(A) 2

2MR

(B) MR2 (C) 2MR2 (D) 2

4MR

Sol.

3. A particle of mass m moves along line PC withvelocity v as shown. What is the angularmomentum of the particle about O ?

[AIEEE 2002]

C

P

O

r

L

l

(A) mvL (B) mvl (C) mvr (D) zeroSol.

4. Initial angular velocity of a circular disc ofmass M is 1. Then two small spheres of mass mare attached gently to two diametrically oppositepoints on the edge of the disc. What is the finalangular velocity of the disc ? [AIEEE 2002]

(A) 1M m

M (B) 1M m

m

(C) 14M

M m (D) 12M

M mSol.



5. Let F be the force acting on a particle having

position vector r and be the torque of this

force about the origin. Then [AIEEE 2003]

(A) r . 0 and F. 0

(B) r . 0 and F. 0

(C) r . 0 and F. 0

(D) r . 0 and F. 0Sol.

6. A particle performing uniform circular motionhas angular momentum L. If its angular frequencyis doubled and its kinetic energy halved, then thenew angular momentum is [AIEEE 2003]

(A) 4L

(B) 2L (C) 4L (D) L/2

Sol.

7. A circular disc X of radius R is made from aniron plate of thickness t, and another disc y ofradius 4R is made from an iron plate of thicknesst/4. Then the relation between the moment ofinertia Ix and IY is [AIEEE 2003]

(A) 32Y xI I (B) 16Y XI I

(C) Y XI I (D) 64Y XI I

Sol.

8. A solid sphere is rotating in free space. If theradius of the sphere is increased keeping masssame, which one of the following will not beaffected ? [AIEEE 2004](A) Moment of inertia(B) Angular momentum(C) Angular velocity(D) Rotational kinetic energySol.

9. One solid sphere A and another hollow sphereB are of same mass and same outer radii. Theirmoment of inertia about their diameters arerespectively IA and IB such that [AIEEE 2004]

(A) A BI I (B) A BI I

(C) A BI I (D) A A

B B

I dI d

Where dA and dB are their densities.Sol.



10.A T shaped object with dimensions shown in

the figure, is lying on a smooth floor. A force Fis applied at the point P parallel to AB, such thatthe object has only the translational motionwithout rotation. Find the location of P withrespect to C. [AIEEE 2005]

A B

C

P2l

l

F

(A) 23

l (B) 32

l (C) 43

l (D) l

Sol.

11. The moment of inertia of uniform semicirculardisc of mass M and radius r about a l ineperpendicular to the plane of the disc throughthe centre is [AIEEE 2005]

(A) 214

Mr (B) 225

Mr (C) 2Mr (D) 212

Mr

Sol.

12.An angular ring with inner and outer radii R1and R2 is rolling without slipping with a uniformangular speed. The ratio of the forces experiencedby the two particles situated on the inner and

outer parts of the ring, 1

2

FF is [AIEEE 2005]

(A) 2

1

RR (B)

2

1

2

RR (C) 1 (D)

1

2

RR

Sol.

13.A thin circular ring of mass m and radius R isrotating about its axis with a constant angularvelocity . Two objects each of mass M areattached gently to the opposite ends of a diameterof the ring. The ring now rotates with an angularvelocity ' = [AIEEE 2006]

(A) ( 2 )m M

m (B) ( 2 )

( 2 )m M

m M

(C) ( )m

m M (D) ( 2 )m

m MSol.

14.Four point masses, each of value m, are placedat the corners of a square ABCD of side l. Themoment of inertia of this system about an axispassing through A and parallel to BD is

[AIEEE 2006]

(A) 22ml (B) 23ml (C) 23ml (D) 2ml



Sol.

15.Angular momentum of the particle rotating witha central force is constant due to [AIEEE 2007](A) constant force(B) constant linear momentum(C) zero torque(D) constant torqueSol.

16.A round uniform body of radius R, mass M andmoment of inertia I, rolls down (without slipping)an inclined plane making an angle with thehorizontal. Then its acceleration is [AIEEE 2007]

(A) 2

sin1 /

gI MR (B) 2

sin1 /

gMR I

(C) 2

sin1 /

gI MR

(D) 2

sin1 /

gMR I

Sol.

17.For the given uniform square lamina ABCD,whose centre is O [AIEEE 2007]

A B

CD

E

F

O

(A) 2 AC EFI I (B) 3AD EFI I

(C) 4AD EFI I (D) 2AD EFI ISol.

18. Consider a uniform square plate of side a andmass m. The moment of inertia of this plate aboutan axis perpendicular to its plane and passingthrough one of its corners is [AIEEE 2008]

(A) 256

ma (B) 2112

ma (C) 2712

ma (D) 223

ma

Sol.



19.A thin uniform rod of length l and mass m isswinging freely about a horizontal axis passingthrough its end. Its maximum angular speed is .Its centre of mass rises to maximum height of

[AIEEE 2009]

(A) 2 21

3l

g (B) 16

lg

(C) 2 21

2l

g (D) 2 21

6l

gSol.

20.A pulley of radius 2 m is rotated about its axisby a force F = (20t – 5t2) N (where t is measuredin seconds) applied tangentially. It the momentof inertia of the pulley about its axis of rotation is10 kg-m2 the number of rotations made by thepulley before its direction of motion if reserved,is [AIEEE 2011](A) more than 3 but less than 6(B) more than 6 but less than 9(C) more than 9(D) less than 3Sol.

21.A thin horizontal circular disc is rotating abouta vertical axis passing through its centre. An insectis at rest at a point near the rim of the disc. Theinsect now moves along a diameter of the disc toreach its other end. During the journey of theinsect, the angular speed of the disc.

[AIEEE 2011](A) continuously decreases(B) continuously increases(C) first increases and then decreases(D) remains unchanged

Sol.

22. A hoop of radius r and mass m rotating withan angular velocity 0 is placed on a roughhorizontal surface. The initial velocity of thecentre of the hoop is zero. What will be thevelocity of the centre of the hoop when it casesto slip ? [JEE Mains 2013]

(1) 2

r 0 (2) r 0 (3) 4

r 0 (4) 3

r 0

Sol.



LEVEL - II JEE ADVANCED

1. Three particles A, B and C, each of mass m,are connected to each other by three masslessrigid rods to form a rigid, equilateral triangularbody of side l. This body is placed on a horizontalfrictionless table (x-y plane) and is hinged to itat the point A so that it can move without frictionabout the vertical axis through A (see figure).The body is set into rotational motion on thetable about A with a constant angular velocity .

y

xA

F

B Cl[JEE’(Scr) 2002]

(a) Find the magnitude of the horizontal forceexerted by the hinge on the body(b) At time T, when the side BC is parallel to thex-axis, a force F is applied on B along BC (asshown). Obtain the x-component and the y-component of the force exterted by the hinge onthe body, immediately after time T.Sol.

2. A particle is moving in a horizontal uniformcircular motion. The angular momentum of theparticle is conserved about the point :

[JEE’(Scr) 2003](A) Centre of the circle (B) Outside the circle(C) Inside the circle(D) Point on circumferenceSol.

3. Two particles each of mass M are connectedby a massless rod of length l. The rod is lying onthe smooth surface. If one of the particle isgiven an impulse MV as shown in the figure thenangular velocity of the rod would be

MMv

(A) v/l (B) 2v/l

(C) v/2 l (D) none [JEE’(Scr) 2003]



Sol.

4. A disc is rolling (without slipping) on a horizontalsurface. C is its center and Q and P are twopoints equidistant from C. Let Vp, VQ and VC bethe magnitude of velocities of points P, Q and Crespectively, then

PC

Q

(A) VQ > VC > VP (B) VQ < VC < VP

(C) VQ = Vp, V VC P12

(D) VQ < VC > VP

[JEE’(Scr) 2004]Sol.

5. A child is standing with folded hands at thecentre of a platform rotating about its centralaxis. The kinetic energy of the system is K. Thechild now stretches his arms so that the momentof inertia of the system doubles. The kineticenergy of the system now is

[JEE’(Scr) 2004](A) 2K (B) K/2(C) K/4 (D) 4KSol.

6. A block of mass m is held fixed against a wallby a applying a hor izontal force F. Which of thefollowing option is incorrect :

2a

F2a

a

(A) friction force = mg(B) F will not produce torque(C) normal will not produce torque(D) normal reaction = F




7. A disc has mass 9m. A hole of radius R/3 is cutfrom it as shown in the figure. The moment of inertiaof remaining part about an axis passing throughthe centre ‘O’ of the disc and perpendicular to theplane of the disc is :

OR

R/3 2R/3

(A) 8 mR2 (B) 4 mR2

(C) 409

2mR (D) 379

2mR


8. A particle moves in circular path with decreasingspeed. Which of the following is correct(A) L is constant

(B) only direction of L is constant

(C) acceleration a is towards the centre(D) it will move in a spiral and finally reach thecentre

[JEE’(Scr) 2005]

Sol.

9. A wooden log of mass M and length L is hingedby a frictionless nail at O. A bullet of mass mstrikes with velocity v and sticks to it. Find angularvelocity of the system immediately after thecollision about O.

v

m

LM

O

[JEE’ 2005]Sol.



10. A cylinder of mass m and radius R rolls downan inclined plane of inclination . Calculate thelinear acceleration of the axis of cylinder.

[JEE’ 2005]Sol.

11. Two identical ladders, each of mass M andlength L are resting on the rough horizontal surfaceas shown in the figure. A block of mass m hangsfrom P. If the system is in equilibrium, find themagnitude and the direction of frictional force atA and B. [JEE’ 2005]

m

P

L

A B

Sol.

12. A solid sphere of mass M, radius R and havingmoment of inertia about an axis passing throughthe centre of mass as I, is recast into a disc ofthickness t, whose moment of inertia about anaxis passing through its edge and perpendicularto its plane remains I. Then, radius of the discwill be [JEE’ 2006](A) 2 15R / (B) R 2 15/

(C) 4 15R / (D) R/4Sol.



13. A solid cylinder of mass m and radius r isrolling on a rough inclined plane of inclination .The coefficient of friction between the cylinderand incline is . Then [JEE’ 2006](A) frictional force is always mg cos (B) friction is a dissipative force(C) by decreasing , frictional force decreases(D) friction opposes translation and supportsrotationSol.

14. A ball moves over a fixed track as shown inthe figure. From A to B the ball rolls withoutslipping. Surface BC is frictionless. KA, KB and KCare kinetic energies of the ball at A, B and C,respe0ctively. Then [JEE’ 2006]

hA

A

B

ChC

(A) hA > hC ; KB > KC (B) hA > hC ; KC > KA(C) hA = hC ; KB = KC (D) hA < hC ; KB > KC

Sol.

15. There is a rectangular plate of mass M kg ofdimensions (a × b). The plate is held in horizontalposition by striking n small balls each of mass mper unit area per unit time. These are striking inthe shaded half region of the plate. The balls arecolliding elastically with velocity v. What is v ?

[JEE’ 2006]b

a

It is given n = 100, M = 3 kg, m = 0.01 kg; b = 2m, a = 1m; g = 10 m/s2.Sol.



Paragraph Q.16 to Q.18 (3 questions)Two discs A and B are mounted coaxially on avertical axle. The discs have moments of inertia Iand 2I respectively about the common axis. DiscA is imparted an initial angular velocity 2 usingthe entire potential energy of a spring compressedby a distance x1. Disc B is imparted an angularvelocity by a spring having the same springconstant and compressed by a distance x2. Boththe discs rotate in the clockwise direction.16. The ratio x1/x2 is [JEE’ 2007](A) 2 (B) 1/2(C) 2 (D) 1/ 2Sol.

17. When disc B is brought in contact with discA, they acquire a common angular velocity in timet. The average frictional torque on one disc bythe other during this period is [JEE’ 2007](A) 2I /(3t) (B) 9I /(2t)(C) 9I /(4t) (D) 3I /(2t)

Sol.

18. The loss of kinetic energy during the aboveprocess is [JEE’ 2007](A) I 2/2 (B) I 2/3(C) I 2/4 (D) I 2/6Sol.



19. A small object of uniform density rolls up acurved surface with an initial velocity v. It reachesup to a maximum height of 3v2 / (4g) with respectto the initial position. The object is [JEE’ 2007]

v

(A) ring (B) solid sphere(C) hollow sphere (D) discSol.

20. STATEMENT-1 If there is no external torqueon a body about its center of mass, then thevelocity of the center of mass remains constantbecauseSTATEMENT-2The linear momentum of an isolated systemremains constant.(A) Statement-1 is True, Statement-2 is True;Statement-2 is a correct explanation forStatement-1(B) Statement-1 is True, Statement-2 is True;Statement-2 is NOT a correct explanation forStatement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True

[JEE 2007]

Sol.

21. STATEMENT-1Two cylinders, one hollow (metal) and the othersolid (wood) with the same mass and identicaldimensions are simultaneously allowed to roll with-out slipping down an inclined plane from the sameheight. The hollow cylinder will reach the bottomof the inclined plane first.STATEMENT-2By the principle of conservation of energy, thetotal kinetic energies of both the cylinders areidentical when they reach the bottom of the in-cline.(A) STATEMENT-1 is True, STATEMENT-2 is True;STATEMENT-2 is a correct explanation for STATE-MENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True;STATEMENT-2 is NOT a correct explanation forSTATEMENT-1(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is True

[JEE-2008]



Sol.

22. If the resultant of all the external forces actingon a system of particles is zero, then from aninertial frame, one can surely say that

[JEE 2009](A) linear momentum of the system does notchange in time(B) kinetic energy of the system does not changein time(C) angular momentum of the system does notchange in time(D) potential energy of the system does notchange in timeSol.

23. A sphere is rolling without slipping on a fixedhorizontal plane surface. In the figure A is thepoint of contact, B is the centre of the sphereand C is its topmost point Then, [JEE 2009]

C

B

A

(A) V V V VC A B C– ( – )2

(B) V V V VC B B A– –

(C) | – | | – |V V V VC A B C2

(D) | – | | |V V VC A B4Sol.



24. A boy is pushing a ring of mass 2 kg andradius 0.5 m with a stick as shown in the figure.The stick applies a force of 2 N on the ring androlls it without slipping with an acceleration of0.3 m/s2. The coefficient of friction between theground and ring is large enough that rolling alwaysoccurs and the coefficient of friction betweenthe stick and the ring is (P/10). The value of Pis?

[JEE 2011]

Ground

stick

Sol.

25. A thin uniform rod, pivoted at O is rotating inthe horizontal plane with constant angular speed, as shown in the figure. At time t = 0, small

insect starts from O and moves with constantspeed with respect to the rod towards the otherend. it reaches the end of the rod at t = T andstops. The angular speed of the system remains

throughout. The magnitude of the torque onthe system about O, as a function of time is bestrepresented by which plot?

O

Z

(A)

0 tT

(B)

0 tT

(C)

0 tT

(D)

0 tT[JEE 2012]

Sol.



26. A small mass m is attached to a masslessstring whose other end is fixed at P as shown inthe figure. The mass is undergoing circular motionin the x-y plane with centre at O and constantangular speed .If the angular momentum of thesystem, calculated about O and P are denoted

by 0L and PL respectively, then.

m

P

O

z

(A) 0L and PL do not vary with time(B) 0L varies with time while PL remains constant(C) 0L remains constant while PL varies with time(D) 0L and PL both vary with time.

[JEE 2012]Sol.

27. A lamina is made by removing a small disc ofdiameter 2R from a bigger disc of uniform massdensity and radius 2R, as shown in the figure.The moment of inertia of this lamina about axespassing through O and P is Io and IP, respec-tively. Both these axes are perpendicular to the

plane of the lamina. The ratio o

P

II

to the nearest

integer is

[JEE 2012]Sol.



28. Consider a disc rotating in the horizontal planewith a constant angular speed about its centreO. The disc has a shaded region on one side ofthe diameter and an unshaded region on the otherside as shown in the figure. When the disc is inthe orientation as shown, two pebbles P and Qare simultaneously projected at an angle towardsR. The velocity of projection is in the y-z planeand is same for both pebbles with respect to thedisc. Assume that (i) they land back on the discbefore the disc has completed 1/8 rotation, (ii)their range is less than half the disc radius, and(iii) remains constant throughout. Then

Q

R

P

y

xO

(A) P lands in the shaded region and Q in theunshaded region(B) P lands in the unshaded region and Q in theshaded region(C) Both P and Q land in the unshaded region(D) Both P and Q land in the shaded region

[JEE 2012]Sol.

Paragraph for Question Nos. 29 to 30The general motion of a rigid body can beconsidered to be a combination of (i) a motion ofits centre of mass about an axis, and (ii) its motionabout an instantaneous axis passing through thecentre of mass. These axes need not bestationary. Consider, for example, a thin uniformdisc welded (rigidly fixed) horizontally at its rimto a massless stick, as shown in the figure. Whenthe disc-stick system is rotated about the originon a horizontal frictionless plane with angularspeed , the motion at any instant can be takenas a combination of (i) a rotation of the centre ofmass of the disc about the z-axis, and (ii) arotation of the disc through an instantaneousvertical axis passing through its centre of mass(as is seen from the changed orientation of pointsP and Q). Both these motions have the sameangular speed in this case.

Now consider two similar systems as shown inthe figure: Case (a) the disc with its face verticaland parallel to x-z plane; case (b) the disc withits face making an angle of 45o with x-y planeand its horizontal diameter parallel to x-axis. Inboth the cases, the disc is welded at point P, andthe systems are rotated with constant angularspeed about the z-axis.

Case (a) Case (b)

29. Which of the following statements about theinstantaneous axis (passing through the centreof mass) is correct ?(A) It is vertical for both the cases (a) and (b).(B) It is vertical for case (a); and is at 45o to thex-z plane and lies in the plane of the disc forcase (b).(C) It is horizontal for case (a); and is at 45o tothe x-z plane and is normal to the plane of thedisc for case (b).(D) It is vertical for case (a); and is at 45o to thex-z plane and is normal to the plane of the discfor case (b).

[JEE 2012]



Sol.

30. Which of the following statements regardingthe angular speed about the instantaneous axis(passing through the centre of mass) is correct

(A) It is 2 for both the cases.

(B) It is for case (a); and 2 for case (b).

(C) It is for case (a); and 2 for case (b).(D) It is for both the cases.

[JEE 2012]

Sol.

31. The figure shows a system consisting of (i) aring of outer radius 3R rolling clockwise withoutslipping on a horizontal surface with angular speed

and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed 2. The ring anddisc are separated by frictionless ball bearings.The system is in the x-z plane. The point P onthe inner disc is at a distance R from the origin,where OP makes an angle of 30o with thehorizontal. Then with respect to the horizontalsurface.

(A) the point O has a linear velocity iR3(B) the point P has a l i near velocity

11 3 ˆˆR i R k4 4

(C) the point P has a l i near velocity

kR43iR

413

(D) the point P has a l i near velocity

kR41iR

433

[JEE 2012]



Sol.

32. Two solid cylinders P and Q of same mass andsame radius start rolling down a fixed inclinedplane from the same height at the same time.Cylinder P has most of its mass concentratednear its surface, while Q has most of its massconcentrated near the axis. Which statement(s)is(are) correct?(A) Both cylinders P and Q reach the ground atthe same time.(B) Cylinder P has larger linear acceleration thancylinder Q.(C) Both cylinders reach the ground with sametranslational kinetic energy.(D) Cylinder Q reaches the ground with largerangular speed.

[JEE 2012]Sol.

33. A uniform circular disc of mass 50 kg andradius 0.4 m is rotating with an angular velocityof 10 rad s-1 about its own axis, which is vertical.Two uniform circular rings, each of mass 6.25 kgand radius 0.2 m, are gently placed symmetricallyon the disc in such a manner that they aretouching each other along the axis of the discand are horizontal. Assume that the friction islarge enough such that the rings are at restrelative to the disc and the system rotates aboutthe original axis. The new angular velocity (in rads-1) of the system is : [JEE 2013]Sol.



ANSWER KEY

Exercise - I OBJECTIVE PROBLEMS (JEE MAIN)

1. D 2. B 3. A 4. D 5. A 6. B

7. A 8. C 9. A 10. C 11. D 12. B

13. D 14. B 15. C 16. D 17. B 18. B

19. A 20. D 21. D 22. D 23. A 24. D

25. A 26. D 27. B 28. C 29. A 30. B

31. B 32. A 33. C 34. C 35. D 36. C

37. B 38. B 39. B 40. C 41. B 42. D

43. B 44. B 45. A 46. D 47. D 48. B

49. A 50. C 51. C 52. D 53. D 54. C

55. C 56. C 57. A 58. D 59. C 60. A

61. D

Exercise - II

1. B 2. C 3. C 4. A 5. B 6. C

7. A 8. C 9. C 10. ABCD 11. B 12. B

13. A 14. C 15. D 16. BC 17. AD 18. BCD

19. ABD 20. ABC 21. ACD 22. C 23. B 24. ACD

25. B 26. C 27. A 28. D 29. B 30. B

31. C 32. B 33. D 34. A 35. C 36. B

37. ACD 38. C 39. BCD 40. A 41. C 42. BC

43. AC 44. BC 45. ABCD 46. BD 47. ABC 48. ABC

49. C 50. BCD 51. ACD 52. ABD 53. B 54. A

55. B 56. C 57. A 58. C 59. D 60. C

61. B 62. B 63. B 64. D 65. D 66. A

67. C



Exercise - III (JEE ADVANCED)

1. MR M R2 2

243

– 2. 145

2mr 3. 2r 4. Ml2

25. 2mR2

6. w sin , when the bob is at the lowest point 7. –14 –i j k10 9

8. P = mg2

cot 9. 32

2Ftm

10. (a) 34

gL

cw( ) (b) Nmg F mg13

163 316

, (c) 16

33

11. 25g

12. w = 5 rad/s 13. at a distance a / 2 3 from the centre down the plane.

14. 12 gm 15. 100

3N 16. 13 W / 12 17.

Rg

2 2

4 18. 2.1 rad/s2 19. 15 2. sec

20. 48 21. gr/bh 22. mgl8

23. 2 24. (i) 10/13 m/s2, (ii) 5000/26 , (iii) 480/13 N

25. M = 2 2 12 2m ghR

– 26.16 kg m2/s 27. 2 2 /kkg m s 28. 0.5 kg – m2/s, 75 J

29. 19.7 rad/s 30. 60° east of south, 30° south of east. 31. (1/2)KE0 32. 3

33. 120 34. 7

102mv 35. 50m/s 36.

107gh

37. 107

g sin 38. 177

mg

39. v v0 0

323

( ), ( ) 40. (a) Ftm

(b) 6Ftm

(c) 2 2 2F t

m (d)

F t2

41. 1316

2Mv 42. 34

2MR



1. 2mg 2. 3 3. 2F/M 4. 2m/s 5. 1/2 ma 6. v = 14

3gR

7. a b ab a–

3 g 8. Fdlx = 3

2mgw

h and Fduy = mg 9. 57/140 MR2 10. (a)

97g

(b) 4

7mg

11. (a) 3

20a

(b) 59a

, (c) 7

12

30a

, (d) 187

PMa

, (e) M ag9

70 12. 300 rad/sec, 150 rad/sec

13. t = 2 5/ sec, q = 4p/5 rad 14. 16 m/s2 15. v = 277

gR 16. (a) 4 37

m s/ , (b) 2007

N

17. – 115 0v 18. 210 19. 1.63 N, 1.224 m

Exercise - IV PREVIOUS YEAR QUESTIONS

LEVEL - I JEE MAIN

1. D 2. A 3. B 4. C 5. D 6. A

7. D 8. B 9. C 10. C 11. D 12. D

13. D 14. C 15. C 16. A 17. C 18. D

19. D 20. A 21. C 22. A

LEVEL - II JEE ADVANCED

1. (a) 3 m 2 l, (b) Fx = F/4, Fy = 3 m 2 l

2. A 3. A 4. A 5. B 6. C 7. B

8. B 9. 3

3mv

m M L( ) 10. a gaxis

23sin

11. f M m g( ) cot2

12. A 13. C,D 14. A,B 15. 10m/s

16. C 17. A 18. B 19. D 20. D

21. D 22. A 23. B,C

24. 0004 25. B 26. C 27. 0003 28. C

29. A 30. D 31. A,B 32. D 33. 0008

STOICHIOMETRY - 1Page # 90


CLASSIFICATION OF MATTER

LAWS OF CHEMICAL COMBINATION

(a) Law of conservation of mass [Lavoisier]In a chemical change total mass remains conserved i.e. mass before the reaction is always equal tomass after the reaction.

H2 + 1/2 O2 H2O ( ) (g) (g) 1 mole 1/2 mole 1 mole

mass before the reaction = 1 × 2 + 1/2 × 32 = 18 gmmass after the reaction = 1 × 18 = 18 gm

Ex. A 15.9g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0g. The twosubstances react, releasing carbon dioxide gas to the atmosphere. After reaction, the contentsof the reaction vessel weigh 29.3g. What is the mass of carbon dioxide given off during thereaction?

Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the finalmass of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is dueto the mass of released carbon dioxide gas.

Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 gm

(b) Law of constant composition [Proust]All chemical compounds are found to have constant composition irrespective of their method of preprationor sources.

In H2O, Hydrogen & oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is Tapwater, river water or seawater or produced by any chemical reaction.

STOICHIOMETRY - 1

STOICHIOMETRY - 1 Page # 91


Ex. The following are results of analysis of two samples of the same or two different compounds ofphosphorus and chlorine. From these results, decide whether the two samples are from thesame or different compounds. Also state the law, which will be obeyed by the given samples.

Amount P Amount ClCompound A 1.156 gm 3.971 gmCompound B 1.542 gm 5.297 gm

Sol. The mass ratio of phosphorus and chlorine in compound A, mP : mCl = 1.156:3.971 = 0.2911:1.000

The mass ratio of phosphorus and chlorine in compound B, mP : mCl = 1.542:5.297 = 0.2911:1.000As the mass ratio is same, both the compounds are same and the samples obey the law of definiteproportion.

(c) Law of multiple proportions [Dalton]When one element combines with the other element to form two or more different compounds, themass of one element, which combines with a constant mass of the other bear a simple ratio to oneanother.

Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show thatthese figures shows the law of multiple proportion.

First oxide Second oxideCarbon 42.9 % 27.3 %Oxygen 57.1 % 72.7% Given

In th first oxide, 57.1 parts by mass of oxygen combine with 42.9 parts of carbon.

1 part of oxygen will combine with 42 957 1

.

. part of carbon = 0.751

Similarly in 2nd oxide

1 part of oxygen will combine with 27 3727

.

. part of carbon = 0.376

The ratio of carbon that combine with the same mass of oxygen = 0.751 : 0.376 = 2 : 1This is a simple whole no ratio this means above data shows the law of multiple proportion.

Ex. Two oxide samples of lead were heated in the current of hydrogen and were reduced to themetallic lead. The following data were obtained

(i) Weight of yellow oxide taken = 3.45 gm; Loss in weight in reduction = 0.24 gm

(ii) Weight of brown oxide taken = 1.227 gm; Loss in weight in reduction = 0.16 gm.

Show that the data illustrates the law of multiple proportion.

Sol. When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the lossin weight in reduction is due to removal of the oxygen present in the oxide, to combine with thehydrogen. Therefore,the composition of the yellow oxide is: oxygen = 0.24 gm and lead = 3.45 – 0.24 = 3.21 gm.

The mass ratio of lead and oxygen, r1 = Pb

O

m 3.21 13.375m 0.244 1.000

and the compositon of the brown oxide is : oxygen = 0.16 gm and lead = 1.227 – 0.16 = 1.067 gm.

The mas ratio of lead and oxygen, r2 = Pb

O

m 1.067 6.669m 0.16 1.000

Now, r1 : r2 = 13.375 : 6.669 = 2.1 (simple ratio) and hence the data illustrates the law of multipleproportion.



(d) Law of reciprocal proportions [Richter]When two elements combine seperately with definite mass of a third element, then the ratio of theirmasses in which they do so is either the same or some whole number multiple of the ratio in which theycombine with each other.This law can be understood easily with the help of the following examples.

H S

O

H S2At. Mass 1 At. Mass 32

At. Mass 16

Let us consider three elements – hydrogen, sulphur and oxygen. Hydrogen combines with oxygen toform H

2O whereas sulphur combines with it to form SO

2. Hydrogen and sulphur can also combine

together to form H2S. The formation of these compounds is shown in fig.In H2O, the ratio of masses of H and O is 2 : 16.In SO2, the ratio of masses of S and O is 32 : 32. Therefore, the ratio of masses of H and S whichcombines with a fixed mass of oxygen (say 32 parts) will be

4 : 32 i.e. 1 : 8 ...(1)When H and S combine together, they form H2S in which the ratio of masses of H and S is

2 : 32 i.e., 1 : 16 ...(ii)The two ratios (i) and (ii) are related to each other as

18

116

: or 2 : 1

i.e., they are whole number multiples of each other.Thus, the ratio masses of H and S which combines with a fixed mass of oxygen is a whole numbermultiple of the ratio in which H and S combine together.

Ex. Methane contains 75 % carbon and 25% hydrogen, by mass. Carbon dioxide contains 27.27 %carbon and 72.73% oxygen, by mass. Water contains 11.11 % hydrogen and 88.89% oxygen,by mass. Show that the data illustrates the law of reciprocal proportion.

Sol. Methane and carbon dioxide, both contains carbon and hence, carbon may be considered as the thirdelement. Now, let the fixed mass of carbon = 1 gm. Then,

the mass of hydrogen combined with 1 gm carbon in methane = 25 175 3 gm

and the mass of oxygen combined with 1 gm carbon in carbon dioxide = 72.73 827.27 3 gm

Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of carbon, r1 = 1 8 1:3 3 8

Now, the mass ratio of hydrogen and oxygen in water, r2 = 11.11 188.89 8

As r1 and r2 are same , the data is according to the law of reciprocal proportion.



(e) Gay Lussac law of combining volumes :When two or more gases react with one another, their volumes bear simple whole number ratio withone another and to the volume of products (if they are also gases) provided all volumes are measuredunder identical conditions of temperature and pressure.When gaseous hydrogen and gaseous chlorine react together to form gaseous hydrogen chlorideaccording to the following equation.

volumestwovolumeone2

volumeone2 )g(HCl2)g(Cl)g(H

It has been observed experimentally that in this reaction, one volume of hydrogen always reacts withone volume of chlorine to form two volumes of gaseous hydrogen chloride. all reactants and productsare in gaseous state and their volumes bear a ratio of 1 : 1 : 2. This ratio is a simple whole numberratio.“These are no longer useful in chemical calculations now but gives an idea of earlier methods ofanalysing and relating compounds by mass.”

Ex. 2.5 ml of a gaseous hydrocarbon exactly requires 12.5 ml oxygen for complete combustion andproduces 7.5 ml carbon dioxide and 10.0 ml water vapour. All the volumes are measured at thesame pressure and temperature. Show that the data illustrates Gay Lussac’s law of volume com-bination.

Sol. Vhydrocarbon : Voxygen : Vcarbon dioxide : Vwater vapour = 2.5 : 12.5 : 7.5 : 10.0

= 1 : 5 : 3 : 4 (simple ratio)

Hence, the data is according to the law of volume combination.

MOLE CONCEPT

Definition of mole : One mole is a collection of that many entities as there are number of atomsexactly in 12 gm of C – 12 isotope.or 1 mole = collection of 6.02 × 1023 species

6.02 × 1023 = NA = Avogadro’s No.1 mole of atoms is also termed as 1 gm-atom, 1 mole of ions is termed as 1 gm-ion and 1 mole ofmolecule termed as 1 gm-molecule.

METHODS OF CALCULATIONS OF MOLE

(a) If no. of some species is given, then no. of moles = Given no

NA

.

(b) If weight of a given species is given, then no of moles = Given wtAtomic wt

for atoms.( ),

or = Given wt

Molecular wtfor molecules

..( )

(c) If volume of a gas is given along with its temperature (T) and pressure (P)

use n = PVRT

where R = 0.0821 lit-atm/mol–K (when P is in atmosphere and V is in litre.)1 mole of any gas at STP (0°C & 1 bar) occupies 22.7 litre.1 mole of any gas at STP (0°C & 1 atm) occupies 22.4 litre.



Atom : Atom is smallest particle which can not be divided into its constituents.

Atomic weight : It is the weight of an atom relative to one twelvth of weight of 1 atom of C-12

RELATIONSHIP BETWEEN GRAM AND AMU

1 amu = 1

12 wt of one C - 12 atom.

for C 1 mole C = 12 gm = 6.023 × 1023 atomswt of 6.023 × 1023 atoms = 12 gm

wt of 1 atom of C = 12N

gmA

(NA Avogadro’s number = 6.23 × 1023)

1 amu = 1

12 wt of one C - 12 atom

= 1

1212NA

gm

1 amu = gmN1A

ELEMENTAL ANALYSIS

For n mole of a compound (C3H7O2)Moles of C = 3nMoles of H = 7nMoles of O = 2n

Ex. Find the wt of water present in 1.61 g of Na2SO4. 10H2O

Sol. Moles of Na2SO4. 10H2O = wt. in gram

molecular wt = 1.61322

= 0.005 moles

Moles of water = 10 × moles of Na2SO4. 10H2O= 10 × 0.05 = 0.05

wt of water = 0.5 × 18 = 0.9 gm Ans.

AVERAGE ATOMIC WEIGHT

= % of isotope X molar mass of isotope.The % obtained by above expression (used in above expression) is by number (i.e. its a mole%)

MOLECULAR WEIGHT

It is the sum of the atomic weight of all the constituent atom.

(a) Average molecular weight = n M

ni i

i

where ni = no. of moles of any compound and m

i = molecular mass of any compound.

Make yourselves clear in the difference between mole% and mass % in question related toabove.



Shortcut for % determination if average atomic weight is given for X having isotopes XA & XB.

100X&X of weightin difference

X of wt –ghtatomic wei AverageXof% BA

BA

Try working out of such a shortcut for XA, XB, XC

EMPIRICAL FORMULA, MOLECULAR FORMULAEmpirical formula : Formula depicting constituent atom in their simplest ratio.Molecular formula : Formula depicting actual number of atoms in one molecule of the compound.Relation between the two : Molecular formula = Empirical formula × n

nMolecular mass

Empirical Formula mass

Check out the importance of each step involved in calculations of empirical formula.Ex. A molecule of a compound have 13 carbon atoms, 12 hydrogen atom, 3 oxygen atoms and

3.02 × 10–23 gm of other element. Find the molecular wt. of compound.

Sol. wt. of the 1 molecule of a compound = 13 × 12NA

+ 12 × 1

NA + 3 ×

16NA

+ 3.02 × 10–23

A

A23–

NN1002.34812156

= 234.18 / NA = 234 amu. Ans.

• Density :

(a) Absolute density(b) Relative density

Absolute density = Mass

volume

Relative density = density of subs ce

density of s dard subs cetan

tan tan

Specific gravity = density of subs cedensity of H O at C

tan

2 4

Vapour density : It is defined only for gas.It is a density of gas with respect to H2 gas at same temp & press

V.D = dgasdH2

= PM RTPM RT

gas

H

//

2

= MM

gas

H2

= M2

V.D = M2

V.D = Molecular wt of gas

Molecular wt of H2 gas

• density of Cl2 gas with respect to O2 gas

= gasO of wtMoleculargas Clof wtMolecular

2

2



• STOICHIOMETRY : Stoichiometry is the calculations of the quantities of reactants and productsinvolved in a chemical reaction. Following methods can be used for solving problems.

(a) Mole Method (For Balance reaction)(b) POAC method } Balancing not required but common sense ------- use it with slight care.(c) Equivalent concept

• CONCEPT OF LIMITING REAGENT.

Limiting Reagent :It is very important concept in chemical calculation. It refers to reactant which is present in minimumstoichiometry quantity for a chemical reaction. It is reactant consumed fully in a chemical reaction. Soall calculations related to various products or in sequence of reactions are made on the basis of limitingreagent.• It comes into picture when reaction involves two or more reactants. For solving any such reactions,first step is to calculate L.R.Calculation of Limiting Reagent.(a) By calculating the required amount by the equation and comparing it with given amount.

[Useful when only two reactant are there](b) By calculating amount of any one product obtained taking each reactant one by one irrespectiveof other reactants. The one giving least product is limiting reagent.(c) Divide given moles of each reactant by their stoichiometric coefficient, the one with least ratiois limiting reagent. [Useful when number of reactants are more than two.]

• PERCENTAGE YIELD :

The percentage yield of product = actual yield

the theoretical maximum yield100

• The actual amount of any limiting reagent consumed in such incomplete reactions is given by [%yield × given moles of limiting reagent] [For reversible reactions]• For irreversible reaction with % yield less than 100, the reactants is converted to product(desired and waste.)

Ex. A compound which contains one atom of X and two atoms of y for each three atoms of z is madeof mixing 5 gm of x, 1.15 × 1023 atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 gmof compound results. Calculate the atomic weight of Y if atomic weight of X and Z are 60 and 80respectively.

Sol. Moles of x = 5

60 =

112

= 0.083

moles of y = 115 10

6 023 100 19

23

23..

.

moles of z = 0.03x + 2y + 3z xy2z3

for limiting reagent, 0.083/1 = 0.083

0192

0 095. . , 01.0303.0

Hence z is limiting reagentwt of xy2z3 = 4.4 gm = moles × molecular wt.

moles of xy2z

3 =

13

0 03. = 0.01

300 + 2 m = 440 2m = 440 – 300 m = 70 Ans.



P O A C Rule : P O A C is the simple mass conservation.

KClO3 KCl + O2

Apply the POAC on K.

moles of K in KClO3 = moles of K in KCl

1 × moles of KClO3 = 1 × moles of KCl

moles of KClO3 = moles of KCl

Apply POAC on O

moles O in KClO3 = moles of O in O2

3 × moles of kClO3 = 2 × moles of O

2

Ex. In the gravimetric determination of phosphorous, an aqueous solution of dihydrogen phosphateion (H

2PO

4–) is treated with a mix of ammonium & magnesium ions to precipitate magnesium

ammonium phosphate MgNH4 PO4.6H2O. This is heated and decomposed to magnesiumPyrophosphate, Mg2P2O7 which is weighted. A solution of H2PO4

– yielded 1.054 gm of Mg2P2O7

what weight of NaH2PO4 was present originally.

NaH2PO4 Mg2P2O7

apply POAC on P

Let wt of NaH2PO4 = w gm

moles of P in NaH2PO4 = moles of P in Mg2P2O7

2232054.1

1120w

w = 1.054 × 2232120

= 1.09 gm Ans.

SOME EXPERIMENTAL METHODS

FOR DETERMINATION OF ATOMIC MASS

Dulong’s and Petit’s Law :

Atomic weight × specific heat (cal/gm°C) 6.4

Gives approximate atomic weight and is applicable for metals only. Take care of units of specific heat.

FOR MOLECULAR MASS DETERMINATION

(a) Victor Maeyer’s process : (for volatile substance)

Procedure : Some known weight of a volatile substance (w) is taken, converted to vapour andcollected over water. The volume of air displaced over water is given (V) and the following expressionsare used.

M = w

PVRT or M

wP P V

RT( – ')

If aq. tension is not given If aq. tension is P

Aqueous tension : Pressure exerted due to water vapours at any given temperature.

This comes in picture when any gas is collected over water. Can you guess why ?



(b) Silver salt method : (for organic acids)

Basicity of an acid : No. of replacible H+ atoms in an acid (H bonded to more electronegative atom isacidic)

Procedure : Some known amount of silver salt (w1 gm) is heated to obtain w2 gm of white shiningresidue of silver. Then if the basicity of acid is n, molecular weight of acid would be

AgnA nAg+ + A–n

Agn A is the salt

wn

M wsalt2

11081

and molecular weight of acid = Msalt – n(108)

This is one good practicle application of POAC.

(c) Chloroplatinate salt method : (for organic bases)

Lewis acid : electron pair acceptor

Lewis base : electron pair donor

Procedure : Some amount of organic base is reacted with H2PtCl6 and forms salt known as chloroplatinate.If base is denoted by B then salt formed.

(i) with monoacidic base = B2H2PtCl6

(ii) with diacidic base = B2(H2PtCl6)2

(iii) with triacidic base = B2(H2PtCl6)3

The known amount (w1 gm) of salt is heated and pt residue is measured. (w

2 gm). If acidity of base is

‘n’ then w

n2

1951

× Msalt = w1 and Mbase = M nsalt – ( )410

2

• For % determination of elements in organic compounds :

• All these methods are applications of POAC

• Do not remember the formulas, derive them using the concept, its easy.

(a) Liebig’s method : (Carbon and hydrogen)

(w) Organic Compound CuO

(w1) CO2 + H2O (w2)

% of C =w

w1

4412

100

% of H = w

w2

182

100

where w1 = wt. of CO2 produced, w2 = wt. of H2O produced,

w = wt, of organic compound taken

(b) Duma’s method : (for nitrogen)

(w) Organic Compound CuO N2 (P, V, T given)

use PV = nRT to calculate moles of N2, n.

% of N = n

w28

100

w = wt of organic compound taken



(c) Kjeldahl’s method : (for nitrogen)

(w) O.C. + H2SO4 (NH4)2SO4 NaOH NH3 + H2SO4 (molarity M and volume (V1) consumed given)

% of N = MV

w1 2 14

100

where M = molarity of H2SO4.

• Some N containing compounds do not give the above set of reaction as in Kjeldahl’s method.

(d) Sulphur :

(w) O.C. + HNO3 H2SO4 + BaCl2 (w1) BaSO4

% of S = 233w 1 × 1 ×

32w

× 100

where w1 = wt. of Ba SO4, w = wt. of organic compound

(e) Phosphorus :

O.C+ HNO3 H3PO4 + [NH3 + magnesia mixture ammonium molybdate] MgNH4 PO4 Mg2P2O7

% of P = 100w

312222w 1

(f) Carius method : (Halogens)

O.C. + HNO3 + AgNO3 AgX

If X is Cl then colour = white

If X is Br then colour = dull yellow

If X is I then colour = bright yellow

• Flourine can’t be estimated by this

% of Xw

M weight of AgXAt wt of X

w1 1

100( . )

( . )

Ex. 0.607 g of a silver salt of a tribasic organic acid was quantitatively reduced to 0.370 g of puresilver. Calculate the molecular weight of the acid (Ag = 108)

Sol. Suppose the tribasic acid is H3A.

H3A Ag3A Ag

acid salt

0.607 g 0.37 g

Since Ag atoms are conserved, applying POAC for Ag atoms,

moles of Ag atoms in Ag3A = moles of Ag atoms in the prduct

3 × moles of Ag3A = moles of Ag in the product

30 607 0 37

108. .

mol. wt. of Ag3A (Ag = 108)

mol. wt. of Ag3A = 531.



mol. weight of tribasic acid (H3A)

= mol wt. of the salt (Ag3A) – 3 × at. wt. of Ag

+ 3 × at. wt. of H

= 531 – 324 + 3 = 210 Ans.

EUDIOMETRY[For reactions involving gaseous reactants and products]

• The stoichiometric coefficient of a balanced chemical reactions also gives that ratio of volumes inwhich gasesous reactants are reacting and products are formed at same temperature and pressure.The volume of gases produced is often given by mentioning certain solvent which absorb containgases.Solvent gas(es) absorbKOH CO2, SO2, Cl2Ammon Cu2Cl2 COTurpentine oil O3

Alkaline pyrogallol O2

water NH3, HCl

CuSO4/CaCl2 H2OAssumption : On cooling the volume of water is negligible

Ex. 7.5 mL of a hydrocarbon gas was exploded with excess of oxygen. On cooling, it was found tohave undergone a contraction of 15 mL. If the vapour density of the hydrocarbon is 14, determineits molecular formula. (C = 12, H = 1)

Sol. CxHy + (x + y4

) O2 X CO2 + y2

H2O

7.5 ml 15on cooling the volume contraction = 15 mli.e. The volume of H2O (g) = 15 mlV.D. of hydrocarbon = 14

Molecular wt. of CxHy = 2812x + y = 28 ...(1)

From reaction

7.5 y2

= 15 y = 4

12 x + 4 = 2812x = 24 x = 2

Hence Hydrocalbon is C2H4.

CONCENTRATION OF SOLUTION

Concentration of solution can be expressed in any of the following ways.

(a) % by wt amount of solute dissolved in 100 gm of solution

4.9% H2SO4 by wt.

100 gm of solution contains 4.9 gm of H2SO4



(b) % by volume volume of solute dissolved in 100 ml of solution

x% H2SO4 by volume

100 ml of solution contains x ml H2SO4

(c) % wt by volume wt. of solute present in 100 ml of solution

(d) % volume by wt. volume of solute present in 100 gm of solution.

CONCENTRATION TERMS

• Molarity (M) : No. of moles of solute present in 1000 ml of solution.

molarity (M) = (lit) solution of volumesolute of moles

M = l)solution(m of volume solute of m.moles

MOLALITY (m)

No. of moles of solute present in 1000 gm of solvent

m = moles of solute

wt.of solvent in kg m = m.moles of solute wt.of solvent in gm

NORMALITY (N)

No of gm equivalents of solute present in 1000 ml of solution

N = gm equivalents of solute

volume of solution(lit) = m. equivalent of solute

volume of solution in (ml)

FORMALITY (f)

The formality is the no. of gm -formula weights of the ionic solute present in 1000 ml of solution.

f = (lit)solution of volume wtformula gm in wt

MOLE FRACTION

The mole fraction of a perticular component in a solution is defined as the number of moles of thatcomponent per mole of solution.

If a solution has nA mole A & nB mole of B.

mole fraction of A (XA) = n

n nA

A B

mole fraction of B (XB) = n

n nB

A B

XA + XB = 1

• Parts per million (ppm) : = solvent of Masssolute of Mass

× 106 610 solution of Mass

solute of Mass



VOLUME STRENGTH OF H2O2

Strength of H2O2 is represented as 10V, 20V, 30V etc.

20V H2O2 means one litre of this sample of H2O2 on decomposition gives 20 It of O2 gas at S.T.P.

Decomposition of H2O2 is given as

H2O2 H2O + 12 2O

1 mole12

× 22.4 It O2 at S.T.P..

= 34 g = 11.2 It O2 at S.T.P.

• To obtain 11.2 litre O2 at S.T.P. at lest 34 gm H2O2 must be decomposed

• for 20 It O2, we should decompose atleast 34

112. × 20 gm H2O2

1 It solution of H2O2 contains 34

11220

. gm H2O2

1 It solution of H2O2 contains 34

1122017.

equivalents of H2O2 ( )E MH O2 2 2

342

17

• Normality of H2O2 = 34

1122017

205 6. .

Normality of H2O2(N) = Volume, strength of H2O2

5 6.

MN

v fN

H OH O H O

2 2

2 2 2 2

2.IInd Method :

H2O2 H2O + 12 2O

From law of equivalence

gm eq. of O2 = gm eq. of H2O2

gm eq. of O2 = moles × n factor of O2, = 20

22 44

. =

205 6.

gm. eq. of H2O2 = 205 6.

and the volume of H2O2 is 1 lit.

this means 1 lit of H2O2 have 205 6.

gm eq.

i.e. Normality N = 205 6.



NORMALITY OF H2O2

= volume str Oength of H2 2

5 6.

• Molarity of H2O2(M) = 2.11

OH of strength Volume, 22

Strength (in g/ ) : Denoted by S

Strength = molarity × mol. wt.

= molarity × 34strength = Normality × Eq. weight.

= Normality × 17

Ex. A bottle labeled with “12V H2O2” contain 700 ml solution. If a sdudent mix 300 ml water in itwhat is the g/litre strenth & normality and volume strength o final solution.

Sol. N = 125 6.

meq. of H2O2 = 125 6

700.

let the normality of H2O2 on dilution is N

meq. before dilution = meq. after dilution

N × 1000 = 125 6

700.

N = 125 6.

× 7

10 = 1.5 M =

152.

strength gm/lit = 152

34.

= 25.5

volume strength = N × 5.6 = 8410

= 8.4 V Ans.

Strength of Oleum

Oleum is SO3 dissolved in 100% H2SO4. Sometimes, oleum is reported as more then 100% by weight,say y% (where y > 100). This means that (y – 100) grams of water, when added to 100 g of givenoleum sample, will combine with all the free SO3 in the oleum to give 100% sulphuric acid.

Hence weight % of free SO3 in oleum = 80 100

18( – )y

Ex. Calculate the percentage of free SO3 in an oleum (considered as a solution of SO

3 in H

2SO

4) that

is labelled '109% H2SO4'.

Sol. '109% H2SO4' refers to the total mass of pure H2SO4, i.e., 109 g that will be formed when 100 g of oleumis diluted by 9 g of H2O which (H2O) combines with all the free SO3 present in oleum to form H2SO4

H2O + SO3 H2SO4

1 mole of H2O combines with 1 mole of SO3

or 18 g of H2O combines with 80 g of SO

3

or 9 g of H2O combines with 40 g of SO

3.

Thus, 100 g of oleum contains 40 g of SO3 or oleum contains 40% of free SO3.



Ex. A 62% by mass of an aqueous solution of acid has specific gravity 1.8. This solution is dilutedsuch that the specific gravity of solutin became 1.2. Find the % by wt of acid in new solutiuon.

Sol. density = mass

volume

1.8 = 100

volume of solnvolume of solution =

10018.

Let x gm water is added in soluion

then d = mass

volume

12 10010018

.

.

x

x

1210018

12 100..

. x x

2003

12 100. x x

0.2 x = 100 – 200

3 =

1003

x = 100

3 0 2. =

10006

= 5003

= 166.67

mass of new solution = 100 + 166.67 = 266.67

266.67 gm solution contains 62 gm of acid

% by mass = 62

266 67100

. = 23.24 %

RELATION SHIP BETWEEN MOLARITY, MOLALITY & DENSITY OF SOLUTION

Let the molarity of solution be 'M', molality be 'm' and the density of solution be d gm/m.

Molarity implies that there are M moles of solute in 1000 ml of solution wt of solution = density ×volume

= 1000 d gm wt of solute = MM1

where M1 is the molecular wt of solute

wt of solvent = (1000d – MM1) gm

(1000d – MM1) gm of solvent contains M moles of solute

1000 gm of solvent have = mole1000MM–d1000

M

1 = Molality



no. of moles of solute present in 1000 gm of solvent = 1000

1000 1

Md MM– = Molality

on simplyfying1000

Mm1Md 1

RELATION SHIP BETWEEN MOLALITY & MOLE FRACTION

consider a binary solution consisting of two components A (Solute) and B (Solvent).

Let xA & x

B are the mole fraction of A & B respectively.

xA = n

n nA

A B, xb =

nn n

B

A B

If molality of solution be m then

mn

mass of solventA 1000 =

nAn MB B

1000

where MB is the molecular wt of the solvent B

mxx M

A

B B

1000

molality = BM

1000B of fraction mole Aof fraction mole

m = solvent of wt.molecular1000

solvent of fraction molesolute of fraction mole

Ex. An aqueous solution is 1.33 molal in methanol. Determine the mole fraction of methanol & H2O

Sol. molality = 1000solvent of mol.wtsolvent of fraction mole

solute of fraction mole

1.33 = 1000Mx

x

BB

A,

B

A

xx

10001833.1

, 23 941000

. xx

A

B

xA = 0.02394 x

B, x

A + x

B = 1

1.02394 xB = 1

xB1

102394. = 0.98, xA = 0.02 Ans.

Second Method : Let wt of solvent = 1000 gm molality = 1.33

= moles of solute

mole fraction of solute = solvent of molessolute of molessolute of moles

, 18/100033.133.1

181000+m

m

mole fraction of solute = 0.02

mole fraction of solvent = 1 – 0.02 = 0.98



Ex. The density of 3 M solution of sodium thiosulphate (Na2S2O3) is 1.25 g/mL. Calculate

(i) amount of sodium thiosulphate

(ii) mole fraction of sodium thiosulphate

(iii) molality of Na+ and S2O32– ions

Sol. (i) Let us consider one litre of sodium thiosulphate solution.

wt. of the solution = density × volume (mL)

= 1.25 × 1000 = 1250 g.

wt. of Na2S2O3 present in 1 L of the solution

= molarity × mol. wt.

= 3 × 158 = 474 g. Ans.

wt. % of Na2S2O3 = 474

1250100 = 37.92%

(ii) Wt. of solute (Na2S

2O

3) = 474 g.

Moles of solute = 474158

3 Ans.

Wt. of solvent (H2O) = 1250 – 474 = 776 g

Moles of solvent = 77618

4311.

mole fraction of Na2S2O3 = 3

3 43110 063

..

(iii) Molality of Na2S2O3 = moles of N S Oa

wt. of solvent in grams2 2 3 1000 =

3776

1000 3 865.

1 mole of Na2S2O3 contains 2 moles of Na+ ions and 1 mole of S2O32– ions.

molality of Na+ = 2 × 3.865 = 7.73 m

Molality of S2O32– = 3.865 m. Ans.



Solved Objective

Ex.1 8 litre of H2 and 6 litre of Cl2 are allowed to react to maximum possible extent. Find out thefinal volume of reaction mixture. Suppose P and T remains constant throughout the course ofreaction -(A) 7 litre (B) 14 litre (C) 2 litre (D) None of these.

Sol. (B)H2 + Cl2 2 HCl

Volume before reaction 8 lit 6 lit 0Volume after reaction 2 0 12

Volume after reaction= Volume of H2 left + Volume of HCl formed = 2 + 12 = 14 lit

Ex.2 Naturally occurring chlorine is 75.53% Cl35 which has an atomic mass of 34.969 amu and 24.47%Cl37 which has a mass of 36.966 amu. Calculate the average atomic mass of chlorine-(A) 35.5 amu (B) 36.5 amu (C) 71 amu (D) 72 amu

Sol. (A)Average atomic mass

= 100

massatomicitsisotopeof%massatomicitsisotopeof%

III

= 75 53 34 969 24 47 36 96

100. . . .x x

= 35.5 amu.

Ex.3 Calculate the mass in gm of 2g atom of Mg-(A) 12 gm (B) 24 gm (C) 6 gm (D) None of these.

Sol. (D) 1 gm atom of Mg has mass = 24 gm 2 gm atom of Mg has mass = 24 x 2 = 48 gm.

Ex.4 In 5 g atom of Ag (At. wt. of Ag = 108), calculate the weight of one atom of Ag -(A) 17.93 × 10–23gm (B) 16.93 × 10–23 gm(C) 17.93 × 1023 gm (D) 36 × 10–23 gm

Sol. (A) N atoms of Ag weigh 108 gm

1 atom of Ag weigh = 108N

= 108

6 023 1023. x = 17.93 × 10–23 gm.

Ex.5 In 5g atom of Ag (at. wt. = 108), calculate the no. of atoms of Ag -(A) 1 N (B) 3N (C) 5 N (D) 7 N.

Sol. (C) 1 gm atom of Ag has atoms = N 5 gm atom of Ag has atoms = 5N.

Ex.6 Calculate the mass in gm of 2N molecules of CO2 -(A) 22 gm (B) 44 gm (C) 88 gm (D) None of these.

Sol. (C) N molecules of CO2 has molecular mass = 44. 2N molecules of CO2 has molecular mass = 44 x 2 = 88 gm.

Ex.7 How many carbon atoms are present in 0.35 mol of C6H12O6 -(A) 6.023 × 1023 carbon atoms (B) 1.26 × 1023 carbon atoms(C) 1.26 × 1024 carbon atoms (D) 6.023 × 1024 carbon atoms



Sol. (C) 1 mol of C6H12O6 has = 6 N atoms of C

0.35 mol of C6H12O6 has = 6 × 0.35 N atoms of C= 2.1 N atoms = 2.1 × 6.023 × 1023 = 1.26 × 1024 carbon atoms

Ex.8 How many molecules are in 5.23 gm of glucose (C6H12O6) -(A) 1.65 × 1022 (B) 1.75 × 1022 (C) 1.75 × 1021( D) None of these

Sol. (B) 180 gm glucose has = N molecules

5.23 gm glucose has = 180

10023.623.5 23

= 1.75 × 1022 molecules

Ex.9 What is the weight of 3.01 × 1023 molecules of ammonia -(A) 17 gm (B) 8.5 gm (C) 34 gm (D) None of these

Sol. (B) 6.023 × 1023 molecules of NH3 has weight = 17 gm 3.01 × 1023 molecules of NH3 has weight

= 23

23

10023.61001.317

= 8.50 gm

Ex.10 How many significant figures are in each of the following numbers -(a) 4.003 (b) 6.023 × 1023 (c) 5000(A) 3, 4, 1 (B) 4, 3, 2 (C) 4, 4, 4 (D) 3, 4, 3

Sol. (C)

Ex.11 How many molecules are present in one m l of water vapours at STP -(A) 1.69 × 1019 (B) 2.69 × 10–19 (C) 1.69 × 10–19 (D) 2.69 × 1019

Sol. (D) 22.4 litre water vapour at STP has

= 6.023 × 1023 molecules 1 × 10–3 litre water vapours at STP has

=4.2210023.6 23

× 10–3 = 2.69 × 10+19

Ex.12 How many years it would take to spend Avogadro's number of rupees at the rate of 1 millionrupees in one second -(A) 19.098 × 1019 years (B) 19.098 years(C) 19.098 × 109 years (D) None of these

Sol. (C) 106 rupees are spent in 1sec.

6.023 × 1023 rupees are spent in = 6

23

1010023.61

sec

= 36524606010

10023.616

23

years , = 19.098 × 109 year



Ex.13 An atom of an element weighs 6.644 × 10–23 g. Calculate g atoms of element in 40 kg-(A) 10 gm atom (B) 100 gm atom (C) 1000 gm atom (D) 104 gm atom

Sol. (C) weight of 1 atom of element

= 6.644 × 10–23 gm weight of 'N' atoms of element

= 6.644 × 10–23 × 6.023 × 1023 = 40 gm 40 gm of element has 1 gm atom.

40 x 103 gm of element has 40

1040 3 , = 103 gm atom.

Ex.14 Calculate the number of Cl– and Ca+2 ions in 222 g anhydrous CaCl2 -(A) 2N ions of Ca+2 4 N ions of Cl– (B) 2N ions of Cl– & 4N ions of Ca+2

(C) 1N ions of Ca+2 & 1N ions of Cl– (D) None of these.Sol. (A)

mol. wt. of CaCl2 = 111 g 111 g CaCl2 has = N ions of Ca+2

222g of CaCl2 has 111

222N

= 2N ions of Ca+2

Also 111 g CaCl2 has = 2N ions of Cl–

222 g CaCl2 has = 111

222N2 ions of Cl–

= 4N ions of Cl– .

Ex.15 The density of O2 at NTP is 1.429g / litre. Calculate the standard molar volume of gas-(A) 22.4 lit. (B) 11.2 lit (C) 33.6 lit (D) 5.6 l it.

Sol. (A) 1.429 gm of O2 gas occupies volume = 1 l itre.

32 gm of O2 gas occupies = 32

1429. ,= 22.4 litre/mol.

Ex.16 Which of the following will weigh maximum amount-(A) 40 g iron (B) 1.2 g atom of N(C) 1 × 1023 atoms of carbon (D) 1.12 litre of O2 at STP

Sol. (A)(A) Mass of iron = 40 g(B) Mass of 1.2 g atom of N = 14 × 1.2 = 16.8 gm

(D) Mass of 1 × 1023 atoms of C = 23

23

10023.610112

= 1.99 gm.

(D) Mass of 1.12 litre of O2 at STP = 4.22

2.132 = 1.6 g

Ex.17 How many moles of potassium chlorate to be heated to produce 11.2 li tre oxygen -

(A) 12

mol (B) 13

mol (C) 14

mol (D) 23

mol.



Sol. (B)2 KClO3 2KCl + 3O2

Mole for reaction 2 2 33 × 22.4 litre O2 is formed by 2 mol KClO3

11.2 litre O2 is formed by 4.2232.112

= 13

mol KClO3

Ex.18 Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone (CaCO 3).(A) 104.4 kg (B) 105.4 kg (C) 212.8 kg (D) 106.4 kg

Sol. (D) 100 kg impure sample has pure

CaCO3 = 95 kg 200 kg impure sample has pure CaCO3

= 100

20095 = 190 kg. CaCO3 CaO + CO2

100 kg CaCO3 gives CaO = 56 kg.

190 kg CaCO3 gives CaO = 100

19056 = 106.4 kg.

Ex.19 The chloride of a metal has the formula MCl3. The formula of i ts phosphate will be-(A) M2PO4 (B) MPO4 (C) M3PO4 (D) M(PO4)2

Sol. (B) AlCl3 as it is AlPO4

Ex.20 A silver coin weighing 11.34 g was dissolved in nitric acid. When sodium chloride was added tothe solution all the silver (present as AgNO3) was precipitated as si lver chloride. The weight ofthe precipitated silver chloride was 14.35 g. Calculate the percentage of silver in the coin -(A) 4.8 % (B) 95.2% (C) 90 % (D) 80%

Sol. (B)Ag + 2HNO3 AgNO3 + NO2 + H2O108AgNO3 + NaCl AgCl + NaNO3

143.5 143.5 gm of silver chloride would be precipitated by 108 g of silver. or 14.35 g of silver chloride would be precipitated 10.8 g of si lver. 11.34 g of silver coin contain 10.8 g of pure silver.

100 g of silver coin contain 10 81134

..

× 100 = 95.2 %.



SOLVED SUBJECTIVE

Ex.1 Calculate the following for 49 gm of H2SO4

(a) moles (b) Molecules (c) Total H atoms (d) Total O atoms

(e) Total electrons

Sol. Molecular wt of H2SO4 = 98

(a) moles = wt in gm

molecular wt4998 =

12

mole

(b) Since 1 mole = 6.023 × 1023 molecules.

12

mole = 6.023 × 1023 × 12

molecules = 3.011 × 1023 molecules

(c) 1 molecule of H2SO4 Contains 2 H atom

3.011 × 1023 of H2SO4 contain 2 × 3.011 × 1023 atoms = 6.023 × 1023 atoms

(d) 1 molecules of H2SO4 contains 4 O atoms

3.011 × 1023 molecular of H2SO4 contains = 4 × 3.011 × 1023 = 12.044 × 1023

(e) 1 molecule of H2SO4 contains 2H atoms + 1 S atom + 4 O atom

this means 1 molecule of H2SO4 Contains (2 + 16 + 4 × 8) e–

So 3.011 × 1023 molecules have 3.011 × 1023 × 50 electrons = 1.5055 × 1025 e–

Ex.2 Calculate the total ions & charge present in 4.2 gm of N–3

Sol. mole = wt in gmIonic wt =

4 214.

= 0.3

total no of ions = 0.3 × NA ions

total charge = 0.3 NA × 3 × 1.6 × 10–19

= 0.3 × 6.023 × 1023 × 3 × 1.6 × 10–19 , = 8.67 × 104 C Ans.

Ex.3 Find the total number of iron atom present in 224 amu iron.

Sol. Since 56 amu = 1 atom

therefore 224 amu = 1

56 × 224 = 4 atom Ans.

Ex.4 A compound containing Ca, C, N and S was subjected to quantitative analysis and formula massdetermination. A 0.25 g of this compound was mixed with Na2CO3 to convert all Ca into 0.16 g CaCO3.A 0.115 gm sample of compound was carried through a series of reactions until all its S was changedinto SO4

2– and precipitated as 0.344 g of BaSO4. A 0.712 g sample was processed to liberated all of itsN as NH3 and 0.155 g NH3 was obtained. The formula mass was found to be 156. Determine theempirical and molecular formula of the compound.

Sol. Moles of CaCO3 = 016100.

= Moles of Ca

Wt of Ca = 016100.

× 40



Mass % of Ca = 0 16100

401000 25

25 6.

..

Similarly Mass % of S = 0 344233

32 1000 115

41..

Similarly Mass % of N = 0 155

1714

0 712100.

. = 17.9

Mass % of C = 15.48

Now :

Elements Ca S N C

Mass % 25.6 41 17.9 15.48

Mol ratio 0.64 1.28 1.28 1.29

Simple ratio 1 2 2 2

Empirical formula = CaC2N2S2,

Molecular formula wt = 156 , n × 156 = 156 n = 1

Hence, molecular formula = CaC2N2S2

Ex.5 A polystyrne having formula Br3C6H3(C3H8)n found to contain 10.46% of bromine by weight. Find thevalue of n. (At. wt. Br = 80)

Sol. Let the wt of compound is 100 gm & molecular wt is M

Then moles of compound = 100M

Moles of Br = 100M

× 3

wt of Br = 100M

× 3 × 80 = 10.46

M = 2294.45 = 240 + 75 + 44 n , Hence n = 45 Ans.

Ex.6 A sample of clay was partially dried and then analysed to 50% silica and 7% water. The original claycontained 12% water. Find the percentage of silica in the original sample.

Sol. In the partially dried clay the total percentage of silica + water = 57%. The rest of 43% must be some

impurity. Therefore the ratio of wts. of silica to impurity = 5043

. This would be true in the original sample

of silica.

The total percentage of silica + impurity in the original sample is 88. If x is the percentage of silica,

xx88

5043–

; x = 47.3% Ans.

Ex.7 A mixture of CuSO4.5H

2O and MgSO

4. 7H

2O was heated until all the water was driven-off. if 5.0 g of

mixture gave 3 g of anhydrous salts, what was the percentage by mass of CuSO4.5H2O in the originalmixture ?

Sol. Let the mixture contain x g CuSO4.5H2O

120246

x–55.159

5.249x

= 3 x = 3.56

Mass percentage of CuSO4. 5H

2O = 100

556.3

= 71.25 % Ans.



Ex .8 367.5 gm KClO3 (M = 122.5) when heated, How many litre of oxygen gas is proudced at S.T.P.

Sol. KClO3 KCl + O2

Applying POAC on O, moles of O in KClO3 = moles of O in O2

3 × moles of KClO3 = 2 × moles of O2

3 × 367 5122 5

.

. = 2 × n, n =

32

× 367 5122 5

.

.Volume of O2 gas at S.T.P = moles × 22.4

= 32

367 5122 5

22 4..

. = 9 × 11.2 = 100.8 lit Ans.

Ex.9 0.532 g of the chloroplatinate of a diacid base on ignition left 0.195 g of residue of Pt. Calculatemolecular weight of the base (Pt = 195)

Sol. Suppose the diacid base is B.

B + H2PtCl6\ BH2PtCl6 Pt

diacid acid chloroplatinate

base 0.532 g 0.195 g

Since Pt atoms are conserved, applying POAC for Pt atoms,

moles of Pt atoms in BH2PtCl6 = moles of Pt atoms in the product

1 × moles of BH2PtCl6 = moles of Pt in the product

0 532 01951956

. .mol. wt. of BH2PtCl

mol. wt. of BH2PtCl6 = 532

From the formula BH2PtCl6, we get

mol. wt. of B = mol. wt. of BH2PtCl6 – mol. wt. of H2PtCl6= 532 – 410 = 122. Ans.

Ex.10 10 mL of a gaseous organic compound containing. C, H and O only was mixed with 100 mL of oxygenand exploded under conditions which allowed the water formed to condense. The volume of the gasafter explosion was 90 mL. On treatment with potash solution, a further contraction of 20 mL in volumewas observed. Given that the vapour density of the compound is 23, deduce the molecular formula. Allvolume measurements were carried out under the same conditions.

Sol. CxHyOz + xy z4 2

– O2 xCO2 + y2

H2O

10 ml

after explosion volume of gas = 90 ml

90 = volume of CO2 gas + volume of unreacted O

2

on treatment with KOH solution volume reduces by 20 ml. This means the volume of CO2 = 20 ml

the volume of unreacted O2 = 70 ml

volume of reacted O2 = 30 ml

V.D of compoud = 23

molecular wt 12x + y + 16z = 46 ...(1)

from equation we can write



104 2

30x y z– , x + y4

– z2

= 3

4x + y – 2z = 12 ...(2)

& 10x = 20 x = 2

from eq. (1) & (2) ; z = 1 & y = 6; Hence C2H6O Ans.

Ex.11 A sample of coal gas contained H2, CH4 and CO. 20 mL of this mixture was exploded with 80 mL ofoxygen. On cooling, the volume of gases was 68 mL. There was a contraction of 10 mL. When treatedwith KOH. Find the composition of the original mixture.

Sol. H2 + CH4 + CO; at H2 = x ml

CH4 = y ml; CO = (20 – x – y) ml

H2 + CH4 + CO + O2 CO2 + H2O

x y 20 – x – y

on cooling the volume of gases = 68 ml = volume of CO2 + unreacted O2

volume contraction due to KOH = 10 ml

this means volume of CO2 = 10 ml

volume of unreacted O2 = 58 ml

volume of reacted O2 = 80 – 58 = 22 ml

Applying POAC on C; y + 20 – x – y = volume of CO2, 20 – x = 10 x = 10

Applying POAC on H; 2x + 4y = 2x moles of H2O; moles of H2O = x + 2y

Applying POAC on O

1 × moles of CO + 2 × moles of O2 = 2 × moles of CO2 + 1 × moles of H2O

1 × 20 – x – y + 2 × 22 = 2 × 10 + x + 2y

20 – x – y + 44 = 20 + x + 2y; 2x + 3y = 44

3y = 44 – 20 = 24; y = 8 ml; x = 10 ml; volume of CO = 20 – x – y = 2 ml Ans.



Class Room Problems

Problem.1 From the following reaction sequenceCl2 + 2KOH KCl + KClO + H2O3KClO 2KCl + KClO34KClO3 3KClO4 + KCl

Calculate the mass of chlorine needed to produce100 g of KClO4.Sol. 205.04 gm

Problem.2 Calculate the weight of FeO produced from2 g VO and 5.75 g of Fe2O3. Also report the limitingreagent. VO + Fe2O3 FeO + V2O5Sol. 5.175 gm

Problem.3 A polystyrene, having formula Br3C6H3(C8H8)n was prepared by heating styrene withtribromobenzoyl peroxide in the absence of air. If itwas found to contain 10.46% bromine be weight, findthe value of n.Sol. 19

Problem.4 5 mL of a gaseous hydrocarbon wasexposed to 30 mL of O2. The resultant gas, on coolingis found to measure 25 mL of which 10 mL are absorbedby NaOH and the remainder by pyrogallol. Determinemolecular formula of hydrocarbon. All measurementsare made at constant pressure and temperature.Sol. C2H4

Problem.5 A gaseous alkane is exploded with oxygen.The volume of O2 for complete combustion to CO2formed is in the ratio of 7 : 4. Deduce molecularformula of alkane.Sol. C2H6

Problem.6 A sample of gaseous hydrocarbonoccupying 1.12 litre at NTP, when completely burnt inair produced 2.2 g CO2 and 1.8 g H2O. Calculate theweight of hydrocarbon taken and the volume of O2 atNTP required for its combustion.Sol. 0.8 gm, 2.24 L



Problem.7 16 mL of a gaseous aliphatic compoundCnH3nOm was mixed with 60 mL O2 and sparked. Thegas mixture on cooling occupied 44 mL. After treatmentwith KOH solution, the volume of gas remaining was12 mL. Deduce the formula of compound. Allmeasurements are made at constant pressure androom temperature.Sol. C2H6O

Problem.8 In what ratio should you mix 0.2M NaNO3and 0.1M Ca(MO3)2 solution so that in resultingsolution, the concentration of –ve ion is 50% greaterthan the concentration of +ve ion ?Sol. 1/2

Problem.9 How much BaCl2 would be needed to make250 mL of a solution having same concentration ofCl– as the one containing 3.78 g of NaCl per 100 mL ?Sol. 16.8 gm

Problem.10 What is the purity of conc. H2SO4solution (specific gravity 1.8 g/mL), if 5.0 mL of thissolution is neutralized by 84.6 mL of 2.0 N NaOH ?Sol. 92.12%

Problem.11 A sample of H2SO4 (density 1.787 g mL–1) islabelled as 86% by weight. What is molarity of acid ?What volume of acid has to be used to make 1 litre of0.2 M H2SO4 ?Sol. 15.68 M, 0.013 L

Problem.12 Mole fraction of I2 in C6H6 is 0.2.Calculate molality of I2 in C6H6.Sol. 3.2



Problem.13 A drop (0.05 mL) of 12 M HCl is spreadover a thin sheet of aluminium foil (thickness 0.10 mmand density of Al = 2.70 g/mL). Assuming whole ofthe HCl is used to dissolve Al, what will be the maximumarea of hole produced in foi l ?Sol. 5.4×10–3

Problem.14 What would be the molarity of solutionobtained by mixing equal volumes of 30% by weightH2SO4 (d = 1.218 g mL–1) and 70% by weight H2SO4(d = 1.610g mL–1) ? If the resulting solution has density1.425 g/mL, calculate its molality.Sol. 7.67 M

Problem.15 A mixture of Al and Zn weighing 1.67 gwas completley dissolved in acid and evolved 1.69litre of H2 at NTP. What was the weight of Al in originalmixture ?Sol. 1.21 gm

Problem.16 A mixture of FeO and Fe3O4 when heatedin air to constant weight, gains 5% in its weight. Findout composition of mixture.Sol. 20.25

Problem.17 25.4 g of I2 and 14.2 g of Cl2 are madeto react completely to yield a mixture of ICl and ICl3.Calculate mole of ICl and ICl3 formed.

Sol. 0.1,0.1

Problem.18 A mixture of HCOOH and H2C2O4 isheated with conc. H2SO4. The gas produced iscollected and on treating with KOH solution the volume

of the gas decreases by 61

th. Calculate molar ratio

of two acids in original mixture.

Sol. 1/4



Problem.19 For the reaction, N2O5(g) 2NO2(g)+ 0.5O2(g), calculate the mole fraction of N2O5(g)decomposed at a constant volume and temperature,if the initial pressure is 600 mm Hg and the pressureat any time is 960 mm Hg. Assume ideal gas behaviour.

Sol. 0.375

Problem.20 0.22 g sample of volatile compound,containing C, H and Cl only on combustion in O2 gave0.195 g CO2 and 0.0804 g H2O. If 0.120 g of thecompound occupied a volume of 37.24 mL at 105ºand 768 mm of pressure, calculate molecular formulaof compound.

Sol. C2H4Cl2

Problem.21 2.0 g sample containing Na2CO3 andNaHCO3 loses 0.248 g when heated to 300ºC, thetemperature at which NaHCO3 decomposes to Na2CO3,and H2O. What is % of Na2CO3 in mixture ?

Sol. 66.4%

Problem.22 10 mL of a solution of KCl containingNaCl gave on evaporation 0.93 g of the mixed saltwhich gave 1.865 g of AgCl by the reaction withAgNO3. Calculate the quantity of NaCl in 10 mL ofsolution.

Sol. 0.67 gm

Problem.23 A sample of CaCO3 and MgCO3 weighed2.21 g is ignited to constant weight of 1.152 g. Whatis the composition of mixture ? Also calculale thevolume of CO2 evolved at 0ºC and 76 cm of pressure.

Sol. 46% 1.19

Problem.24 2.0 g of a mixture of carbonate,bicarbonate and chloride of sodium on heatingproduced 56 mL of CO2 at NTP. 1.6 g of the samemixture required 25 mL of N HCl solution forneutra l ization . Calculate percentage of eachcomponent present in mixture.

Sol. 69.56%

Problem.25 Igniting MnO2 in air converts i tquantitatively to Mn3O4. A sample of pyrolusite hasMnO2 80%, SiO2 15% and rest having water. Thesample is heated in air to constant mass. What isthe % of Mn in ignited sample ?

Sol. 59.37



Problem.26 A solid mixture 5 g consits of lead nitrateand sodium nitrate was heated below 600ºC untilweight of residue was constant. If the loss in weightis 28%, find the amount of lead nitrate and sodiumnitrate in mixture.

Sol. 3.32 gm

Problem.27 Determine the formula of ammonia formthe folloiwng data :

(i) Volume of ammonia = 25 mL.

(ii) Volume on addition of O2 after explosion = 71.2 mL.

(iii) Volume after explosion and reaction with O2 oncooling = 14.95 mL.

(iv) Volume after being absorbed by alkal inepyrogallol = 12.5 mL.

Sol.

Problem.28 A mixture of ethane (C2H6) and ethene(C2H4) occupies 40 litre at 1.00 atm and at 400 K.The mixture reacts completely with 130 g of O2 toproduce CO2 and H2O. Assuming ideal gas behaviour,calculate the mole fractions of C2H4 and C2H6 in themixture.

Sol. 0.63, 0.37

Problem. 29 0.50 g of a mixture of K2CO3 and Li2CO3required 30 mL of 0.25 N HCl solution for neturalization.What is % composition of mixture ?

Sol.

Problem.30 A mixture in which the mole ratio of H2and O2 is 2 : 1 is used to prepare water by thereaction,

2H2(g) + O2(g) 2H2O(g)

The total pressure in the container is 0.8 atm at20ºC before the reaction. Determine the final pressureat 120ºC after reaction assuming 80% yield of water.

Sol. 0.7865



OBJECTIVE PROBLEMS (JEE MAIN)EXERCISE – ISingle correct1. For the reaction

2x + 3y + 4z 5wInitially if 1 mole of x, 3 mole of y and 4 mole ofz is taken. If 1.25 mole of w is obtained then %yield of this reaction is(A) 50% (B) 60%(C) 70% (D) 40%

Sol.

2. A solution of A (MM = 20) and B (MM = 10),[Mole fraction XB = 0.6] having density 0.7 gm/ml then molarity and molality of B in this solutionwill be ________________ and ______________respectively.(A)30M,75m (B) 40M,75m(C) 30M,65m (D) 50M,55m

Sol.

3. 125 ml of 8% w/w NaOH solution (sp. gravity 1)is added to 125 ml of 10% w/v HCl solution. Thenature of resultant solution would be _________(A)Acidic (B) Basic(C) Neutral (D) None

Sol.

4. Ratio of masses of H2SO4 and Al2 (SO4)3 isgrams each containing 32 grams of S is__________.(A)0.86 (B) 1.72(C) 0.43 (D) 2.15

Sol.

5. The vapour density of a mixture of gas A(Molecular mass = 40) and gas B (Molecular mass= 80) is 25. Then mole % of gas B in the mixturewould be(A)25% (B) 50% (C) 75%(D) 10%

Sol.

6. For the reaction2A + 3B + 5C 3DInitially if 2 mole of A, 4 mole of B and 6 mole ofC is taken, With 25% yield, moles of D whichcan be produced are _____________.(A)0.75 (B) 0.5(C) 0.25 (D) 0.6

Sol.

7. Fill in the blanks in the following table.Compound Grams Grams Molality

Compd Waterof Compd Na2CO3

______ 2500.0125(A)0.331 (B) 0.662(C) 0.165 (D) 0.993

Sol.



8. Equal volumes of 10% (v/v) of HCl is mixed with10% (v/v) NaOH solution. If density of pure NaOHis 1.5 times that of pure HCl then the resultantsolution be :(A) basic (B) neutral(C) acidic (D) can’t be predicted.

Sol.

9. A definite amount of gaseous hydrocarbon wasburnt with just sufficient amount of O2. Thevolume of all reactants was 600 ml, after theexplosion the volume of the products [CO2(g)and H2O(g)] was found to be 700 ml under thesimilar conditions. The molecular formula of thecompound is :(A) C3H8 (B) C3H6

(C) C3H4 (D) C4H10

Sol.

10. One gram of the silver salt of an organic dibasicacid yields, on strong heating, 0.5934 g of silver.If the weight percentage of carbon in it 8 timesthe weight percentage of hydrogen and half theweight percentage of oxygen, determine themolecular formula of the acid. [Atomic weight ofAg = 108](A) C4H6O4 (B) C4H6O6

(C) C2H6O2 (D) C5H10O5

Sol.

11. One mole mixture of CH4 & air (containing 80%N2 20% O2 by volume) of a composition suchthat when underwent combustion gave maximumheat (assume combustion of only CH4). Thenwhich of the statements are correct, regardingcomposition of initial mixture. (X presents molefraction)

(A) 24 OCH X,

111

X = 118

X,112

2N

(B) 21

X,81

X,83

X24 N2OCH

(C) 32

X,61

X,61

X24 N2OCH

(D) Data insufficientSol.

12. C6H5OH(g) + O2(g) CO2(g) + H2O(l)Magnitude of volume change if 30 ml of C6H5OH(g) is burnt with excess amount of oxygen, is(A) 30 ml (B) 60 ml(C) 20 ml (D) 10 ml

Sol.

13. 10 ml of a compound containing ‘N’ and ‘O’ ismixed with 30 ml of H

2 to produce H

2O (l) and 10

ml of N2(g). Molecular formula of compound ifboth reactants reacts completely, is(A) N2O (B) NO2

(C) N2O3 (D) N2O5

Sol.



14. Similar to the % labelling of oleum, a mixture ofH3PO4 and P4O10 is labelled as (100 + x) % wherex is the maximum mass of water which can reactwith P4O10 present in 100 gm mixture of H3PO4

and P4O10. If such a mixture is labelled as 127%Mass of P4O10 is 100 gm of mixture, is(A) 71 gm (B) 47 gm(C) 83gm (D) 35 gm

Sol.

15. Mass of sucrose C12H22O11 produced by mixing84 gm of carbon, 12 gm of hydrogen and 56 lit.O2 at 1 atm & 273 K according to given reaction,is C(s) + H2(g) + O2(g) C12H22O11(s)(A) 138.5 (B) 155.5(C) 172.5 (D) 199.5

Sol.

16. If 50 gm oleum sample rated as 118% is mixedwith 18 gm water, then the correct option is(A) The resulting solution contains 18 gm of waterand 118 gm H2SO4

(B) The resulting solution contains 9 gm of waterand 59 gm H2SO4

(C) The resulting solution contains only 118 gmpure H2SO4

(D) The resulting solution contains 68 gm of pureH2SO4

Sol.

17. In the quantitative determination of nitrogenusing Duma’s method, N2 gas liberated from 0.42gm of a sample of organic compound wascollected over water. If the volume of N2 gas

collected was 11

100 ml at total pressure 860 mm

Hg at 250 K, % by mass of nitrogen in the organiccompound is[Aq. tension at 250K is 24 mm Hg and R = 0.08 Latm mol–1 K–1]

(A) %3

10(B) %

35

(C) %3

20(D) %

3100

Sol.

18. 40 gm of a carbonate of an alkali metal oralkaline earth metal containing some inertimpurities was made to react with excess HClsolution. The liberated CO2 occupied 12.315 lit.at 1 atm & 300 K. The correct option is(A) Mass of impurity is 1 gm and metal is Be(B) Mass of impurity is 3 gm and metal is Li(C) Mass of impurity is 5 gm and metal is Be(D) Mass of impurity is 2 gm and metal is Mg

Sol.

19. The percentage by mole of NO2 in a mixtureNO2(g) and NO(g) having average molecular mass34 is :(A) 25% (B) 20%(C) 40% (D) 75%

Sol.



20. The minimum mass of mixture of A2 and B4 requiredto produce at least 1 kg of each product is :(Given At. mass of ‘A’ = 10; At mass of ‘B’ = 120)5A

2 + 2B

4 2AB

2 + 4A

2B

(A) 2120 gm (B) 1060 gm(C) 560 gm (D) 1660 gm

Sol.

21. The mass of CO2 produced from 620 gm mixtureof C2H4O2 & O2, prepared to produce maximumenergy is(A) 413.33 gm (B) 593.04 gm(C) 440 gm (D) 320 gm

Sol.

22. Assuming complete precipitation of AgCl,calculate the sum of the molar concentration ofall the ions if 2 lit of 2M Ag2SO4 is mixed with 4 litof 1 M NaCl solution is :(A) 4M (B) 2M(C) 3M (D) 2.5 M

Sol.

23. 12.5 gm of fuming H2SO

4 (labelled as 112%) is

mixed with 100 lit water. Molar concentration ofH+ in resultant solution is :[Note : Assume that H2SO4 dissociate completelyand there is no change in volume on mixing]

(A) 700

2(B)

3502

(C) 350

3(D)

7003

Sol.

24. 74 gm of sample on complete combustion gives132 gm CO2 and 54 gm of H2O. The molecularformula of the compound may be(A) C5H12 (B) C4H10O(C) C3H6O2 (D) C3H7O2

Sol.

25. The % by volume of C4H10 in a gaseous mixtureof C4H10, CH4 and CO is 40. When 200 ml of themixture is burnt in excess of O2. Find volume (inml) of CO2 produced.(A) 220 (B) 340(C) 440 (D) 560

Sol.

26. What volumes should you mix of 0.2 M NaCl arid0.1 M CaCl2 solution so that in resulting solutionthe concentration of positive ion is 40% lesserthan concentration of negative ion. Assumingtotal volume of solution 1000 ml.(A) 400 ml NaCl, 600 ml CaCl2(B) 600 ml NaCl, 400 ml CaCl2(C) 800 ml NaCl, 200 ml CaCl2(D) None of these

Sol.



27. An iodized salt contains 0.5% of Nal. A personconsumes 3 gm of salt everyday. The number ofiodide ions going into his body everyday is(A) 10–4 (B) 6.02 × 10–4

(C) 6.02 × 1019 (D) 6.02 × 1023

Sol.

28. The pair of species having same percentage(mass) of carbon is :(A) CH3COOH and C6H12O6

(B) CH3COOH and C2H5OH(C) HCOOCH3 and C12H22 O11

(D) C6H12O6 and C12H22O11

Sol.

29. 200 ml of a gaseous mixture containing CO, CO2

and N2 on complete combustion in just sufficient

amount of O2 showed contraction of 40 ml. Whenthe resulting gases were passed through KOHsolu ti on i treduces by 50 % then calculate the volume ratio

of 22 NCOCO V:V:V in original mixture.

(A) 4 : 1 : 5 (B) 2 : 3 : 5(C) 1 : 4 : 5 (D) 1 : 3 : 5

Sol.

30. Density of a gas relative to air is 1.17. Find themol. mass of the gas. [M

air = 29 g/mol]

(A) 33.9 (B) 24.7(C) 29 (D) 22.3

Sol.

31. Weight of oxygen in Fe2O3 and FeO is in thesimple ratio for the same amount of iron is :(A) 3 : 1 (B) 1 : 2(C) 2 : 1 (D) 3 : 1

Sol.

32. Two elements X (atomic mass 16) and Y (atomicmass 14) combine to form compounds A, B andC. The ratio of different masses of Y whichcombines with a fixed mass of X in A, B andC is 1:3:5. If 32 parts by mass of X combineswith 84 parts by mass of Y in B, then in C,16 parts by mass of X will combine with___parts by mass of Y.(A) 14 (B) 42(C) 70 (D) 84

Sol.

33. In a textile mill, a double-effect evaporatorsystem concentrates weak liquor containing4% (by weight) caustic soda to produce a lyecontaining 25% solids (by weight). Calculatethe weight of the water evaporate per 100-kgfeed in the evaporator.(A) 125.0 g (B) 50.0 kg(C) 84.0 kg (D) 16.0 kg

Sol.

34. Zinc ore (zinc sulphide) is treated with sulphuricacid, leaving a solution with some undissolvedbits of material and releasing hydrogen sulphidegas. If 10.8g of zinc ore is treated with 50.0ml of sulphuric acid (density 1.153 g/ml), 65.1gof solution and undissolved material remains. Inaddition, hydrogen sulphide (density 1.393 g/L) is evolved. What is the volume (in liters) ofthis gas?(A) 4.3 (B) 3.35(C) 4.67 (D) 2.40



Sol.

35. A sample of an ethanol-water solution has avolume of 54.2 cm3 and a mass of 49.6g. Whatis the percentage of ethanol (by mass) in thesolution? (Assume that there is no change involume when the pure compounds are mixed.)The density of ethanol is 0.80 g/cm3 and thatof water is 1.00 g/cm3.(A) 18.4% (B) 37.1%(C) 33.95% (D) 31.2%

Sol.

36. A student gently drops an object weighing15.8 g into an open vessel that is full ofethanol, so that a volume of ethanol spills outequal to the volume of the object. Theexperimenter now finds that the vessel and itscontents weigh 10.5 g more than the vesselfull of ethanol only. The density of ethanol is0.789 g/cm3. What is the density of theobject?(A) 6.717 gm/cm3 (B) 4.182 gm/cm3

(C) 1.563 gm/cm3 (D) 2.352 gm/cm3

Sol.

37. A person needs on average of 2.0 mg ofriboflavin (vitamin B2) per day. How many gmof butter should be taken by the person perday if it is the only source of riboflavin? Buttercontains 5.5 microgram riboflavin per gm.(A) 363.6 gm (B) 2.75 mg(C) 11 gm (D) 19.8 gm

Sol.

38. A sample of clay contains 40% silica and 15%water. The sample is partially dried by which itloses 5 gm water. If the percentage of waterin the partially dried clay is 8, calculate thepercentage of silica in the partially dried clay.(A) 21.33% (B) 43.29%(C) 75% (D) 50%

Sol.

39. The density of quartz mineral was determinedby adding a weighed piece to a graduatedcylinder containing 51.2ml water. After thequartz was submersed, the water level was65.7 ml. The quartz piece weighed 38.4g. Whatwas the density of quartz?(A) 1.71 gm/ml (B) 1.33 gm/ml(C) 2.65 gm/ml (D) 1.65gm/ml

Sol.

40. Which has maximum number of atoms of oxygen(A) 10 ml H2O(l)(B) 0.1 mole of V2O5(C) 12 gm O3(g)(D) 12.044 × 1022 molecules of CO2

Sol.

41. Mass of one atom of the element A is 3.9854× 10–23. How many atoms are contained in 1gof the element A?(A) 2.509 × 1023 (B) 6.022 × 1023

(C) 12.044 × 1023 (D) NoneSol.

42. The number of atoms present in 0.5 g-atomsof nitrogen is same as the atoms in(A) 12 g of C (B) 32 g of S(C) 8 g of oxygen (D) 24g of Mg



Sol.

43. A graph is plotted for an element, by puttingits weight on X-axisand the corresponding number of number ofatoms on Y-axis. Determinethe atomic weight of the element for which thegraph is plotted.

(A) infinite (B) 40(C) 0.025 (D) 20

Sol.

44. The O18/O16 ratio in some meteorites is greaterthan that used to calculate the average atomicmass of oxygen on earth. The average massof an atom of oxygen in these meteorites is___ that of a terrestrial oxygen atom?(A) equal to (B) greater than(C) less than (D) None of these

Sol.

45. If isotopic distribution of C12 and C14 is 98.0%and 2.0% respectively, then the number of C14

atoms in 12 gm of carbon is(A) 1.032 ×1022 (B) 1.20 ×1022

(C) 5.88 ×1023 (D) 6.02 ×1023

Sol.

46. At one time there was a chemical atomic weightscale based on the assignment of the value16.0000 to naturally occurring oxygen. Whatwould have been the atomic weight, on sucha table, of silver, if current information hadbeen available? The atomic weights of oxygenand silver on the present table are 15.9994and 107.868.(A) 107.908 (B) 107.864(C) 107.868 (D) 107.872

Sol.

47. Two isotopes of an element Q are Q97 (23.4%abundance) and Q94 (76.6% abundance). Q97

is 8.082 times heavier than C12 and Q94 is7.833 times heavier than C12. What is theaverage atomic weight of the element Q?(A) 94.702 (B) 78.913(C) 96.298 (D) 94.695

Sol.

48. The element silicon makes up 25.7% of theearth's crust by weight, and is the secondmost abundant element, with oxygen being thefirst. Three isotopes of silicon occur in nature:Si28 (92.21%), which has an atomic mass of27.97693 amu; Si29 (4.70%), with an atomicmass of 28.97649 amu; and Si30 (3.09%), withan atomic mass of 29.97379 amu. What is theatomic weight of silicon?(A) 28.0856 (B) 28.1088(C) 28.8342 (D) 29.0012

Sol.

49. The average atomic mass of a mixturecontaining 79 mol % of 24 Mg and remaining21 mole % of 25 Mg and 26Mg, is 24.31. %mole of 26Mg is(A) 5 (B) 20(C) 10 (D) 15

Sol.

50. The oxide of a metal contains 30% oxygen byweight. If the atomic ratio of metal and oxygenis 2 : 3, determine the atomic weight of metal.(A) 12 (B) 56(C) 27 (D) 52

Sol.



OBJECTIVE PROBLEMS (JEE ADVANCED)EXERCISE – II1. The average mass of one gold atom in a sample

of naturally occuring gold is 3.257 × 10–22 g. Usethis to calculate the molar mass of gold.

Sol.

2. A plant virus is found to consist of uniformcylindrical particles of 150 Å in diameter and 5000Å long. The specific volume of the virus is 0.75cm3/g. If the virus is considered to be a singleparticle, find its molecular weight. [V = r2l]

Sol.

MOLE3. Calculate(a) Number of nitrogen atoms in 160 amu of NH4NO3

(b) Number of gram-atoms of S in 490 kg H2SO4

(c) Grams of Al2(SO4)3 containing 32 amu of S.

Sol.

4. A chemical compound “dioxin” has been verymuch in the news in the past few years. (it isthe by-product of herbicide manufacture and isthrough to be quite toxic.) Its formula isC12H4Cl4O2. If you have a sample of dirt (28.3 g)that contains 8.78 × 10–8 moles of dioxin,calculate the percentage of dioxin in the dirtsample ?

Sol.

5. Nitrogen (N), phosporus (P), and potassium (K)are the main nutrients in plant fertil izers.According to an industry convention, thenumbers on the label refer to the mass % of N,P2O5, and K2O, in the order. Calculate the N : P :K ratio of a 30 : 10 : 10 fertilizer in terms ofmoles of each elements, and express it as x : y :1.0.

Sol.

EMPIRICAL & MOLECULAR FORMULA6. Polychlorinated biphenyls, PCBs, known to be

dangerous environmental pollutants, are a groupof compounds with the general empirical formulaC12HmCl10–m, where m is an integer. What is thevalue of m, and hence the empirical formula ofthe PCB that contains 58.9% chlorine by mass ?

Sol.

7. Given the following empirical formulae andmolecular weights, compute the true molecularformulae : Empirical Formula Molecular weight

(a) CH2 84(b) CH2O 150(c) HO 34(d) HgCl 472(e) HF 80



Sol.

8. What is the empirical formula of a compound0.2801 gm of which gave on complete combustion0.9482 gm of carbon dioxide and 0.1939 gm ofwater ?

Sol.

9. A 5.5 gm sample of an organic compound gaveon quantitative analysis 1.4 gm of N and 3.6 gmof C and 0.5 gm of H. If Molecular mass of thecompound is 55 then calculate E.F. and M.F.

Sol.

PROBLEMS RELATED WITH MIXTURE10. One gram of an alloy of aluminium and magnesium

when heated with excess of dil. HCl formsmagnesium chloride, aluminium chloride andhydrogen. The evolved hydrogen collected overmercury at 0ºC has a volume of 1.12 liters at 1atm pressure. Calculate the composition of thealloy.

Sol.

11. A sample containing only CaCO3 and MaCO3 isignited to CaO and MgO. The mixture of oxidesproduced weight exactly half as much as theoriginal sample. Calculate the percentages ofCaCO3 and MgCo3 in the sample.

Sol.

12. Determine the percentage composition of amixture of anhydrous sodium carbonate andsodium bicarbonate from the following data :wt of the mixture taken = 2gLoss in weight on heating = 0.11 gm

Sol.

13. A 10 g sample of a mixture of calcium chlorideand sodium chloride is treated with Na2CO3 toprecipitate calcium as calcium carbonate. ThisCaCO3 is heated to convert all the calcium toCaO and the final mass of CaO is 1.12gm.Calculate % by mass of NaCl in the originalmixture.

Sol.

14. A mixture of Ferric oxide (Fe2O3) and Al is usedas a solid rocket fuel which reacts to give Al2O3and Fe. No other reactants and products areinvolved. On complete reaction of 1 mole ofFe2O3, 200 units of energy is released.

(i) Write a balance reaction representing the abovechange.

(ii) What should be the ratio of masses of Fe2O3 andAl taken so that maximum energy per unit massof fuel is released.

(iii) What would be energy released if 16 kg of Fe2O3reacts with 2.7 kg of Al.

Sol.



LIMITING REACTANT15. Titanium, which is used to make air plane engines

and frames, can be obtained from titaniumtetrachloride, which in turn is obtained fromtitanium oxide by the following process :3 TiO2 (s) + 4C(s) + 6Cl2 (g) 3TiCl4(g) + 2CO2(g) +2Co(g)A vessel contains 4.32 g TiO2, 5.76 g C and;6.82 g Cl2, suppose the reaction goes tocompletion as written, how many gram of TiCl4

and be produced ? (Ti = 48)

Sol.

More than one correct :16 Two gases A and B which react according to the

equation

)g()g()g()g( dDcCbBaA

to give two gases C and D are taken (amountnot known) in an Eudiometer tube (operating ata constant Pressure and temperature) to causethe above.If on causing the reaction there is no volumechange observed then which of the followingstatement is/are correct.(A) (a + b) = (c + d)(B) average molecular mass may increase ordecrease if either of A or B is present in limitedamount.(C) Vapour Density of the mixture will remainsame throughout the course of reaction.(D) Total moles of all the component of mixturewill change.

Sol.

17. A mixture of C3H8 (g) O2 having total volume 100ml in an Eudiometry tube is sparked & it isobserved that a contraction of 45 ml is observedwhat can be the composition of reacting mixture.(A) 15 ml C

3H

8 & 85 ml O

2

(B) 25 ml C3H8 & 75 ml O2

(C) 45 ml C3H8 & 55 ml O2

(D) 55 ml C3H8 & 45 ml O2

Sol.

18. An aqueous solution consisting of 5 M BaCl2,58.8% w/v NaCl solution & 2m Na2 X has a densityof 1.949 gm/ml. Mark the option(s) whichrepresent correct molarity (M) of the specifiedion.[Assume 100% dissociation of each salt andmolecular mass of X2– is 96](A) [Cl–] = 20 M(B) [Na+] = 11 M(C) [Total anions] = 20.5 M(D) [Total cations]=15 M

Sol.

19. A mixture of 100 ml of CO, CO2 and O2 wassparked. When the resulting gaseous mixture waspassed through KOH solution, contraction involume was found to be 80 ml, the compositionof initial mixture may be (in the same order)(A) 30 ml, 60 ml, 10 ml(B) 30 ml, 50 ml, 20 ml(C) 50 ml, 30 ml, 20 ml(D) 30 ml, 40 ml, 30 ml

Sol.



20. Given following series of reactions :(I) NH3 + O2 NO + H2O(II) NO + O2 NO2

(III) NO2 + H2O HNO3 + HNO2

(IV) HNO2 HNO3 + NO + H2OSelect the correct option (s) :(A) Moles of HNO3 obtained is half of moles ofAmmonia used if HNO2 is not used to produceHNO3 by reaction (IV)

(B) %6

100 more HNO3 will be produced if HNO2 is

used to produce HNO3 by reaction (IV) than ifHNO2 is not used to produce HNO3 by reaction(IV)

(C) If HNO2 is used to produce HNO

3 then th

41

of total HNO3 is produced by reaction (IV)(D) Moles of NO produced in reaction (IV) is50% of moles of total HNO3 produced.

Sol.

21. Solution(s) containing 40 gm NaOH is/are(A) 50 gm of 80% (w/w) NaOH(B) 50 gm of 80% (w/v) NaOH [dsoln = 1.2 gm/ml](C) 50 gm of 20 M NaOH [dsoln = 1 gm/ml](D) 50 gm of 5m NaOH

Sol.

22. The incorrect statement(s) regarding 2M MgCl2

aqueous solution is/are (dsolution = 1.09 gm/ml)(A) Molality of Cl is 4.44 m(B) Mole fraction of MgCl

2 is exactly 0.035

(C) The conc. of MgCl2 is 19% w/v(D) The conc. of MgCl

2 is 19 × 104 ppm

Sol.

23. An organic compound is burnt with excess of O2to produce CO

2 (g) and H

2O(l), which results in

25% volume contraction. Which of the followingoption(s) satisfy the given conditions.(A) 10 ml C3H8 + 110 ml O2

(B) 20 ml C2H6O + 80 ml O2

(C) 10 ml C3H6O2 + 50 ml O2

(D) 40 ml C2H2O4 + 60 ml O2

Sol.

24. A sample of H2O2 solution labelled as 56 volumehas density of 530 gm/L. Mark the correctoption(s) representing concentration of samesolution in other units. (Solution contains onlyH2O and H2O2)

(A) 6M22OH

(B) % 17vw

(C) Mole fraction of H2O2 = 0.25

(D) 72

1000m

22OH

Sol.

25. Solution(s) containing 30 gm CH3COOH is/are(A) 50 gm of 70% (w/v) CH3COOH [dsol = 1.4gm/ml](B) 50 gm of 10 M CH3COOH [dsol = 1 gm/ml](C) 50 gm of 60% (w/w) CH3COOH(D) 50 gm of 10 m CH

3COOH

Sol.



26. ‘2V’ ml of 1 M Na2SO4 is mixed with ‘V’ ml of 2MBa(NO3)2 solution.(A) Molarity of Na+ ion in final solution can’t becalculated as V is not known.

(B) Molarity of BaSO4 in final solution is M32

(C) Molarity of –3NO in final solution is M

34

(D) Molarity of –3NO in final solution is M

32

Sol.

Match the Column

27. One type of artifical diamond (commonly calledYAG for yttrium aluminium garnet) can berepresented by the formula Y3Al5O12 [Y = 89, Al= 27]

Column I Column IIElement Weight percentage(A) Y (P) 22.73 %(B) Al (Q) 32.32 %(C) O (R) 44.95 %

Sol.

28. The recommended daily dose is 17.6 milligramsof vitamin C (ascorbic acid) having formulaC6H8O6.Match the following. Given : NA = 6 × 1023

Column I Column II(A) O-atoms present (P) 10–4 mole(B) Moles of vitamin C in 1gm (Q) 5.68 × 10–3

of vitamin C(C) Moles of vitamin C in 1gm (R) 3.6 × 1020

should be consumed daily

Sol.

29. Column I Column II(A) 10 M MgO (P) Wsolvent = 120 gm (dsol = 1.20 gm/ml) per 100 ml of solution. Solute: MgO, Solvent:H2O

(B) 40% w/v NaOH (Q) Wsol = 150 gm (dsol. = 1.6 gm/ml) per 100 gm solvent Solute:NaOH,Solvent:H2O

(C) 8 m CaCO3 (R) Wsolute = 120 gm per Solute:CaCO

3,Solvent:H

2O 100 gm of solvent

(D) 0.6 mol fraction of ‘X’ (S) Wsolvent = 125 gm (molecular mass = 20) per 100 gm of solute in ‘Y’ (molecular mass 25) Solute : X, Solvent : Y

Sol.

30. Bunty & Bubbly have two separate containersone having N2 gas & other H2 gas : It is knownthat N

2 & H

2 react to give N

2H

2(l) and/or N

2H

4

(g) depending upon the ratio in which N2 & H2are taken & that N2H2 reacts with H2 to giveN2H4. Formation of 1 mole of N2H4 requires 30units of energy & formation of 1 mole of N2H2(l)release 30 units of energy. From this informationmatch Column I (representing composition ofgases taken) with Column II (representing theobservation)

Column I Column II(Composition of gases)(Observation)(A) 40 lit N2 & 30 lit H2 (P) Contraction by 22.4 lit (same tem perature & pressure)

(B) 11.2 lit of N2 & H2 (Q) Contraction by 20 lit. taken at 1 atm & 273 K in a ratio such that max. release of energy is observed

(C) 11.2 lit of N2 & 30 lit (R) Contraction by 60 lit of H2 (same temperature & pressure)

(D) 10 lit of N2 & more

than 22.4 lit of H2 (S) Contraction by 11.2 lit (same temperature & pressure)



Sol.

31. Br2 reacts with O2 in either of the following waysdepending upon supply of O2.

Br2 + OBrO21

22 , 3222 OBrO23

Br

Match composition of the final mixture for initialamount of reactants.

Column I Column II(Initial reactants) (Final product)(A) 320 gm Br2 is mixed (P) 1 mole Br2O3

with 64 gm of O2

(B) 160 gm Br2 is mixed (Q) 21 mole (Br O),2

with 8 gm of O2 21

mole(Br )2

(C) 80 gm Br2 is mixed with (R) 1 mole (Br2O), 1 32 gm of O2 mole (Br2O3)

(D) 160 gm Br2 is mixed with (S) 21

mole (Br2O3),

48 gm of O2 41

mole (O2)

Sol.

COMPREHENSION

32. A 4.925 g sample of a mixture of CuCl2 and CuBr2

was dissolved in water and mixed thoroughly witha5.74 g portion of AgCl. After the reaction thesolid, a mixture of AgCl, and AgBr, was filtered,washed, and dried. Its mass was found to be6.63 g.(a) % By mass of CuBr2 in original mixture is(A) 2.24 (B) 74.5(C) 45.3 (D) None(b) % By mass of Cu in original mixture is(A) 38.68 (B) 19.05(C) 3.86 (D) None(c) % by mole of AgBr in dried precipitate is(A) 25 (B) 50(C) 75 (D) 60

(d) No. of moles of Cl – ion present in thesolution after precipitation are(A) 0.06 (B) 0.02(C) 0.04 (D) None

Sol.

33. NaBr, used to produce AgBr for use inphotography can be self prepared as follows :Fe + Br2 FeBr2 ...(i)FeBr2 + Br2 Fe3Br8 ...(ii) (not balanced)Fe3Br8 + Na2CO3 NaBr + CO2 + Fe3O4...(iii)(not balanced)(a) Mass of iron required to produce 2.06 × 103

kg NaBr(A) 420 gm (B) 420 kg(C) 4.2 × 105 kg (D) 4.2 × 108 gm(b) If the yield of (ii) is 60% & (iii) reaction is70% then mass of iron required to produce 2.06× 103 kg NaBr(A) 105kg (B) 105 gm(C) 103 kg (D) None(c) If yield of (iii) reaction is 90% then mole ofCO

2 formed when 2.06 × 103 gm NaBr is formed :

(A) 20 (B) 10(C) 40 (D) None



Sol.

34. N2O5 and H2O can react to form HNO3, accordingto given reactionN2O5 + H2O 2HNO3

the concentration of a mixture of HNO3 andN2O5(g) can be expressed similar to oleum. Thenanswer the following question.(a) Find the percentage labelling of a mixturecontaining 23 gm HNO3 and 27 gm N2O5.(A) 104.5% (B) 109%(C) 113.5% (D) 118%(b) Find the maximum and minimum value ofpercentage labelling :(A) 133.3% (B) 116.66%, 0%(C) 116.66%, 100% (D) None(c) Find the new labelling if 100 gm of this mixture(original) is mixed with 4.5 gm water

(A) 100 + 15.4

(B) 100 + 045.1

5.4

(C) 100 + 5.1045.4

(D) 100 + 09.15.4

Sol.

35. For a gaseous reaction,2A(g) 3B(g) + C(g)Whose extent of dissociation depends ontemperature is performed in a closed container,it is known that extent of dissociation of A isdifferent in different temperature range. With ina temperature range it is constant. (Temperaturerange T0 – T1, T1 – T2, T2 – T ). A plot of P v/s Tis drawn under the given condition. Given : tan55 = 1.42, tan 50 = 1.19, tan 60 = 1.73

T0 T1 T2

60°

55°

50°T(k)

(a) If 1ii T–T is the degree of dissociation of Athen in the temperature range Ti Ti + 1

(A) i0 T–T is lowest (B) i0 T–T is highest

(C) 1T–T2 (D) 0T–T2

(b) If initially 1 mole of A is taken in a 0.0821 lcontainer then [R = 0.0821 atm lit/k]

(A) 19.0i0 T–T (B) 095.0

10 T–T

(C) 42.021 T–T (D) 73.0

21 T–T

Sol.

36. A 10 ml mixture of N2, a alkane & O

2 undergo

combustion in Eudiometry tube. There wascontraction of 2 ml, when residual gases arepassed through KOH. To the remaining mixturecomprising of only one gas excess H2 was added& after combustion the gas produced is absorbedby water, causing a reduction in volume of 8 ml.(a) Gas produced after introduction of H2 in themixture ?(A) H

2O (B) CH

4(C) CO2 (D) NH3(b) Volume of N2 present in the mixture ?(A) 2 ml (B) 4 ml(C) 6 ml (D) 8 ml(c) Volume of O 2 remained after the fi rstcombustion ?(A) 4 ml (B) 2 ml(C) 0 (D) 8 ml(d) Identify the hydrocarbon.(A) CH4 (B) C2H6(C) C3H8 (D) C4H10



Sol.

37. An evacuated glass vessel weighs 50 gm whenempty, 148.0 g when completely filled with liquidof density 0.98 gml–1 and 50.5 g when filled withan ideal gas at 760 mm at 300 K. Determine themolecular weight of the gas. [JEE ‘98,3]

Sol.

38. At 100° C and 1 atmp, if the density of liquidwater is 1.0 g cm–3 and that of water vapour is0.0006 g cm–3, then the volume occupied bywater molecules 1 L of steam at that temperatureis : [JEE ‘2001 (Scr), 1]

(A) 6 cm3 (B) 60 cm3

(C) 0.6 cm3 (D) 0.06 cm3

Sol.

Each of the questions given below consists ofStatement-I and Statement-II. Use the followingKey to choose the appropriate answer.

(A) If both statement-1 and statement-2 arecorrect, and statement-2 is the correct explana-tion of statement-1

(B) If both statement-1 and statement-2 are cor-rect, and statement-2 is not the correct explanation ofstatement-1

(C) If statement-1 is correct and statement-2 isincorrect

(D) If statement-1 is incorrect and statement-2is correct

39 Statement-1 : Molarity of pure water is55.5 M.

Statement-2 : Molarity is temperaturedependent parameter

Sol.

40. Statement-1 : 1 g-atom of sulphur containsAvogadro number of sulphur molecules

Statement-2 : Atomicity of sulphur is eight.

Sol.

41. Statement-1 : The number of O atoms in1 gm. of O2, 1 gm O3 and 1 gm of atomicoxygen is same.

Statement-2 : Each of the speciesrepresents 1/16 gm-atom of oxygen.

Sol.

42. Statement-1 : The ratio by volume of H2 :Cl2 : HCl in a reaction H2(g) + Cl2(g) 2HCl(g)is 1 : 1 : 2.

Statement-2 : Substances always react insuch a way that their volume ratio is insimple integers.

Sol.

43. Statement-1 : 0.2 N H2SO4 solution hasmolarity equal to 0.2 M.

Statement-2 : H2SO4 is a diabasic acid.

Sol.



Integer Type

44. If one mole of ethanol (C2H5OH) completelyburns to form carbon dioxide and water, theweight of carbon dioxide formed is about -

Sol.

45. How many grams are contained in 1gm-atomof Na

Sol.

46. The mass of CaCO3 produced when carbondioxide is passed in excess through 500 ml of0.5 M Ca(OH)2 wi ll be-

Sol.

47. The mass of oxygen that would be requiredto produce enough CO, which completelyreduces 1.6 kg Fe2O3 (at. mass Fe = 56) is-

Sol.

48. 1.5 gm of divalent metal displaced 4 gm ofcopper (at. wt. = 64) from a solution ofcopper sulphate. The atomic weight of themetal is-

Sol.

49. Assuming that petrol is iso-octane (C8H18) andhas density 0.8 gm/ml, 1.425 litre of petrol oncomplete combustion will consume oxygen -

Sol.

50. A mixture containing 100 gm H2 and 100 gmO2 is ignited so that water is formed accordingto the reaction, 2H2 + O2 2H2O; Howmuch water will be formed -

Sol.



SUBJECTIVE PROBLEMS (JEE ADVANCED)EXERCISE – III1. 4222 BAB2A and 4322 BAB2A

23

Two substance A2 & B2 react in the above mannerwhen A2 is limited it gives A2B4, when in excessgives A

3B

4. A

2B

4 can be converted to A

3B

4 when

reacted with A2. Using this information calculatethe composition of the final mixture whenmentioned amount of A

2 & B

2 are taken

(i) If 4 moles of A2 & 4 moles of B2 is taken inreaction container

(ii) If 21

moles of A2 & 2 moles of B2 is taken in

reaction container

(iii) If 45

moles of A2 & 2 moles of B

2 is taken

Sol.

2. How much minimum volume of 0.1 M aluminiumsulphate solution should be added to excesscalcium nitrate to obtain atleast 1 gm of eachsalt in the reaction.

43323342 CaSO3)NO(Al2)NO(Ca3)SO(Al

Sol.

3. A sample of fuming sulphuric acid containingH2SO4, SO3 and SO2 weighing 1.00 g is found torequire 23.47 ml of 1.00 M alkali (NaOH) forneutralisation. A separate sample shows thepresence of 1.50% SO2. Find the percentage of“free” SO3, H2SO4 and “combined” SO3 in thesample.

Sol.

4. Chloride samples are prepared for analysis byusing NaCl, KCl and NH4Cl separately or asmixture. What minimum volume of 5% by weightAgNO3 solution (sp gr., 1.04 g ml–1) must be addedto a sample of 0.3 g in order to ensure completeprecipitation of chloride in every possible case ?

Sol.

5. One litre of milk weighs 1.035 kg. The butter fatis 10% (v/v) of milk has density of 875 kg/m3.The density of fat free skimed milk is ?

Sol.

6. 10 mL of gaseous organic compound contain C,H and O only was mixed with 100 mL of O2 andexploded under identical conditions and thencooled. The volume left after cooling was 90mL. On treatment with KOH a contraction of 20Ml was observed. If vapour density of compoundis 23, derive molecular formula of the compound.

Sol.



7. For a hypothetical chemical reaction representedby

)g(D)g(C)g(A3 , the following informations

are known.Information(i) At t = 0, only 1 mole of A is present and thegas has V.D. = 60.(ii) At t = 30 min, the gaseous mixture consist ofall three gases and has a vapour density = 75.(iii) Molecular Mass of C = 200Calculate(a) Molecular weight of A and D.(b) Moles of each specie at t = 30 min.

Sol.

8. 523 PClClPCl

HCl3POHOH3PCl 3323

A sample containing very large amount of PCl3was exposed to a sample of “Chlorinated water”having Cl2 dissolved in H2O so that the abovetwo reactions occurred. It was observed thatration of mass of PCl5 to mass of H3PO3 was 417: 246. From this information calculate.(i) ratio of moles of PCl5 to moles of H3PO3.(i i) ratio of moles of Cl2 : H2O present inchlorinated water(iii) Molality (m) of Chlorine in Chlorinated water.

Sol.

9. A mixture of H2, N2 & O2 occupying 100 mlunderwent reaction so as to from H2O2(l) andN2H2(g) as the only products, causing the volumeto contract by 60 ml. The remaining mixture waspassed through pyrogallol causing a contractionof 10ml. To the remaining mixture excess H2 wasadded and the above reaction was repeated,causing a reduction in volume of 10 ml. Identifythe composition of the initial mixture in mol %.(No other products are formed)

Sol.

10. Consider the following set reactions

CH3

CH3

HOCCH

HOCCH

HOCCH

||

||

||

O

O

O

n times

resideuWhiteSaltSilverAgNO)excess(3

If 0.1 moles of silver salt is taken & wt. of residueobtained is 54 gms then what will be the molecularmass of

3

n

|||3 CH

CH.......CHCHCH

Br Br BrSol.

11. 124 gm of mixture containing NaHCO3, AlCl3 &KNO3 requires 500 ml, 8% w/w NaOH solution[dNaOH = 1.8 gm/ml] for complete neutralisation.On heating same amount of mixture, it showsloss in weight of 18.6 gm. Calculate % compositionof mixture by moles. Weak base formed doesn’tinterfere in reaction. Assume KNO3 does notdecompose under given conditions.

Sol.



12. A mixture of three gases an alkane (generalformula Cn H2n + 2), an alkene (general formulaCxH2x) and O2 was subjected to sparking to causecombustion of both the hydrocarbon at 127ºC.After the reaction three gases were present andnone of the hydrocarbon remained. On passingthe gases through KOH (absorb CO 2), anincrement in mass of KOH solution by 132 gmwas observed. The remaining gases were passedover white anhydrous CuSO4 and the weight ofblue hydrated CuSO4 crystals was found to be72 gm more than that of white anhydrous CuSo4.Given that initially total 10 moles of the threegases were taken and moles of alkane andalkene were equal and if molecular mass ofalkene molecular mass of alkane = 12 i.e. (Malkene

– Malkane = 12), then answer the followingquestions. (Show calculations)(a) Which three gases are remained after thecombustion reactions.(b) What are the number of moles of productgases.(c) What is the molecular formula of the twohydrocarbon.(d) What is the number of moles of each of thetwo hydrocarbons and O2 gas taken initially.

Sol.

13. KOH2K2OH 2yield%40

22 II

50% yield

2 2 4 2 4

2 4 4 2 2

H O 2KMnO 3H SOK SO 2MnSO 3O 4H O

100 ml of H2O sample was divided into two parts.First part was treated with KI. And KOH formedrequired 200 ml of M/2 H2SO4 for completeneutralisation. Other part was treated with justsufficient KMnO4 yielding 6.74 lit. of O2 at 1 atm& 273 K. Calculate(a) Moles of KOH produced(b) Moles of KMnO4 used(c) Total moles of H2O2 used in both reaction(d) Volume strength of H2O2 used.

Sol.

14. The vapours of organic compound was burnt inoxygen. Equal volume of both gaseous substancewere taken at same pressure and temperature.After the reaction, the system was returned tothe original condition and it turn out that itsvolume has not changed. The product ofcombustion contain 50% CO2 (g) and 50% H2O(g)by volume and no other gas. Find the molecularweight of organic compound (in gram/mol) inquestion.

Sol.

15. “Prussian blue” can be prepared by the followingreactions.

I. 2442 HFeSOSOHFe

II. OH)SO(FeO21

SOHFeSO 23422424

III. 4236464342 SOK])CN(Fe[Fe])CN(Fe[K)SO(Fe

Calculate number of moles of Fe4 [Fe(CN)6]3

produced, if(i) 50 moles of Fe and 30 moles of H

2SO

4 are

used with sufficient amount of other reactants.(ii) 50 moles of Fe, 70 moles of H2SO4 and 30moles of K4[Fe(CN)6] are used with sufficientamount of other reactants.(iii) 400 moles of Fe are used with sufficientamount of other reactants (assuming the yieldof I, II & III reactions are 50%, 40% and 60%respectively).



Sol.

16. A chemist wants to prepare diborane by thereaction 6LiH + 8BF3 6LiBF4 + B2H6

If he starts with 2.0 moles each of LiH & BF3.How many moles of B2H6 can be prepared.

Sol.

17. Carbon reacts with chlorine to form CCl4. 36 gmof carbon was mixed with 142 g of Cl2. Calculatemass of CCl4 produced and the remaining massof reactant.

Sol.

MISCELLANEOUS PROBLEM

18. P4S3 + 8O2 P4O10 + 3SO2

Calculate minimum mass of P4S3 is required toproduce atleast 1 gm of each product.

Sol.

19. By the reaction of carbon and oxygen, a mixtureof CO and CO 2 is obtained. What is thecomposition by mass of the mixture obtained when20 grams of O2 reacts with 12 grams of carbon ?

Sol.

20. The chief ore of Zn is the sulphide, ZnS. The oreis concentrated by froth floatation process andthen heated in air to convert ZnS to ZnO.

2%75

2 SO2ZnO2O3ZnS2

OHZnSOSOHZnO 24%100

42

242%80

24 OSOH2Zn2OH2ZnSO2

(a) What mass of Zn will be obtained from asample of ore containing 291 kg of ZnS.(b) Calculate the volume of O2 produced at 1atm & 273 K in part (a).

Sol.

21. In a determination of P an aqueous solution ofNaH2PO4 is treated with a mixture of ammoniumand magnesium ions to precipitate magnesiumammonium phosphate Mg(NH

4)PO

4, 6H

2O. This is

heated and decomposed to magnesiumpyrophosphate, Mg2P2O7 which is weighed. Asolution of NaH

2PO

4 yielded 1.054 g of Mg

2P

2O

7.

What weight of NaH2PO4 was present originally ?Sol.

22. A mixture of nitrogen and hydrogen. In the ratioof one mole of nitrogen to three moles ofhydrogen, was partially converted into NH3 sothat the final product was a mixture of all thesethree gases. The mixture was to have a densityof 0.497 g per litre at 25ºC and 1.00 atm. Whatwould be the mass of gas in 22.4 liters at 1atmand 273 K ? Calculate the % composition of thisgaseous mixture by volume.

Sol.



23. 1gm sample of KClO3 was heated under suchconditions that a part of it decomposed accordingto the equation (1) 2KClO3 –– 2 KCl + 3O2

and remaining underwent change according tothe equation. (2) 4KClO

3 –– 3 KClO

4 + KCl

If the amount of O2 evolved was 112 ml at 1 atmand 273 K., calculate the % by weight of KClO4

in the residue.Sol.

24. In one process for waterproofing, a fabric isexposed to (CH3)2SiCl2 vapour. The vapour reactswith hydroxyl groups on the surface of the fabricor with traces of water to form the waterproofingfilm [(CH3)2SiO]n, by the reaction

n232223 ]SiO)CH[(OnHnCl2nOH2SiCl)CH(n

where n stands for a large integer. Thewaterproofing film is deposited on the fabric layerupon layer. Each layer is 6.0 Å thick [thethickness of the (CH3)2SiO group]. How much(CH3)2 SiCl2 is needed to waterproof one side ofa piece of fabric, 1.00 m by 3.00 m, with a film300 layers thick ? The density of the film is1.0 g/cm3.

Sol.

CONCENTRATION TERMS25. Calculate the molarity of the following solutions

(a) 4g of caustic soda is dissolved in 200 mL ofthe solution.(b) 5.3 g of anhydrous sodium carbonate isdissolved in 100 mL of solution(c) 0.365 g of pure HCl gas is dissolved in 50 mLof solution.

Sol.

26. Density of a solution containing 13% by mass ofsulphuric acid is 1.09 g/mL. Then molarity ofsolution will be.

Sol.

27. The density of a solution containing 40% by massof HCl is 1.2 g/mL. Calculate the molarity of thesolution.

Sol.

28. 15g of methyl alcohol is present in 100 mL ofsolution. If density of solution is 0.90 g mL–1.Calculate the mass percentage of methyl alcoholin solution

Sol.

29. Units of parts per million (ppm) or per billion (ppb)are often used to describe the concentrationsof solutes in very dilute solutions. The units aredefined as the number of grams of solute permillion or per billion grams of solvent. Bay of Bengalhas 1.9 ppm of lithium ions. What is the molalityof Li+ in this water ?

Sol.



30. A 6.90 M solution of KOH in water contains 30%by mass of KOH. What is density of solution ingm/ml.

Sol.

31. A solution of specific gravity 1.6 is 67% byweight. What will be % by weight of the solutionof same acid if it is diluted to specific gravity1.2 ?

Sol.

32. Find out the volume of 98% w/w H2SO4 (density =1.8 gm/ml) must be diluted to prepare 12.5 litresof 2.5 M sulphuric acid solution.

Sol.

33. Determine the volume of diluted nitric acid (d =1.11 gmL–1, 19% w/v HNO3) That can be preparedby diluting with water 50 mL of conc. HNO3 (d =1.42 g ML–1, 69.8 % w/v).

Sol.

34. (a) Find molarity of Ca2+ and NO3– in 2 MCa(NO3)2

aqueous solution of density 1.328 g/mL.(b) Also find mole fraction of solvent in solution.

Sol.

35. Calculate molality (m) of each ion present in theaqueous solution of 2M NH4 Cl assuming 100%dissociation according to reaction.

)aq(Cl)aq(NH)aq(ClNH 44

Given : Density of solution = 3.107 gm/ml.Sol.

36. 500 ml of 2 M NaCl solution was mixed with 200ml of 2 M NaCl solution. Calculate the final volumeand molarity of NaCl in final solution if finalsolution has density 1.5 gm/ml.

Sol.

EXPERIMENTAL METHODS

37. What is the percentage of nitrogen in an organiccompound 0.14 gm of which gave by Dumasmethod 82.1 c.c. of nitrogen collected over waterat 27ºC and at a barometric pressure of 774.5mm ? (aqueous tension of water at 27ºC is 14.5mm)

Sol.



38. 0.20 gm of an organic compound was treated byKjeldahl’s method and the resulting ammonia waspassed into 50 cc of M/4 H2 SO4. The residualacid was then found to require 40 cc of M/2NaOH for neutralisation. What is the percentageof nitrogen in the compound ?

Sol.

39. 0.252 gm of an organic compound gave oncomplete combustion 0.22 gm of carbon dioxideand 0.135 gm of water. 0.252 gm of the samecompound gave by Carius method 0.7175 gm ofsilver chloride. What is the empirical formula ofthe compound ?

Sol.

40. 0.6872 gm of an organic compound gave oncomplete combustion 1.466 gm of carbon dioxideand 0.4283 gm of water. A given weight of thecompound when heated with nitric acid and silvernitrate gave an equal weight of silver chloride.0.3178 gm of the compound gave 26.0 cc ofnitrogen at 15ºC and 765 mm pressure. Deducethe empirical formula of the compound ?

Sol.

41. 0.80 g of the chloroplatinate of a mono acidbase on ignition gave 0.262 g of Pt. Calculatethe mol wt of the base.

Sol.

42. The molecular mass of an organic acid wasdetermined by the study of its barium salt. 2.562g of salt was quantitatively converted to freeacid by the reaction 30 ml of 0.2 M H2 SO4, thebarium salt was found to have two moles of waterof hydration per Ba+2 ion and the acid is monobasic. What is molecular weight of anhydrousacid ? (At mass of Ba = 137)

Sol.

SOME TYPICAL CONCENTRATION TERMS

43. Calculate composition of the final solution if 100gm oleum labelled as 109% is added with(a) 9 gm water (b) 18 gm water(c) 120 gm water

Sol.

44. For ‘44.8 V’ H2O2 solution having d = 1.136 gm/ml calculate(i) Molarity of H2O2 solution.(ii) Mole fraction of H2O2 solution.

Sol.



45. An oleum sample is labelled as 118%, Calculate(i) Mass of H

2SO

4 in 100 gm oleum sample.

(ii) Maximum mass of H2SO4 that can be obtainedif 30 gm sample is taken.(iii) Composition of mixture (mass of components)if 40 gm water is added to 30 gm given oleumsample.

Sol.

EUDIOMETRY

46. 10ml of a mixture of CO, CH4 and N2 explodedwith excess of oxygen gave a contraction of 6.5ml. There was a further contraction of 7 ml,when the residual gas treated with KOH. Volumeof CO, CH4 and N2 respectively is

Sol.

47. When 100 ml of a O2 – O3 mixture was passedthrough turpentine, there was reduction ofvolume by 20 ml. If 100 ml of such a mixture isheated, what will be the increase in volume ?

Sol.

48. 60 ml of a mixture of nitrous oxide and nitricoxide was exploded with excess of hydrogen. If38 ml of N2 was formed, calculate the volume ofeach gas in the mixture.

Sol.

49. When a certain quantity of oxygen was ozonisedin a suitable apparatus, the volume decreasedby 4 ml. On addition of turpentine the volumefurther decreased by 8 ml. All volumes weremeasured at the same temperature and pressure.From these data, establish the formula of ozone.

Sol.

50. 10 ml of ammonia were enclosed in an eudiometerand subjected to electric sparks. The sparkswere continued till there was no further increasein volume. The volume after sparking measured20 ml. Now 30 ml of O2 were added and sparkingwas continued again. The new volume thenmeasured 27.5 ml. All volume were measuredunder identical conditions of temperature andpressure. V.D. of ammonia is 8.5. Calculate themolecular formula of ammonia. Nitrogen andHydrogen are diatomic.

Sol.

PREVIOUS YEARSEXERCISE – IVJEE MAINLEVEL – I

1. The weight of 2.01 × 1023 molecules of CO is -[AIEEE-2002]

(A) 9.3 gm (B) 7.2 gm(C) 1.2 gm (D) 3 gm

Sol.

2. In an organic compound of molar mass 108 gmmol–1 C, H and N atoms are present in 9 : 1 : 3.5by weight. Molecular formula can be –

[AIEEE-2002](A) C6H8N2 (B) C7H10N(C) C5H6N3 (D) C4H18N3

Sol.

3. Number of atoms in 560 gm of Fe (atomic mass56 g mol–1) is – [AIEEE-2003](A) is twice that of 70 gm N(B) is half that of 20 gm H(C) both are correct(D) None is correct

Sol.

4. 6.02 ×1020 molecules of urea are present in100 ml of its solution. The concentration of ureasolution is - [AIEEE-2004](A) 0.001 M (B) 0.01 M(C) 0.02 M (D) 0.1 M(Avogadro constant, NA = 6.02 ×1023 mol–1)

Sol.

5. How many moles of magnesium phosphate,Mg3(PO4)2 will contain 0.25 mole of oxygen at-oms ? [AIEEE 2006](A) 3.125 × 10–2 (B) 1.25 × 10–2

(C) 2.5 × 10–2 (D) 0.02Sol.

6. In the reaction, [AIEEE 2007]2Al(s)+6HCl(aq) 2Al3+

(aq)+6Cl(aq)+3H2(g),(A) 6L HCl(aq) is consumed for every 3L H2(g)produced(B) 33.6 L H2(g) is produced regardless oftemperature and pressure for every mole Althat reacts(C) 67.2 L H2(g) at STP is produced for everymole Al that reacts(D) 11.2 L H2(g) at STP is produced for everymole HCl (aq) consumed

Sol.

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]


JEE ADVANCEDLEVEL I – I

1. How many moles of e– weigh one Kg

(A) 6.023 × 1023 (B) 108.91

× 1031

(C) 5410

108.9023.6

(D) 810

023.6108.91

[JEE ‘2002 (Scr), 1]Sol.

2. Calculate the molarity of pure water using itsdensity to be 1000 kg m–3. [JEE’ 2003]

Sol.

3. One gm of charcoal absorbs 100 ml 0.5 MCH3COOH to form a monolayer, and there by themolarity of CH3COOH reduces to 0.49. Calculatethe surface area of the charcoal adsorbed byeach molecule of acetic acid. Surface area ofcharcoal = 3.01 × 102 m2/gm. [JEE’ 2003]

Sol.

4. Calculate the amount of Calcium oxide requiredwhen it reacts with 852 gm of P4O10.[JEE 2005]

Sol.

5. 20% surface sites have adsorbed N2. On heating

N2 gas evolved from sites and were collected at

0.001 atm and 298 K in a container or volume is2.46 cm3. Density of surface sites is 6.023 ×1014/cm2 and surface area is 1000 cm2, find outthe no. of surface sites occupied per moleculeof N2. [JEE 2005]

Sol.

6. Dissolving 120 g of urea (mol. wt. 60) in 1000 gof water gave a solution of density 1.15 g/mL.The molarity of the solution is:[JEE 2011](A) 1.78 M (B) 2.00 M(C) 2.05 M (D) 2.22 M

Sol.

7. Reaction of Br2 with Na2CO3 in aqueous solutiongives sodium bromide and sodium bromate withevolution of CO2 gas. The number of sodium bro-mide molecules involved in the balanced chemi-cal equation is : [JEE 2011]

Sol.

8. A decapeptide (Mol. wt. 796) on complete hy-drolysis gives glycine (Mol. Wt. 75), alanine andphenylalanine. Glycine contributes 47.0% to thetotal weight tto the hydrolysed products. Thenumber of glycine units present in thedecapeptideis : [JEE 2011]

Sol.

ANSWER-KEY

Answer Ex–I OBJECTIVE PROBLEMS (JEE MAIN)

1. A 2. A 3. A 4. A 5. A 6. A 7. A 8. A

9. A 10.B 11. A 12. B 13. C 14. A 15. B 16. B

17. A 18. B 19. A 20. A 21. C 22. B 23. A 24. C

25. C 26. D 27. C 28. A 29. C 30. A 31.A 32.C

33. C 34.D 35. B 36. D 37. A 38. B 39.C 40. C

41. D 42.C 43. B 44. B 45. B 46. D 47.D 48. A

49. C 50.B

Answer Ex–II OBJECTIVE PROBLEMS (JEE ADVANCED)

1. 196.169 gm 2. 7.092 × 107 3. (a) 4 ; (b) 5000 moles; (c) 1.89 × 10–22gm

4. 1.0 × 10–4% 5. 10 : 0.66 : 1 6. m = 4, C6H2Cl3

7. (a) C6H12, (b) C5H10O5, (c) H2O2, (d) Hg2 Cl2, (e) H4F4 8. CH 9. C3H5N, C3H5N

10. Al = 60%; Mg = 40% 11. CaCO3 = 28.4%; MgCO3 = 71.6%

12. NaHCO3 = 14.9%; Na2CO3 = 85.1%

13. %NaCl = 77.8% 14. (i) Fe2O3 + 2Al Al2O3 + 2Fe; (ii) 80 : 27 ; (iii) 10,000 units 15. 9.12

16. A,C 17. A, B 18. A,B,C 19. A,B 20. A,C,D 21. A,C 22. B,D 23. A,C

24. B,D 25. B,C 26. C 27. (A) R, (B) P, (C) Q 28. (A) R, (B) Q, (C) P

29. (A) Q; (B) P; (C) S; (D) R 30. (A) R; (B) S; (C) P; (D) Q 31. (A) R, (B) Q, (C) S, (D) P

32. (a) C; (b) A; (c) B (d) A 33. (a) B; (b) C; (c) B 34. (a) B; (b) C; (c) B

35. (a) A; (b) A 36. (a) D; (b) B; (c) C; (d) A 37. 123 g/mol 38. C 39. B

40. D 41. A 42. C 43. D 44. 88 45. 23 46. 25

47. 480 48. 24 49. 125 50. 113



Answer Ex–III SUBJECTIVE PROBLEMS (JEE ADVANCED)

1. (i) A2 = 1, A

3B

4 =2; (ii) B

2 =1 A

2B

4=1/2; (iii) A

3B

4 = 0.5 A

2B

4 = 0.5 2. 24.51 ml

3. H2SO4 = 35.38%, Free SO3 = 63.1%, combined SO3 = 28.89% 4. 18.38 ml

5. 1.052 gm/ml 6. C2H6O 7. (a) mA = 120, mD = 160; (b) 52

nA , 51nC

, 51

nD

8. (i) 2/3 ; (ii) 2/9; (iii) kg/mol1000189

2Cl2

m 9. N2 = 30 ml, H2 = 40 ml 10. 495

11. AlCl3 = 33.33 ; NaHCO3 = 50 ; KNO3 = 16.67 12.(a) CO2, H2O and O2 ; (b) nCO2 = 3, nH2O = 4 ;

(c) C2H4 and CH4 are the H.C; (d) nO2 = 8

13. (a) 0.2; (b) 0.4 moles ; (c) 0.45 ; (d) 50.4 ‘V’ 14. 30 15. (a) 5, (b) 10, (c) 12

16. 0.25 mole 17. wc = 24gm ; WCCl4 = 154 gm 18. 1.1458 19. 21 : 11

20. (a) 117 kg; (b) 20.16 × 103 lit. 21. 1.14 gm 22. 12.15 gm, N2 = 14.28% H2 = 42.86 %, NH3 = 42.86%

23. 59.72% 24. 0.9413 gram 25. (a) 0.5 M, (b) 0.5 M, (c) o.2 M 26. 1.445 27. 13.15

28. 16.66% 29. 2.7 × 10–4 30. 1.288 31. 29.77 32. 1.736 litre33. 183.68 ml

34. (a) [Ca2+] = 2 molar [NO3–] = 4 molar; (b) 0.965 35. 0.6667, 0.6667 36. 2M

37. 66.67% 38. 35% 39. CH3Cl 40. C7H10NCl 41. 92.70 42. 128

43. (a) pure H2SO

4 (109 gm); (b) 109 gm H

2SO

4, 9 gm H

2O; (c) 109 gm H

2SO

4, 111 gm H

2O

44. (i) 4M; (ii) 0.06 45. (i) 20 gm H2SO4; (ii) 35.4 gm H2SO4; (iii) H2SO4 = 35.4 gm, H2O = 34.6 gm

46. 5 ml, 2ml, 3ml 47. 10 ml 48. NO = 44 ml; N2O = 16 ml 49. O

350. NH

3



Answer Ex–IV PREVIOUS YEARS PROBLEMS

JEE MAINLEVEL – I

1. A 2. A 3. C 4. B 5. A 6. D


1. D 2. 55.5 mol L–1 3. 5 × 10–19 m2 4. 1008 gm

5. 0002 6. C 7. 0005 8. 0006



Page # 149TRIGONOMETRIC RATIOS & IDENTITIES


A. BASIC TRIGONOMETRIC IDENTITIES

(a) sin² + cos² = 1 ; 1 sin 1 ; 1 cos 1 R

(b) sec² tan² = 1 ; sec 1 R

(c) cosec² cot² = 1 ; cosec 1 R

Important Trigonometric Ratios :(a) sin n = 0 ; cos n = (-1)n ; tan n = 0where n I

(b) sin2

)1n2( = ( 1)n & cos

2)1n2(

= 0 where n I

Trigonometric Functions Of Allied Angles :If is any angle , then , 90 ± , 180 ± , 270 ± , 360 ± etc. are called ALLIED ANGLES .(a) sin ( ) = sin ; cos ( ) = cos ; tan ( ) = – tan

(b) sin (90°- ) = cos ; cos (90° ) = sin ; tan (90° ) = cot

(c) sin (90°+ ) = cos ; cos (90°+ ) = sin ; tan (90°+ ) = cot

(d) sin (180° ) = sin ; cos (180° ) = cos ; tan (180° ) = tan

(e) sin (180°+ ) = sin ; cos (180°+ ) = cos ; tan (180°+ ) = tan

(f) sin (270° ) = cos ; cos (270° ) = sin ; tan (270° ) = cot

(g) sin (270°+ ) = cos ; cos (270°+ ) = sin ; tan (270°+ ) = – cot

Ex.1 Express 1·2 radians in degree measure.

Sol. 1·2 radians = 1·2 ×180

degrees = 1·2 ×7/22

180[ =

722

(approx).]

=22

71802·1 = 68·7272 = 68º (·7272 × 60)’ = 68º (43·63)’

= 68º 43’ (·63 × 60)” = 68º 43’ 37·8”

Ex.2 Calculate sin if cos = –119

and 23, .

Sol. For any angle belonging to the indicated interval sin is negative, and therefore sin = – 2cos1

= –2

1191 = –

11102

.

Ex.3 Calculate tan if cos = –55

and 23, .

Sol. For any angle belonging to the indicated interval tan is positive and cos is negative, and

therefore tan =cos

cos1 2

= 2.

TRIGONOMETRIC RATIOS & IDENTITIES

Page # 150 TRIGONOMETRIC RATIOS & IDENTITIES


Ex.4 Given that 5 cos2 2 sin 2 = 0 54

74

, then find the value of cot2

.

Sol. Making a quadratic equation in sin2

(sin + 1) (5 sin 3) = 0 sin = 1 sin =35 not possible as

54

74

=32

,2

=34

cot34

= 1

Ex.5 Prove that 3(sin x – cos x)4 + 4(sin6x + cos6x) + 6(sin x + cos x)2 = 13

Sol. L.H.S. = 3[(sin x – cos x)2]2 + 4[(sin2 x)3 + (cos2 x)3)]+ 6(sin2 x+ cos2x + 2 sin x cos x)

= 3 (sin2 x + cos2 x – 2 sin x cos x)2 + 4(sin2 x + cos2 x) (sin4x + cos4x – sin2 x cos2x)]+ 6(sin2 x + cos2x + 2 sin x cos x)

= 3(1 – 2 sin x cos x)2 + 4 [(sin4 x + cos4 x ) – sin2 x cos2 x] + 6 (1 + 2 sin x . cos x)= 3 (1 + 4 sin2x cos2x – 4 sin x cos x) + 4 [(sin2x + cos2x)2

– 2sin2 x cos2x – sin2 x cos2x] + 6 + 12 sinx cos x

= 3 + 12sin2x cos2x – 12 sin x cos x + 4 (1 – 3 sin2 x cos2x) + 6 + 12 sin x cos x= 3 + 12 sin2x cos2x + 4 – 12sin2x cos2x + 6 = 13

Ex.6 Simplify the expression xtanba

xsina

ab1

xsina

ab

.ab

1 2

2 where b > a > 0.

Sol. After a few simple manipulations, this expression (for brevity denote it by P) can be rewritten

P = xsinbxcosa

xtanbaxsin

xsin)ab(a

xtanbaxsin22

2

2

2

Some students handle this as follows:

xcosxsinbxcosa

xcosxsinbaxtanba

22

2

22

and get a wrong answer: P = tan x. In this transformation what we actually have to simplify is the

expression xcos2 which is equal to |cos x|. And so the final result is P = sinx / |cos x|.



Ex.7 If tan =

212

12

1 where (0, 2 ), find the possible values of .

Sol. Let tan = x =

212

12

1 =

x21

x2 + 2x – 1 = 0 x =2

82 = ( 12 ) 12 is not b/w (0, 2 )

tan = 12 =8

or8

9

Ex.8 If ba

1b

cosa

sin 44

, prove that 33

8

3

8

)ba(1

bcos

asin

Sol. Given ba

1b

cosa

sin 44

or, b(a + b) sin4 + a(a + b) (1 – sin2 )2 = ab.or, b(a + b) sin4 + a(a + b) (1 + sin4 – 2sin2 ) = abor, (a + b)2 sin4 – 2a (a + b) sin2 + a2 + ab = abor, (a + b)2 sin4 – 2(a + b) sin2 . a + a2 = 0or, [(a + b) sin2 – a]2 = 0

or, (a + b) sin2 – a = 0 sin2 = ba

acos2 =

bab

Now, 3

8

asin

+ 3

8

bcos

= 34

4

a.)ba(a

+ 34

4

b)ba(b

= 4)ba(a

+ 4)ba(b

= 4)ba(ba

= 3)ba(1

Ex.9 If –2

< x <2

and y = log10(tan x + sec x). Then the expression E =21010 yy

simplifies to one of the

six trigonometric functions. find the trigonometric function.

Sol. y = log10(tan x + sec x), y = log10 xcosxsin1

E = 21010 yy

=2

xsin1xcos

xcosxsin1

= )xsin1(xcos2xcosxsin2xsin1 22

= )xsin1(xcos2xsin2xsin2 2

= )xsin1(xcos2)xsin1(xsin2

= tan x



B. TRIGONOMETRIC FUNCTIONS OF SUM OR DIFFERENCE OF TWO ANGLES

(a) sin (A ± B) = sinA cosB ± cosA sinB

(b) cos (A ± B) = cosA cosB sinA sinB

(c) sin²A sin²B = cos²B cos²A = sin (A+B) . sin (A B)

(d) cos²A sin²B = cos²B sin²A = cos (A+B) . cos (A B)

(e) tan (A ± B) = BtanAtan1BtanAtan (f) cot (A ± B) = AcotBcot

1BcotAcot

Factorisation Of The Sum Or Difference Of Two sines Or cosines :

(a) sinC + sinD = 2 sin2

DC cos

2DC

(b) sinC sinD = 2 cos2

DC sin

2DC

(c) cosC + cosD = 2 cos2

DC cos

2DC

(d) cosC cosD = 2 sin2

DC sin

2DC

Transformation Of Products Into Sum Or Difference Of sines & cosines :

(a) 2 sinA cosB = sin(A+B) + sin(A B) (b) 2 cosA sinB = sin(A+B) sin(A B)(c) 2 cosA cosB = cos(A+B) + cos(A B) (d) 2 sinA sinB = cos(A B) cos(A+B)

Ex.10 Suppose x and y are real numbers such that tan x + tan y = 42 and cot x + cot y = 49. Find the valueof tan(x + y).

Sol. tan x + tan y = 42 and cot x + cot y = 49

tan(x + y) = ytanxtan1ytanxtan

now, cot x + cot y = 49 ytan1

xtan1

= 49 ytan·xtanxtanytan

= 49

tan x · tan y =49

ytanxtan =

4942

= 76

tan (x + y) = )76(142

= 7142

= 294 Ans.

Ex.11 If x sin = y sin23

= z sin43

then :

(A) x + y + z = 0 (B) xy + yz + zx = 0 (C) xyz + x + y + z = 1 (D) none

Sol. x sin = y 12

32

sin cos xy

=3

2 cot

12

similarly xz =

32

cot 12 on adding

xz +

xy

= 1 xy + yz + zx = 0 Ans. B



Ex.12 Find satisfying the equation, tan 15° · tan 25° · tan 35° = tan , where (0, 15°).Sol. LHS = tan 15° · tan (30° – 5°) · tan (30° + 5°)

let t = tan 30° and m = tan 5°

tan = tan 15° ·tm1mt

·tm1mt

= tan )5(3 · 22

22

mt1mt

= 2

2

2

3

m3m31·

m31mm3

= 2

2

2

2

m3)m31(·

)m31()m3(m

= m = tan 5°. Hence = 5°3/1tº30tant

;xtan31

xtanxtan3x3tan

22

3

Ex.13 If tan A & tan B are the roots of the quadratic equation, a x2 + b x + c = 0 then evaluatea sin2 (A + B) + b sin (A + B) . cos (A + B) + c cos2 (A + B).

Sol. tan A + tan B = ba

; tan A . tan B = ca

tan (A + B) =baca1

= b

c a

Now E = cos2 (A + B) [a tan2 (A + B) + b tan (A + B) + c]

=1

1 22

bc a( )

a bc a

bc a

c2

2

2

( ) =

( )( )

c ab c a

2

2 2 b

c aa

c ac

2

1

=( )

( )c a

b c a

2

2 2 b c

c ac

2

2( )E = c

Ex.14 Show that cos2A + cos2(A + B) + 2 cosA cos(180° + B) · cos(360° + A + B) is independent of A. Hencefind its value when B = 810°.

Sol. cos2A + cos2(A + B) – [2 cosA · cosB · cos (A + B)]cos2A + cos2(A + B) – [ {cos(A + B) + cos(A – B) } cos (A + B) ]cos2A + cos2(A + B) – cos2(A + B) – (cos2A – sin2B)= sin2B which is independent of A now, sin2(810°) = sin2(720° + 90°) = sin290° = 1 Ans.

Ex.15 Simplify: cos x · sin(y – z) + cos y · sin(z – x) + cos z · sin (x – y) where x, y, z R.Sol. (1/2)[sin(y – z + x) + sin(y – z – x) + sin(z – x + y) + sin(z – x – y) + sin(x – y + z) + sin(x – y – z)] = 0

C. MULTIPLE ANGLES AND SUB-MULTIPLE ANGLES

(a) sin 2A = 2 sinA cosA ; sin = 2 sin2

cos2

(b) cos 2A = cos²A sin²A = 2cos²A 1 = 1 2 sin²A ;

cos = cos² 2 sin² 2 = 2cos² 2 1 = 1 2sin² 2 .

2 cos²A = 1 + cos 2A , 2sin²A = 1 cos 2A ;

2 cos²2

= 1 + cos , 2 sin²2

= 1 cos .



(c) tan 2A =Atan1

Atan22 ; tan =

222

tan1tan2

(d) sin 2A =Atan1

Atan22 , cos 2A =

Atan1Atan1

2

2

(e) sin 3A = 3 sinA 4 sin3A

(f) cos 3A = 4 cos3A 3 cosA (g) tan 3A = Atan31

AtanAtan32

3

Important Trigonometric Ratios

(i) sin 15° or sin12

=2213

= cos 75° or cos 125

;

cos 15° or cos12

=2213 = sin 75° or sin

125

;

tan 15° =1313

=2 3 = cot 75° ; tan 75° =1313

= 2 3 = cot 15°

(ii) sin 8 =2

22 ; cos 8 =

222

; tan 8 = 2 1 ; tan 83

= 2 1

(iii) sin10 or sin 18° =4

15 & cos 36° or cos 5 = 4

15

Ex.16 If cot = 1/2, then find the values of sin2 and cos2 .

Sol. sin 2 = 2tan1tan2

=412·2

= 54

; cos 2 = 2

2

tan1tan1

=4141

= – 53

Ex.17 Prove that tan8tan

= (1 + sec2 ) (1 + sec4 ) (1 + sec8 ).

Sol. RHS = 8cos8cos1

4cos4cos1

2cos2cos1

= 8cos4cos2cos4cos22cos2cos2 222

=8cos

cos]4cos2coscos8[ =

8cos

cossin

8sin

Ex.18 If x = 7.5° then find the value of xsinx3sinx3cosxcos

.

Sol.xsinx3sinx3cosxcos

=x2cosxsin2xsinx2sin2

= tan 2x = tan (2 × 7.5) = tan 15° = 2 – 3 Ans.



Ex.19 Prove the identity,

cos32

4 + sin (3 8 ) sin (4 12 ) = 4 cos 2 cos 4 sin 6 .

Sol. LHS : sin 4 + sin 8 + sin 12 = 2 sin 8 cos 4 + sin 8 = 2 sin 8 cos 4 + 2 sin 4 cos 4

= 2 cos 4 [sin 8 + sin 4 ] = 2 cos 4 [2 sin 6 cos 2 ] = 4cos 2 cos 4 sin 6

Ex.20 Calculate 4 sin 61 cos 3

1 .

Sol. 4sin 61 cos 3

1 = 2 31

61sin

31

61sin

= 2 6sin

22sin = 2 2

1)2(

2sin = 2cos (–2) – 1 = 2 cos 2 – 1.

Thus, 4 sin 61 cos 3

1 = 2 cos 2 – 1.

Ex.21 If cos =2 12cos

cos then find the value of tan

2 cot

2 (0 < < and 0 < < )

Sol.1

cos=

22 1

coscos

11

coscos

=3 1

1cos

cos (Componendo & dividendo)

tan2

2 = 3 tan2

2 tan2

2 cot2

2 = 3 Ans. 3

Ex.22 Calculate cos2

if sin =54

and ,23

.

Sol. First of all we seek cos . Since cos is negative for any angle belonging to the indicated interval,

we have cos = – 2sin1 = – 53

.

Since ,2

3, it follows that 2

,4

32 . For any angle

2 belonging to this interval cos

2 is

also negative, and therefore cos2

= –55

2cos1

. Thus cos 55

2.



Ex.23 Calculate tan2

if cos 2 =327

and 43, .

Sol. Since cos is negative for an angle belonging to the indicated interval,

we have cos =2

2cos1 = –

839

.

Since 43, , it follows that

2 8

3,2 . For any angle

2 belonging to this interval tan

2

is negative, and therefore cos1cos1

2tan tan

2 = –

5398

.

Ex.24 The figure (not drawn to scale) shows a regular octagon ABCDEFGH with diagonal AF = 1. Find the

numerical value of the side of the octagon.

Sol. = 22.5° ( AOB = 45º)

tan 22.5° = 12·

2x

A B

C

D

EF

G

HO

x = tan 22.5° = 12

Ex.25 If 3tantan

tan=

31

, find the value of 3cotcot

cot.

Sol.3tantan

tan=

31

3 tan = tan – tan 3 2 tan + tan 3 = 0

2 tan + 2

3

tan31tantan3

= 0, 2(1 – 3 tan2 ) + 3 – tan2 = 0 tan2 = 75

now,tan

tan31tantan3)tan31(

tantan3tan3tan

3tan3cotcot

cot

2

32

3

=)tan31tan3)(tan31(tan

)tan31)(tan3(tan222

22=

)tan1(2tan3

2

2 = )75(12

)75(3 = 12·2

16 =

32

Ans.

Alternatively: Prove that3tantan

tan +

3cotcotcot

= 1 now proceed



Ex.26 In a kite ABCD, AB = AD and CB = CD. If A = 108° and C = 36° then the ratio of the area of ABD to

the area of CBD can be written in the formc

36tanba 2 where a, b and c are relatively prime positive

integers. Determine the ordered triple (a, b, c).Sol. Since the triangles ABD and CBD have a common base,

hence the ratio of their areas equals the ratio of their heights.

Since tan 36° = xh

, then h = x tan 36°.

|||ly tan 72° =xk

then k = x tan 72°.

Hence,kh

=72tanx36tanx

=

36tan136tan2

36tan

2

= 236tan1 2

Then ordered triple (a, b, c) is (1,1, 2)

Ex.27 If , , and be the roots of the equation, 2 cos 2 2 cos + 1 = 0, all lying in the interval [0, 2 ]then find the value of the product, cos . cos . cos . cos .

Sol. 4 cos2 2 cos 1 = 0 cos =2 4 16

8 =

1 54

cos =5 14

or cos = 5 14

= sin 10 = cos 510 10 = cos 10

6

=5

or95

; 5

3 or

57

Hence P = cos 5 cos35 cos

75 cos

95 =

116

Ex.28 If sin x, sin22x and cos x · sin 4x form an increasing geometric sequence, find the numerical value of

Sol. Given sin x, sin22x and cos x · sin 4x are in G.P. (r > 1 as G.P. is increasing)sin42x = (sin x) (cos x) (sin 4x) 16 sin4x cos4x = sin x cos x sin 4x16 sin3x cos3x = sin 4x (sin x 0, cos x 0)16(sin x cos x)3 = 2 sin 2x · cos 2x (sin 2x)3 = sin 2x · cos 2xsin22x = cos 2x (sin 2x 0), 1 – cos22x = cos 2x, y2 + y – 1 = 0

cos 2x = 2

51; cos 2x cannot be

215

hence rejected cos 2x = 2

51

sin x =2

x2cos1 =

22

151 =

253

= 2215

cos 2x = 2

15r =

xsinx2sin 2

= 4 sin x cos2x = 2 sin x(1 + cos 2x)

r =2

15 ·

215

=22

4 = 2



Ex.29 Prove using induction or otherwise that, 2 cos2n = 2 2 2 2 2...... cos

where R. H. S. contains n radical signs and (0 , ).

Sol. 2 cos2

= 2 1( cos )

2 cos22 = 2 1 2cos = 2 2 1( cos )

2 cos 23 = 2 1 22cos = 2 2 22cos = 2 2 2 1( cos ) and so on.

In the same way 2 cos2n = 2 2 2 2 2...... cos

Similarly 2 sin2n = 2

1

22 1cos n

= 2 22 1cos n

= 2 2 2 2 2 2...... cos where R. H. S. contains n radical signs

Ex.30 Show that 7

21

157cos

156cos

155cos

154cos

153cos

152cos

15cos .

Sol. We have sin152

= 2 sin15

cos 15

, sin154

= 2 sin152

cos 152

,

sin158

= 2 sin154

cos 154

, sin15

16 = 2 sin

158

cos 158

.

Multiplying the equalities and noting that sin15

16 = – sin

15, cos

158

= – cos157

.

cos 15 . cos 152

. cos 154

. cos 157

= 421

Further cos 21

155

. and sin156

= 2 sin153

cos 153

, sin15

12 = 2 sin

156

cos 156

.

Hence cos153

. cos156

= 221

. The rest is obvious.



D. CONDITIONAL IDENTITIES

tan (A+B+C) = AtanCtanCtanBtanBtanAtan1CtanBtanAtanCtanBtanAtan

If A+B+C = then(a) tanA + tanB + tanC = tanA tanB tanC

(b) tan2A

tan2B

+ tan2B

tan2C

+ tan2C

tan2A

= 1

(c) sin2A + sin2B + sin2C = 4 sinA sinB sinC

(d) sinA + sinB + sinC = 4 cosA2

cosB2

cosC2

(e) cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C

(f) cos A + cos B + cos C = 1 + 4 sin2A

sin2B

sin 2C

Ex.31 If A + B + C = , prove thatCtan.Btan

Atan = tan A 2 cot A .

Sol. L H S = CtanBtanAtan

CtanBtanAtan 222

= Atan

BtanAtan2)CtanBtanA(tan 2

[ tan A = tan A]

= tan A 2 CtanBtanAtanAtanCtanCtanBtanBtanAtan

= tan A 2 cot A]

Ex.32 If A + B + C = and cot = cot A + cot B + cot C, show that ,sin (A ) . sin (B ) . sin (C ) = sin3 .

Sol. Given cot = cot A + cot B + cot C or cot cot A = cot B + cot C

orsin ( )sin sin

AA

=sin ( )sin sin

B CB C

= sin

sin sinA

B C or sin (A ) =

sinsin sin

2 AB C

sin (1)

similarly sin (B ) =sin

sin sin

2 BC B

sin (2) sin (C ) =sin

sin sin

2 CA B

sin (3)

Multiplying (1) , (2) and (3) we get the result

Ex.33 Find whether a triangle ABC can exists with the tangents of its interior angle satisfying, tan A = x, tanB = x + 1 and tan C = 1 – x for some real value of x. Justify your assertion with adequate reasoning.

Sol. In a triangle Atan = Atan (to be proved)x + x + 1 + 1 – x = x(1 + x)(1 – x)2 + x = x – x3; x3 = – 2; x = – 21/3

Hence tanA = x < 0 and tanB = x + 1 = 1 – 21/3 < 0Hence A and B both are obtuse. Which is not possible in a triangle. Hence no such triangle can exist.



Ex.34 Prove that(a) sin3 A cos (B – C) + sin3 B cos (C – A) + sin3 C cos (A – B) = 3 sin A sin B sin C;(b) sin3 A sin (B – C) + sin3 B sin (C – A) + sin3 C sin (A – B) = 0

if A + B + C = .Sol. (a) We have

sin3 A cos (B – C) = sin2 A sin A cos (B – C) =

=21

sin2 A {sin (A + B – C) + sin (A – B + C)}.

But since A + B + C = , we have sin3 A cos (B – C) =21

sin2 A (sin 2C + sin 2B)

= sin2 A (sin B cos B + sin C cos C)

= sin2 A sin B cos B + sin2 A sin C cos C + sin2 B sin C cos C + sin2 B sin A cos A +sin2 C sin A cos A + sin2 C sin B cos B == sin A sin B (sin A cos B + cos A sin B)+ sin A sin C (sin A cos C + cos A sin C)+ sin B sin C (sin B cos C + cos B sin C)= sin A sin B sin (A + B) + sin A sin C sin (A + C) + sin B sin C sin (B + C) = 3 sin A sin B sin C.

(b) We have

sin3 A sin (B – C) = sin2 A sin A sin (B – C) = sin2 A sin (B + C) sin (B – C)

= 21

sin2 A {cos 2C – cos 2B) = sin2 A(sin2 B – sin2 C)

= sin2 A sin2 B sin2C Bsin1

Csin1

22 = sin2 A sin2 B sin2 C

× Asin

1Bsin

1Csin

1Asin

1Bsin

1Csin

1222222 = 0

Ex.35 Given the product p of sines of the angles of a triangle & product q of their cosines, find the cubicequation, whose coefficients are functions of p & q & whose roots are the tangents of the angles ofthe triangle.

Sol. Given sinA sinB sinC = p ; cosA cosB cosC = qHence tanA tanB tanC = tanA + tanB + tanC = p/qHence equation of cubic is

x3 – qp

x2 + tanA tan Bx – qp

= 0 ...(i)

now tan tan sin sin cos sin sin cos sin sin coscos cos cos

A B A B C B C A C A BA B C

We know that A + B + C = cos(A+B+C) = –1; cos(A+B) cosC – sin(A+B) sinC = –1( cosA cosB – sinA sin B) cosC – sinC (sinA cosB + cosA sinB) = –11+ cosA cosB cosC= sinA sinB cosC + sinB sinC cosA + sinC sinA cosB

dividing by cosA cosB cosC1 q

qA Btan tan

Hence (i) becomes qx3 – px2 + (1 + q)x – p = 0



E. MAXIMUM & MINIMUM VALUES OF TRIGONOMETRIC FUNCTIONS

(a) Min. value of a2 tan2 + b2 cot2 = 2ab

(b) Max and Min. value of acos + bsin are 22 ba and – 22 ba(c) If f( ) = acos( ) + bcos( ) where a, b, and are known quantities then

– )cos(ab2ba 22 f( ) )cos(ab2ba 22

(d) If 2,0 and = (constant) then the maximum values of the expression

cos cos , cos + cos , sin + sin and sin sinoccurs when /2

(e) If 2,0 and = (constant) then the minimum values of the expression

sec + sec , tan + tan , cosec + cosec occurs when /2.(f) If A, B, C are the anlges of a triangle then maximum value of

sinA + sinB + sinC and sinA sinB sinC occurs when A = B = C = 60º(g) In case a quadratic in sin or cos is given then the maximum or minimum values can be

interpreted by making a perfect square

Ex.36 Find the minimum vertical distance between the graphs of y = 2 + sin x and y = cos x.

Sol. dmin = min(2 + sin x – cos x) = min[2 + 2 sin 4x ] = 2 – 2 at x =

47

Ex.37 If a sin2x + b lies in the interval [–2, 8] for every x R then find the value of (a – b).Sol. f (x) = a sin2x + b

f (x) has a maximum value of 8 which occurs when sin2x = 1a + b = 8 ....(1)

|||ly f (x) has a minimum value of – 2 which occurs where sin x = 0b = – 2 ....(2)

from (1) and (2) a = 10; b = – 2 a – b = 12 [Ans. 12]

Ex.38 Find the greatest value of c such that system of equationsx2 + y2 = 25; x + y = c has a real solution.

Sol. put x = 5 cos y = 5 sin

5(cos + sin ) = c; but (cos + sin )max = 2 and (cos + sin )min = – 2

hence, 25cmax Ans.

Ex.39 Find the minimum and maximum value of f (x, y) = 7x2 + 4xy + 3y2 subjected to x2 + y2 = 1.Sol. Let x = cos and y = sin

y = f ( ) = 7 cos2 + 4 sin cos + 3 sin2 = 3 + 2 sin 2 + 2(1 + cos 2 )

= 5 + 2(sin 2 + cos 2 ) but – 2 (sin 2 + cos 2 ) 2

ymax = 5 + 22 and ymin = 5 – 22



Ex.40 If 1, 2, ...... , n are real numbers, show that,(cos 1 + cos 2 + ...... + cos n)2 + (sin 1 + ......+ sin n)2 n2 .

Sol. L H S = (cos2 1 + sin2 1) + ....... + (cos2 n + sin2 n) + 2 cos ( 1 2)nC2 terms

n + 2n n( )1

2 = n2

Ex.41 Show that the expression cos (sin + 22 sinsin ) always lies between the values of ± 2sin1 .

Sol. Let y = cos (sin + 22 sinsin )

or, y – cos sin = cos ( 22 sinsin )

or, (y – cos sin )2 = cos2 (sin2 + sin2 )or, y2 – 2ysin cos + cos2 = cos2 sin2 + cos2 sin2

or, y2 – 2ysin cos + cos2 = cos2 + cos2 . sin2

[Here we have added cos2 on both sides to get 1 + sin2 ]or, y2 – 2y sin cos + cos2 = cos2 (1 + sin2 )or, y2.sec2 – 2y tan + 1 = 1 + sin2 (dividing by cos2 )or, y2tan2 – 2ytan + 1 = (1 + sin2 ) – y2 (sec2 = 1 + tan2 )or, (ytan – 1)2 = (1 + sin2 ) – y2

square of a real number 0 1 + sin2 – y2 0

or, y2 – ( 2sin1 )2 0 y lies between – 2sin1 and 2sin1 .

F. SUMMATION OF TRIGONOMETRIC SERIES

Sum of sines or cosines of n angles

sin + sin ( + ) + sin ( + 2 ) + ...... + sin )1n( =2

2n

sin

sin sin

21n

cos + cos ( + ) + cos ( + 2 ) + ...... + cos )1n( =2

2n

sin

sin cos 2

1n

Ex.42 Find the sum of the series, cos 1n2 + cos 1n23

+ cos 1n25

+ ........ upto n terms.

Do not use any direct formula of summation.

Sol. Let = 1n2S = cos + cos 3 + cos 5 + ........ cos (2n – 1)(2 sin ) S = 2 sin [cos + cos 3 + cos 5 + ........ cos (2n – 1) ]T1 = sin 2 – 0; T2 = sin 4 – sin 2 T3 = sin 6 – sin 4 Tn = sin 2n – sin 2(n – 1)

(2 sin ) S = sin2n S =

1n2sin2

1n2n2sin

=21



Ex.43 Given35

1kk5sin = tan

nm

, where angles are measured in degrees, and m and n are relatively prime

positive integers that satisfynm

< 90, find the value of (m + n).

Sol. LHS: S = sin 5 + sin 10 + sin 15 + .......... + sin 170 + sin 175

S 25sin2 = 2 sin

25

[sin 5 + sin 10 + ......... + sin 175]

T1 = cos25

– cos2

15; T2 = cos

215

– cos2

25.........; T35 = cos

2345

– cos2

355

25

sin2 · S = cos25

– cos2

355 = 2 sin

2180

· sin2

175 = 2 sin

2175

S =

25sin

2175sin

=

2590cos

2175sin

=

2175cos

2175sin

= tan 2175

= tan nm

m = 175 and n = 2 m + n = 177

Ex.44 Find the sum of the series ,cot 2 x . cot 3 x + cot 3 x . cot 4 x + ...... + cot (n + 1) x . cot (n + 2) x .

Sol. cot x = cot [ (n + 2) x (n + 1) x ] = x)2n(cotx)1n(cot1x)1n(cot.x)2n(cot

or cot x [ cot (n + 1) x cot (n + 2) x ] = cot (n + 2) x . cot (n + 1) x + 1Hence cot (n + 1) x cot (n + 2) x = cot x [ cot (n + 1) x cot (n + 2) x ] 1 Put n = 1 , 2 , 3 , ...... , n and adding we get sum of the series

= cot x [ cot 2 x cot (n + 2) x ] n2

Ex.45 Let f (x) denote the sum of the infinite trigonometric series, f (x) = 1n

nn 3xsin

3x2sin .

Find f (x) (independent of n) also evaluate the sum of the solutions of the equation f (x) = 0 lying in theinterval (0, 629).

Sol. f (x) =1n

nn 3xsin

3x2sin =

1nnn 3

xsin3

x2sin221

= 1n

1nn 3xcos

3xcos

21

now substituting n = 1, 2, 3, 4........

f (x) = xcos3xcos

21

+ 3xcos

3xcos

21

2 + 23 3xcos

3xcos

21

.......... + 1nn 3xcos

3xcos

21

f (x) = xcos3xcos

21Lim nn = 2

1[1 – cos x] now f (x) = 0 cos x = 1 x = 2n , n I

sum of the solutions in (0, 629), S = 2[ + 2 + 3 + ....... + 100 ] = 2 · 5050 = 10100



Ex.46 Evaluate 89

1n2)n(tan1

1.

Sol. S = 22222 )89(tan11

)88(tan11........

)3(tan11

)2(tan11

)1(tan11

reversing the sum

S = 2222 )89(cot11

)88(cot11..............................

)2(cot11

)1(cot11

2S =89

1n22 )n(cot1

1)n(tan1

1 =

89

1n2

2

2 )n(tan1)n(tan

)n(tan11

=89

1n1 = 1 + 1 + ....... + 1 = 89 S = 44.5

G. ELIMINATION

Ex.47 Eliminate between the equation a sec + b tan + c = 0 and p sec + q tan + r = 0.Sol. Given a sec + b tan + c = 0 ...(1)

and p sec + q tan + r = 0 ...(2)Solving (1) and (2) by cross multiplication method, we have

pbaq1

arpctan

qcbrsec

sec2 – tan2 = 1

22

pbaqarpc

pbaqqcbr

= 1 or, (br – qc)2 – (pc – ar)2 = (aq – pb)2

Ex.48 If is eliminated from the equations, a cos + b sin = c & a cos2 + b sin2 = c, show that theeliminant is, (a b)2 (a c) (b c) + 4 a2 b2 = 0 .

Sol. a cos + b sin = c ..............(1)a cos2 + b sin2 = c ..............(2)

From (2) sin2 =c ab a and cos2 =

b cb a

Now squaring (1) a2 cos2 + b2 sin2 + 2 ab sin cos = c2

a2b cb a

+ b2c ab a

c2 = 2 ab b cb a

c ab a

or a2 (b c) + b2 (c a) c2 (b a) = 2 ab b c c a

(a b) (b c) (c a) = 2 ab b c c a

(a b)2 (b c)2 (c a)2 = 4 a2 b2 (b c) (c a)(a b)2 (b c) (c a) = 4 a2 b2 Result



Miscellaneous Questions

Ex.49 Prove that tan7

. tan 72

. tan73

= 7

Sol. Let =7

7 =

or, 4 + 3 = or, tan(4 ) = tan( – 3 ) or, tan4 = –tan3

or, 42

3

tantan61tan4tan4

= – 2

3

tan31tantan3

or, 2

3

42

3

z31zz3

zz61z4z4

[where tan = z (suppose)]

or, (4 – 4z2) (1 – 3z2) = –(3 – z2)(1 – 6z2 + z4) or 12z4 – 16z2 + 4 = –(–z6 + 9z4 – 19z2 + 3)

or, z6 – 21z4 + 35z2 – 7 = 0 ...(1)

This is cubic equation in z2 i.e. in tan2 , the roots of this equation are therefore tan2

7, tan2

72

and tan2

73

From (1), product of the roots = 7

tan2 7

. tan2

72

. tan2

73

= 7 tan 7

. tan7

2 . tan

73

= 7 Hence the result.

Ex.50 In triangle ABC, cos A . cos B + cos B . cos C + cos C . cos A = 1 – 2 cos A . cos B . cos C. Prove thatit is possible if and only if ABC is equilateral.

Sol. cos A . cos B = 1 – 2 cos A . cos B . cos C = 1 – cos C (cos (A+ B) + cos (A – B) )= 1 – cos C (cos (A – B) – cosC) = 1 + cos (A + B) cos (A – B) +cos2 C= 1 + cos2 A – sin2 B + cos2C = cos2 A + cos2 B +cos2C = cos2A.Thus we have, 2 cos2 A – 2 cos A . cos B = 0

(cos A – cos B)2 + (cos B – cos C)2 + (cos C – cos A)2 = 0 cos A = cos B = cos C A = B = CThus triangle ABC is equilateral

Now if is equilateral A = B = C =3

cosA cos B =43

and 1 – 2 cos A cos B cos C

= 1 – 43

82

. Hence the given expression is true if and only if ABC is equilateral.



JEE MAINEXERCISE – I

1. If tan +cot =a then the value of tan4 +cot4 =(A) a4 + 4a2 +2 (B) a4 – 4a2 + 2(C) a4 – 4a2 – 2 (D) None of theseSol.

2. If a cos + b sin = 3 & a sin – b cos = 4 thena2 + b2 has the value =(A) 25 (B) 14 (C) 7 (D) None of theseSol.

3. The value of tan 1º tan 2º tan 3º ..... tan 89º is(A) 1 (B) 0 (C) (D) 1/2Sol.

4.

x2

3tan.2

xcos

x2

7sinx2

3cos2

xtan 3

when simplified reduces to :(A) sinx cosx (B) – sin2x (C) –sinx cosx (D) sin2x

Sol.

5. The expression

3 )3(sin2

3sin 44–2 )5(sin

2sin 66

is equal to(A) 0 (B) 1 (C) 3 (D) sin 4 + sin 6Sol.

6. cos (540º – ) – sin (630º – ) is equal to(A) 0 (B) 2 cos (C) 2 sin (D) sin –cosSol.

7. The value of sin( ) sin ( ) cosec2 is equal to(A) –1 (B) 0 (C) sin (D) None of these



Sol.

8. If sin sin – cos cos + 1 = 0, then the valueof 1 + cot tan is(A) 1 (B) –1 (C) 2 (D) None of theseSol.

9. The value of º69sinº51cosº39cosº21sin

º66sinº6sinº6cosº24sin is

(A) –1 (B) 1 (C) 2 (D) None of theseSol.

10. If 3 sin = 5 sin , then

2tan

2tan

is equal to

(A) 1 (B) 2 (C) 3 (D) 4Sol.

11. In a triangle ABC if tan A < 0 then :(A) tan B . tan C > 1 (B) tan B . tan C < 1(C) tan B . tan C = 1 (D) None of theseSol.

12. If tan A – tan B = x and cot B – cot A = y, thencot (A – B) is equal to

(A) x1

y1

(B) y1

x1

(C) y1

x1

(D) None of these

Sol.

13. If tan 25º=x, thenº115tanº155tan1

º115tanº155tanis equal to

(A) x2x1 2

(B) x2x1 2

(C) 2

2

x1x1

(D) 2

2

x1x1

Sol.

14. If A + B = 225º, then the value of

Bcot1Bcot.

Acot1Acot

is

(A) 2 (B) 1/2 (C) 3 (D) 1/3



Sol.

15. The value of tan 3A – tan 2A – tan A is equal to(A) tan 3A tan 2A tan A(B) – tan 3A tan 2A tan A(C) tan A tan 2A – tan 2A tan 3A – tan 3A tan A(D) None of theseSol.

16. tan 203º + tan 22º + tan 203º tan 22º =(A) –1 (B) 0 (C) 1 (D) 2Sol.

17. The value of º15tan1º15tan1

2

2

is

(A) 1 (B) 3 (C) 23

(D) 2

Sol.

18. If A lies in the third quadrant and 3 tan A – 4 = 0,then 5 sin 2A + 3 sinA + 4 cosA is equal to

(A) 0 (B) –524

(C) 5

24(D)

548

Sol.

19. º80sin

º10sinº50sinº70sin8º20cos2 is equal to

(A) 1 (B) 2 (C) 3/4 (D) None of theseSol.

20. If cos A = 3/4, then the value of16cos2 (A/2) – 32 sin (A/2) sin (5A/2) is(A) –4 (B) –3 (C) 3 (D) 4Sol.

21. The value of the expression

109cos1

107cos1

103cos1

10cos1 is

(A) 1/8 (B) 1/16 (C) 1/4 (D) 0



Sol.

22. The numerical value of sin 12º . sin 48º . sin 54ºis equal to(A) 1/2 (B) 1/4 (C) 1/16 (D) 1/8Sol.

23. If = 2 , then

(A) tan 2

+ tan 2

+ tan 2

= tan 2

tan 2

tan 2

(B) tan 2

tan 2

+ tan 2

tan 2

+ tan 2

tan 2

= 1

(C) tan 2

+ tan 2

+ tan 2

= – tan 2

tan 2

tan 2

(D) tan 2

tan 2

+ tan 2

tan 2

+ tan 2

tan 2

= 0

Sol.

24. cos0+cos 7 +cos 72

+cos 73

+cos 74

+cos 75

+cos 76

=

(A) 1/2 (B) –1/2 (C) 0 (D) 1Sol.

25. A regular hexagon & a regular dodecagon areinscribed in the same circle. If the side of the

dodecagon is ( 3 –1), then the side of the hexagon is

(A) 2 +1 (B) 2

13(C) 2 (D) 2

Sol.



26. In a right angled triangle the hypotenuse is 2 2times the perpendicular drawn from the oppositevertex. Then the other acute angles of the triangle are

(A) 6

&3

(B) 8

3&8

(C) 4

&4

(D) 103&

5Sol.

27. If ,2 then the value of

sin1 – sin1 is equal to

(A) 2 cos 2

(B) 2 sin 2

(C) 2 (D) None of these

Sol.

28. The value of cot x + cot(60º + x) + cot (120º + x)is equal to(A) cos 3x (B) tan 3x

(C) 3 tan 3x (D) xtanxtan3

xtan933

2

Sol.

29. If x 23, then

4 cos2 x2sinxsin42x

424 is always equal to

(A) 1 (B) 2 (C) –2 (D) None of theseSol.



30. The expression xcos10x3cos5x5cos

10x2cos15x4cos6x6cos

is equal to(A) cos 2x (B) 2 cos x (C) cos2 x (D) 1 + cos xSol.

31. If cos (A – B) = 3/5 and tan A tan B = 2,

(A) cosA cosB = – 51

(B) sinA sinB = – 52

(C) cos (A + B) = – 51

(D) sin A cos B = 54

Sol.

32. If A + B + C = 2

3, then cos 2A + cos 2B + cos 2C

is equal to(A) 1–4cos A cosB cosC (B) 4 sinA sinB sinC(C) 1+2 cosA cosB cosC (D) 1–4 sinA sinB sinC

Sol.

33. For –2

< <2

,2coscos1

2sinsin lies in the interval

(A) (– ) (B) (–2, 2) (C) (0, ) (D) (–1, 1)Sol.

34. If 0 < x < and cos x + sin x = 21

, then tan x is

(A) 3

)74((B) –

3)74(

(C) 4

)71((D)

4)71(



Sol.

35. Let be such that 3 .

If sin + sin = – 6521

and cos + cos = –6527

, then

the value of cos 2

is

(A) – 1303

(B) 1303

(C) 656

(D) – 656

Sol.

36. The value of the expressioncos 1° cos 2° ......... cos 179° equals

(A) 0 (B) 1 (C) 1/ 2 (D) – 1

Sol.

37. Which is correct one ?(A) sin 1° < sin 1 (B) sin 1° = sin 1

(C) sin 1° > sin 1 (D) sin 1° = sin 180

Sol.

38. The value of cos 10° – sin 10° is(A) Positive (B) Negative (C) 0 (D) 1Sol.



JEE ADVANCED ( )OBJECTIVEEXERCISE – II

SINGLE CORRECTLEVEL – I1. If tan A and tan B are the roots of the quadraticequation x2 – ax + b = 0, then the value of sin2 (A + B)

(A) 22

2

)b1(aa

(B) 22

2

baa

(C) 2

2

)cb(a

(D) 22

2

)a1(ba

Sol.

2. If A = tan 6º tan 42º and B = cot 66º cot 78º, then(A) A = 2B (B) A = 1/3 B (C) A = B (D) 3A = 2BSol.

3. º250sin3

1º290cos

1 =

(A) 3

32(B)

334

(C) 3 (D) None of these

Sol.

4. If A + B + C = & sin 2CA = k sin

2C

,

then tan 2A

tan 2B

=

(A) 1k1k

(B) 1k1k

(C) 1k

k(D)

k1k

Sol.

5. In any triangle ABC, which is not right angled cos A . cosec B . cosec C is equal to

(A) 1 (B) 2 (C) 3 (D) None of these



Sol.

6. If 3 cos x + 2 cos 3x = cos y, 3 sin x + 2 sin 3x = siny, then the value of cos 2x is(A) –1 (B) 1/8 (C) –1/8 (D) 7/8Sol.

7. If cos + cos = a, sin + sin = b and = 2 ,

then cos

3cos =

(A) a2 + b2 – 2 (B) a2 + b2 – 3(C) 3 – a2 – b2 (D) (a2 + b2) /4Sol.

8. If A + B + C = & cos A = cos B . cos C thentan B . tanC has the value equal to(A) 1 (B) 1/2 (C) 2 (D) 3Sol.

9. The value of tan 16

+ 2 tan 8

+ 4 is equal to

(A) cot8

(B) cot16

(C) cot16

–4 (D) None of these

Sol.

10. The value of cos19

+cos193

+cos195

+...+ cos19

17

is equal to(A) 1/2 (B) 0 (C) 1 (D) None of theseSol.

11. If 4

3 < , then 2sin

1cot2 is equal to

(A) 1 +cot (B) –1 – cot (C) 1 – cot (D) –1 + cot Sol.



12. If f( ) = sin4 + cos2 , then range of f( ) is

(A) 1,21

(B) 43,

21

(C) 1,43

(D) None of these

Sol.

13. If 2 cos x + sin x = 1, then value of 7 cos x + 6 sinx is equal to(A) 2 or 6 (B) 1 or 3 (C) 2 or 3 (D) None of theseSol.

14. If cosec A + cot A = 211

, then tan A is

(A) 2221

(B) 1615

(C) 11744

(D) 43

117

Sol.

15. If 0° < x < 90° & cos x = 103

, then the value of

log10 sin x + log10 cos x + log10 tan x is(A) 0 (B) 1 (C) –1 (D) None of theseSol.

16. If cot + tan = m and cos

1 – cos = n, then

(A) m (mn2)1/3 – n(nm2)1/3 = 1(B) m(m2n)1/3 – n(nm2)1/3 = 1(C) n (mn2)1/3 – m(nm2)1/3 = 1(D) n(m2n)1/3 – m(mn2)1/3 = 1Sol.

17. If 2 sec2 – sec4 – 2 cosec2 + cosec4 = 15/4, then tan is equal to

(A) 2/1 (B) 1/2 (C) 1/2 2 (D) 1/4

Sol.



18. If 25

BcosAcosand

23

BsinAsin

, 0 < A, B < /2,

then tan A + tan B is equal to

(A) 5/3 (B) 3/5 (C) 1 (D) 5/)35(Sol.

19. If 3 sin x + 4 cos x = 5 then 4 sin x – 3 cos x isequal to(A) 0 (B) 1 (C) 5 (D) None of theseSol.

20. If sin 2 = k, then the value of 2

3

2

3

cot1cot

tan1tan

is equal to

(A) kk1 2

(B) kk2 2

(C) k2 + 1 (D) 2 – k2

Sol.

21. If f( ) = sin2 + sin2 32

+ sin2 34

,

then f 15 is equal to

(A) 32

(B) 23

(C) 31

(D) 21

Sol.



MULTIPLE CORRECTLEVEL I – I

1. The value of xcosxcosxsin

3 =

(A) 1+tanx + tan2x –tan3x (B) 1+tan x+tan2x+tan3x(C) 1–tanx + tan2x +tan3x (D) (1 + tan x) sec2xSol.

2. If (sec A + tan A) (sec B + tan B) (sec C + tan C)= (sec A – tan A) (sec B – tan B) (sec C – tan C)then each side is equal to(A) 1 (B) –1 (C) 0 (D) None of theseSol.

3. The value of )º11sinº11(cos)º11sinº11(cos is

(A) –tan 304º (B) tan 56º (C) cot 214º (D) cot 34ºSol.

4. If tan2 = 2 tan2 + 1, then the value ofcos 2 + sin2 is(A) 1 (B) 2 (C) –1 (D) Independent of Sol.

5. The value of cos10

cos102

cos104

cos108

cos10

16 is

(A) 64

5210 (B) –16

)10/cos(

(C) 16

)10/cos((D) –

645210

Sol.

6. If x + y = z, then cos2 x + cos2 y + cos2 z – 2 cos xcos y cos z is equal to(A) cos2 z (B) sin2 z (C) cos (x + y – z) (D) 1



Sol.

7. If tan A + tan B + tan C = tan A . tan B . tan C, then(A) A, B, C may be angles of a triangle(B) A + B + C is an integral multiple of (C) sum of any two of A, B, C is equal to third(D) None of theseSol.

8. In a triangle tan A + tan B + tan C = 6 and tan Atan B = 2, then the values of tanA, tan B and tan C are(A) 1, 2, 3 (B) 2, 1, 3(C) 1, 2, 0 (D) None of theseSol.

9. An extreme value of 1 + 4 sin + 3 cos is(A) – 3 (B) – 4 (C) 5 (D) 6Sol.

10. If the sides of a right angled triangle are{cos2 + cos2 + 2cos( )} and{sin2 + sin2 + 2sin( )},then the length of thehypotneuse is(A) 2 [1 + cos( )] (B) 2 [1 – cos( )]

(C) 4 cos2

2(D) 4 sin2

2Sol.

11. For 0 < < /2, tan + tan 2 + tan 3 = 0 if(A) tan = 0 (B) tan 2 = 0(C) tan 3 = 0 (D) tan tan 2 = 2Sol.



12. (a+2) sin + (2a – 1) cos = (2a+1) if tan =

(A) 3/4 (B) 4/3 (C) 1a

a22 (D)

1aa2

2

Sol.

13. If tan x = ca

b2, (a c)

y = a cos2x + 2b sin x cos x + c sin2xz = a sin2x – 2b sin x cos x + c cos2x, then

(A) y = z (B) y + z = a + c(C) y – z = a – c (D) y – z = (a – c)2 + 4b2

Sol.

14. nn

BcosAcosBsinAsin

BsinAsinBcosAcos

(A) 2 tann 2

BA(B) 2 cotn

2BA

: n is even

(C) 0 : n is odd (D) None of these

Sol.

15. The equation sin6x + cos6x = a2 has real solution if

(A) a (–1,1) (B) a (–1, –1/2)

(C) a 21

21

(D) a (1/2, 1)Sol.

16. If 3 sin =sin (2 + ), then tan ( ) – 2 tan is(A) independent of (B) independent of (C) dependent of both and (D) independent of but dependent of Sol.



JEE ADVANCEDEXERCISE – IIIComprehension # 1

If cos + cos = a and sin + sin = b and isarithmetic mean between and , then sin 2 +

cos 2 = 1 + 22 ba)ba(nb.

where n is some integer then answer the followingquestions :

1. The value of n is(A) 0 (B) 1 (C) 2 (D) – 2Sol.

2. If for n obtained in above question, sinnA = x, thensin A sin 2A sin 3A sin 4A is a polynomial in x, ofdegree(A) 5 (B) 6 (C) 7 (D) 8Sol.

3. If degree of polynoimal obtained in previousquestion is p and (p – 5) + sin x, cos x, tan x are inG.P., then cos9x + cos6x + 3 cos5 x – 1 =(A) –1 (B) 0 (C) 1 (D) None of theseSol.

Comprehension # 2Let p be the product of the sines of the angles oftriangle ABC and q is the product of the cosines ofthe angles.

4. In this triangle tan A + tan B + tan C is equal to

(A) p + q (B) p – q (C) qp

(D) None of these

Sol.

5. tan A tan B + tan B tan C + tan C tan A is equal to

(A) 1 + q (B) qq1

(C) 1 + p (D) pp1

Sol.

6. The value of tan3 A + tan3 B + tan3 C is

(A) 3

23

qpq3p

(B) 3

3

pq

(C) 3

3

qp

(D) 3

3

qpq3p

Sol.



Matrix Match Type1. Column - I Column - II(A) sin 420° cos 390° (P) 0

+ cos (–660°) sin (–330°)(B) tan 315° cot (–405°) (Q) 1

+ cot 495° tan (–585°)

(C) The value of )23tan1)(22tan1()37tan1)(8tan1(

= (R) 2

(D) Value of xsin31

42

is (S) 5

(where [.] represents greatest integer function)Sol.

2. Column - I Column - II(A) If for some real x, then equation (P) 2

x + x1

= 2 cos holds

then cos is equal to (Q) 1(B) If sin + cosec = 2,

then sin2008 + cosec2008 is equal to (R) 0(C) Maximum value of sin4 + cos4 is(D) Least value of 2 sin2 + 3 cos2 is (S) –1Sol.



SUBJECTIVE TYPE1. Eliminate from the relations a sec = 1 – b tan ,a2 sec2 = 5 + b2 tan2

Sol.

2. If tan = –5/12, is not in the second quadrant,

then show that 338181

)(eccos)º270sec()º90tan()º360sin(

Sol.

3. Prove that

4cot2

cos

4cot1

4cot1

2

2

sec 2

9 = cosec 4 .

Sol.



4. Prove than

(i) A2tanA8tan

1A4sec1A8sec

Sol.

(ii) AsinAcosAsinAcos

– AsinAcosAsinAcos

= 2 tan 2A

Sol.

5. If A+B=45º, prove that (1+tan A) (1+tan B) = 2

and hence deduce that tan 22 122º1

Sol.

6. If 0 < < /4, then show that

)4cos1(22 = 2 cos .

Sol.

7. Prove thattan tan (60º + ) tan (60º – ) = tan 3 and hencededuce that tan 20º tan 40º tan 60º tan 80º = 3.Sol.

8. Prove that 4(cos3 20º+cos3 40º)=3(cos 20º+cos 40º)Sol.



9. Prove that

sin2 +sin22 +sin23 +....+sin2 n =2n

–sin2

)1ncos(nsin

Sol.

10. If is the exterior angle of a regular polygon of nsides and is any constant, then prove thatsin + sin ( ) + ....... up to n terms = 0Sol.

11. If x + y + z = 2

show that,

sin 2x + sin 2y + sin 2z = 4 cosx cosy cosz.Sol.

12. If x + y = + z, then prove thatsin2x + sin2y – sin2z = 2 sin x sin y cos z.Sol.

13. If A + B + C = 2S then prove thatcos (S – A) + cos(S – B) + cos (S – C) + cos S = 4

cos 2A

cos 2B

cos 2C

Sol.



14. If A + B + C = 0º then prove thatsin 2A + sin 2B + sin 2C = –4 sin A sin B sin C.Sol.

15. Find the extreme values of

cos x cos x32

cos x3

2

Sol.

16. Find the maximum and minimum values of(i) cos 2x + cos2 xSol.

(ii) cos2 x4 (sin x – cos x)2

Sol.

17. Prove that, sin3x . sin3 x + cos 3 x . cos3 x = cos3 2x.Sol.

18. If tan = tan.tan1tantan

, prove than

sin 2 = 2sin.2sin12sin2sin

.

Sol.

19. Show that :

(i) cot 7 2º1

or tan 82 2º1

= )12)(23(

or 6432



Sol.

(ii) tan 1422º1

= 2 + 632 .

Sol.

20. If sin x + sin y = a & cos x + cos y = b, show that,

sin (x + y) = 22 baab2

and tan 2

yx = ± 22

22

baba4

.

Sol.

21. Calculate the following without using trigonometrictables :(i) tan 9º – tan 27º – tan 63º + tan 81ºSol.



(ii) cosec 10º – 3 sec 10º

Sol.

(iii) 2 2 sin10º º35sin2º5sinº40cos

2º5sec

Sol.

(iv) cot 70º + 4 cos 70ºSol.

(v) tan 10º – tan 50º + tan 70ºSol.

22. If cos ( ) + cos ( ) + cos ( ) = 23

,

prove thatcos + cos + cos = 0, sin + sin + sin = 0Sol.

23. If sinby

cosax

= a2 – b2, 22 sincosby

cossinax

= 0.

Show that (ax)2/3 + (by)2/3 = (a2 – b2)2/3

Sol.



24. If Pn = cosn + sinn and Qn = cosn – sinn , thenshow that Pn – Pn – 2 = – sin2 cos2 Pn – 4

Qn – Qn–2 = –sin2 cos2 Qn – 4 and hence show thatP4 = 1 – 2 sin2 cos2

Q4 = cos2 – sin2

Sol.

25. If sin ( ) = a & sin ( ) = b (0 < /2)then find the value of cos2 ( ) – 4 ab cos ( )Sol.

26. If A + B + C = , Prove thattanB tanC+tanC tanA+tanA tanB=1+secA . sec B . secC.Sol.

27. If tan2 +2 tan . tan 2 =tan2 +2 tan . tan2 ,then prove that each side is equal to 1 ortan = ± tan .



Sol.

28. For all in 2,0 show that cos (sin ) > sin (cos

)Sol.

29. Find the length of an arc of a circle of radius10 cm which subtends an angle of 45° at the centre.Sol.

30. If the arcs of the same length in two circles subtendangles 75° and 120° at the centre, find the ratio of

their radii.Sol.

31. If tan x = 43

, < x < 2

3, find the value of

sin 2x

and cos 2x

.

Sol.

32. Prove that :(i)sec4 A (1 – sin4 A) – 2 tan2 A = 1Sol.

(ii) sin1

)1(seccot2

= sec2 . sec1sin1

Sol.

33. In a ABC, prove that



sin2A

+sin2B

+sin2C

=1+4sin 4A

sin 4B

sin 4C

Sol.

34. Prove that : cos² + cos² ( + ) 2cos cos cos ( + ) = sin²Sol.

35. Prove that : cos 2 = 2 sin² + 4cos ( + ) sin sin + cos 2( + )Sol.

36. Prove that :(a) tan 20° . tan 40° . tan 60° . tan 80° = 3Sol.

(b) 23

167sin

165sin

163sin

16sin 4444

Sol.

37. If X = sin 127

+ sin 12 + sin

123

, Y = cos 127

+ cos 12 + cos

123

then prove that XY

YX

= 2 tan 2 .

Sol.

38. If m tan ( - 30°) = n tan ( + 120°) , show

that cos 2 = )nm(2nm

.

Sol.



39. If A + B + C = , prove that

CtanBtanAtan

= )A(cot2)A(tan .

Sol.

40. Show that1 sinA cosB 2sinA 2sinB

cosA 1 sinB sin(A B) cos A cosB .

Sol.

41. If + = , prove thatcos² + cos² + cos² = 1 + 2 cos cos cos .Sol.

42. In A, B, C denote the angles of a triangle ABCthen prove that the triangle is right angled if and onlyif sin4A + sin4B + sin4C = 0Sol.

43. If = 72

, prove that

tan . tan 2 + tan 2 . tan 4 + tan 4 . tan = 7.Sol.



44. Let k = 1º, then prove that

ksinkcos

k)1ncos(.nkcos1

2

88

0n.

Sol.

45. If cos A = tan B, cos B = tan C and cos C = tan A,then prove that sin A = sin B = sin C = 2 sin 18º.Sol.

46. Prove that the average of the numbersnsinnº, n = 2, 4, 6,...180, is cot 1º.Sol.

47. Show that elliminating x & y from the equations,sin x + sin y = a ; cos x + cos y = b & tan x + tan y = c

gives 2222 a4)ba(ba8

= c.

Sol.



PREVIOUS YEARSEXERCISE – IV

JEE MAINLEVEL – I1. If cos x + cos y + cos = 0 and sin x + sin y + sin

= 0, then cot 2yx

= [AIEEE-2002]

(A) sin (B) cos (C) cot (D) 2 sin Sol.

2. cos 1°. cos 2°. cos 3°.... cos 179° = [AIEEE-2002](A) 0 (B) 1 (C) 2 (D) 3Sol.

3. Let be such that < – < 3 .

If sin + sin = – 6521

and cos + cos = –6527

, then

the value of cos 2

is- [AIEEE-2004]

(A) – 1303

(B) 1303

(C) 656

(D) 65

6

Sol.

4. Let A and B denote the statementsA : cos + cos + cos = 0 [AIEEE-2009]B : sin + sin + sin = 0

If cos +cos +cos =23

,then :

(A) A is false and B is true (B) both A and B are true(C) both A and B are false (D) A is true and B is false



Sol.

5. Let cos ( + ) = 54

and let sin ( – ) = 513 ,

where 0 4

. Then tan 2 = [AIEEE-2010]

(A) 2516

(B) 5633

(C) 1912

(D) 207

Sol.

6. If A = sin2x + cos4x, then for all real x : [AIEEE-2011]

(A) 43

A 1 (B) 1316

A 1

(C) 1 A 2 (D) 34 A

1316

Sol.

7. In a PQR i f 3 sin P + 4 cos Q = 6 and4 sin Q + 3 cos P = 1, then the angle R is equal to :

[AIEEE-2012]

(A) 4 (B) 34 (C)

56 (D) 6

Sol.



JEE ADVANCEDLEVEL I – I1. (a) Let f( ) = sin (sin + sin 3 ). Then f( ) :

[JEE 2000 (Scr.), 1](A) 0 only when 0 (B) 0 for all real (C) 0 for all real (D) 0 only when 0.Sol.

(b) In any triangle ABC, prove that,

cot2A

+ cot2B

+ cot2C

= cot2A

cot 2B

cot2C

.

[JEE 2000 (Mains), 3]Sol.

2. (a) Find the maximum and minimum values of27cos 2x . 81sin 2x.Sol.

(b) Find the smallest positive values of x & y

satisfying, x–y= 4 , cot x+cot y=2. [REE 2000, 3]

Sol.

3. If + = 2

and + = then tan equals [JEE 2001 (Scr.), 1]

(A) 2(tan + tan ) (B) tan + tan (C) tan + 2tan (D) 2tan + tan Sol.



4. If and are acute angles sin = 1/2, cos = 1/3,then + [JEE 2004 (Scr.)]

(A) 2,

3 (B) 32,

2 (C) 65,

32

(D) ,65

Sol.

5. In an equilateral triangle, 3 coins of radii 1 uniteach are kept so that they touch each other and alsothe sides of the triangle. Area of the triangle is

[JEE 2005 (Scr.)]A

B C

(A) 324 (B) 346

(C)4

3712 (D) 4373

Sol.

6. Let (0, /4) and t1 =(tan )tan , t2 =(tan )cot ,t3 = (cot )tan , t4 = (cot )cot , then [JEE 2006, 3](A) t1 > t2 > t3 > t4 (B) t4 > t3 > t1 > t2(C) t3 > t1 > t2 > t4 (D) t2 > t3 > t1 > t4

Sol.



One or more than one is/are correct : [Q.7 (a) & (b)]

7. (a) If 2

xsin4+

3xcos4

= 51

, then [JEE 2009, 4+4]

(A) tan2x = 32

(B) 8

xsin8 +

27xcos8

= 1251

(C) tan2x = 31

(D) 8

xsin8 +

27xcos8

= 125

2

Sol.

(b) For 0 < < /2, the solution(s) of

6

1m4

)1m(eccos cosec 4m

= 24 is (are)

(A) /4 (B) /6 (C) /12 (D) 5 /12Sol.

8. The maximum value of the expression

22 cos5cossin3sin1

is [JEE 2010]

Sol.



9. The positive integer value of n > 3 satisfying the

equation

n3sin

1

n2sin

1

nsin

1 is [JEE 2011]

Sol.

10. Let , [0,2 ] be such that [JEE 2012]

22cos (1 sin ) sin tan cot cos 12 2

3tan(2 ) 0and 1 sin2

Then cannot satisfy

(A) 02

(B) 4

2 3

(C) 4 33 2

(D) 3 22

Sol.



Answer Ex–I JEE MAIN

1. B 2. A 3. A 4. D 5. B 6. A 7. A 8. D

9. A 10. D 11. B 12. C 13. A 14. B 15. A 16. C

17. C 18. A 19. B 20. C 21. B 22. D 23. A 24. D

25. D 26. B 27. A 28. D 29. B 30. B 31. C 32. D

33. A 34. B 35. A 36. A 37. A 38. A

Answer Ex–II JEE ADVANCED (OBJECTIVE)

SINGLE CORRECTLEVEL – I

1. A 2. C 3. B 4. A 5. B 6. A 7. B 8. C

9. B 10. A 11. B 12. C 13. A 14. C 15. C 16. A

17. A 18. D 19. A 20. B 21. B

MULTIPLE CORRECTLEVEL I – I

1. BD 2. AB 3. ABCD 4. D 5. BD 6. CD 7. AB 8. AB

9. BD 10. AC 11. CD 12. BD 13. BC 14. BC 15. BD 16. AB

Answer Ex–III JEE ADVANCED

Comprehension # 1 1. C 2. A 3. B Comprehension # 2 4. C 5. B 6. D

Matrix Match Type1. (A)–(Q) ; (B)–(R) ; (C)–(Q) ; (D)–(P) 2. (A)–(Q, S) ; (B)–(P) ; (C)–(Q) ; (D)–(P)

Subjective Type 1. a2b2 + 4a2 = 9b2 15. – 41

, 41

16. (i) 2, –1 (ii) 2, 0

21. (i) 4 (ii) 4 (iii) 4 (iv) 3 (v) 3 25. 1 – 2a2 – 2b2 29. 2

5 cm

30. r1 : r2 = 8 : 5 31. sin 103

2x

and cos 101

2x



Answer Ex–IV PREVIOUS YEARS

JEE MAINLEVEL – I

1. C 2. A 3. A 4. B 5. B 6. A 7. D


1. (a) C 2. (a) max. = 35 & min. = 3–5 ; (b) x = 125

; y = 6

3. C 4. B

5. B 6. B 7. (a) A, B ; (b) C, D 8. 2 9. 7 10. A,C,D

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MOTION SAMPLE BOOKLET CLASS XI - Motion IIT- · PDF fileNo part of this work herein should be reproduced or used either ... STOICHIOMETRY-I ... (denoted by I) is an important quantity