Acc Sample Physics

8/9/2019 Acc Sample Physics

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7 inch

2.5 inch

2.5 inch2.5 inch

UNIT & DIMENSION

VECTORS

AND

BASIC MATHEMATICS


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I N X

Topic Page No.

UNITS AND DIMENSIONS

1. Physical Quantities 12. Set of fundamental quantities 1

3. Derived physical quantities 2

4. Dimensions and dimensional formula 2

5. Principle of homogeneity 3

6. Uses of dimensional analysis 4

7. Limitations of dimensional analysis 8

9. Order of magnitude calculation 9

10. Solved examples 10V E C TO R S

11. Introductrion 13

12. Representation of vectors 13

13. Terminology of vectors 14

14. Angle between vectors 15

15. Lows of addition and subtraction of vectors 16

16. Unit vector 22

17. Co-ordinate system 23

18. Concept of equilibrium 27

19. Position vector 28

UNIT & DIMENSION

VECTORS

AND

BASIC MATHEMATICS


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20. Displacement vector 29

21. Product of vectors 31

22. Solved Examples 38 BASIC MA THEMATICS

23. Trigonometry 44

24. Calculus 4825. Solved Examples 60


26. Exercises 62

27. Answer Key 68V E C TO R S

28. Exercises 69

29. Answer Key 77

BAS IC MATHE MATICS 30. Exercises 78

31. Answer Key 7932. Hints & Solutions 80


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A C C P H U N I T D I M E NS I O N 1


PHYSICAL QUANTITIESAllquantities that can be measuredare called physical quantities. eg. time, length, mass, force, work done, etc. Inphysics we studyabout physical quantities andtheir inter relationships.

MEASUREMENT Measurement is the comparison of a quantitywith a standardof thesame physical quantity.

UNITSAll physical quantities aremeasuredw.r.t. standardmagnitude of thesame physical quantityandthesestandards are called UNITS. eg. second,meter, kilogram, etc.

So the four basicproperties ofunits are:— 1. Theymust be well defined.2. Theyshould beeasilyavailableand reproducible.3. Theyshouldbe invariablee.g. stepasa unitof lengthis not invariable.4. Theyshouldbe acceptedto all.

SET OF FUNDAMENTAL QUANTITIESA setofphysical quantieswhicharecompletelyindependent ofeach otherandallotherphysical quantities

canbeexpressed in terms of thesephysical quantities is called Setof Fundamental Quantities.

Physical Quantity Units(SI) Units(CGS) Notations

Mass kg (kilogram) g M

Length m (meter) cm L

Time s (second) s T

Temperature K (kelvin) °C Current A (ampere) A I or A

Luminous intensity cd (candela) — cdAmount of substance mol — mol

Physical Quantity Definition(SI Unit)

Length (m) The distance travelled by light in vacuum in458,792,299

1

second is called 1 metre.

Mass (kg) Themassofacylindermadeofplatinum-iridiumalloykeptat InternationalBureau of Weights and Measures is de-finedas1 kilogram.

Time (s) The second is the duration of 9,192,631,770 periods of


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A C C P H U N I T D I M EN S I O N


the radiation corresponding to the transition between thetwo hyperfine levelsof the ground stateof thecesium-133 atom.

ElectricCurrent (A) If equal currents are maintained in the twoparallel infinitelylong

wires ofnegligiblecross-section, so that the force betweenthem is 2 × 10 –7 newtonper metre of the wires, thecurrent inany of the wires is called1Ampere.

ThermodynamicTemperature (K) The fraction16.273

1of the thermodynamic temperature

of triple point ofwater is called1 KelvinLuminous Intensity(cd) 1 candela is the luminous intensityof a blackbody of

surface area 2m000,600

1 placed at the temperature of

freezing platinumand at a pressure of101,325 N/m 2, inthedirection perpendicular to its surface.

Amount of substance (mole) The mole isthe amount ofa substance thatcontainsasmanyelementary entitiesas there arenumberofatoms in0.012 kg of carbon-12.

There are two supplementaryunits too:1. Planeangle(radian) angle = arc / radius

r = / r

2. SolidAngle (steradian)

DERIVED PHYSICAL QUANTITIESThe physical quantities those can be expressed in terms of fundamental physical quantities are calledderivedphysical quantities.eg.speed = distance/time.

DIMENSIONS AND DIMENSIONAL FORMULAAll the physical quantitiesof interest can bederived from the base quantities.

DIMENSIONThe power (exponent)ofbase quantitythatenters into the expression ofa physicalquantity, is called thedimension of thequantityin that base.

To make it clear, consider thephysicalquantity"force".Force = mass × acceleration

=time

time/lengthmass

= mass × length × (time) –2So the dimensionsof force are 1 inmass, 1 in length and –2in time. Thus

[Force] = MLT –2


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Similarlyenergyhasdimensional formula givenby[Energy] = ML 2T –2

i.e. energyhas dimensions, 1 in mass,2 in lengthand -2 in time.Suchanexpressionfor a physicalquantity in termsofbasequantitiesis called dimensional formula.

DIMENSIONAL EQUATIONWhenever thedimensionofa physical quantityis equated with itsdimensional formula,weget a dimen-

sionalequation.

PRINCIPLE OF HOMOGENEITYAccordingto this principle, wecanmultiplyphysical quantitieswith sameordifferentdimensional formu-

lae at our convenience, however no such rule applies to addition and substraction, where only like

physical quantites can only be added or substracted. e.g. If P + Q P & Q both represent same physicalquantity.

Illustration :

Calculate the dimensional formula of energy from the equation E =21

mv2. Sol. Dimensionally, E = mass × (velocity) 2.

Since21

is a number and has no dimension.

or, [E] = M ×2

T

L

= ML 2T –2.

Illustration :

Kinetic energy of a particle moving along elliptical trajectory is given by K = s2 where s is thedistance travelled by the particle. Determine dimensions of .

Sol. K = s2

[ ] =)L(

)TLM(2

22

[ ] = M 1 L0 T –2

[ ] = (M T –2

) Illustration :

The position of a particle at time t, is given by the equation, x(t) = 0v

(1 – e – t ), where v 0 is a

constant and > 0. The dimensions of v 0 & are respectively.(A) M 0 L1 T 0 & T –1 (B) M 0 L1 T –1 & T (C*) M 0 L1 T –1 & T –1 (D) M 1 L1 T –1 & LT –2

Sol. [V 0 ] = [x] [ ] & [ ] [t]= M 0 L0T 0

= M 0 L1T –1 [ ] = M 0 L0T –1

Illustration :The distance covered by a particle in time t is going by x = a + bt + ct 2 + dt 3 ; find the dimensionsof a, b, c and d.


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4 A C C P H U N I T D I M EN S I O N


Sol. The equation contains five terms. All of them should have the same dimensions. Since [x] =length, each of the remaining four must have the dimension of length.Thus, [a] = length = L

[bt] = L, or [b] = LT –1

[ct 2 ] = L, or [c] = LT –2

and [dt 3 ] = L or [d] = LT –3

USES OF DIMENSIONAL ANALYSIS

(I) TOCONVERT UNITS OFAPHYSICALQUANTITYFROM ONE SYSTEMOFUNITSTO

ANOTHER :

It is based on the fact that,

Numericalvalue× unit = constant

So on changing unit,numerical value will also gets changed. Ifn 1 and n 2 are the numerical values of agiven physical quantityandu 1 andu 2 be theunits respectivelyin twodifferent systems ofunits, then

n1u1 = n 2u2c

2

1

b

2

1

a

2

112 T

TLL

MM

nn

Illustrat ion

Young's modulus of steel is 19 × 10 10 N/m2. Express it in dyne/cm 2. Here dyne isthe CGS unit of

force. Sol. The unit of Young's modulus is N/m 2.

This suggest that it has dimensions of 2)cetandis(Force

.

Thus, [Y] = 2L]F[

= 22

LMLT

= ML –1T –2.

N/m2 is in SI units,

So, 1 N/m2 = (1 kg)(1m) –1 (1s) –2

and 1 dyne/cm 2 = (1g)(1cm) –1 (1s) –2

so, 22

cm/dyne1m/ N1

=

g1kg1 1

cm1m1

2 –

s1s1

= 1000 ×100

1× 1 = 10

or, 1 N/m 2 = 10 dyne/cm 2

or, 19 × 10 10 N/m2 = 19 × 10 11 dyne/m 2.

Illustration :

The dimensional formula for viscosity of fluids is,

=M 1 L –1T –1

Find how many poise (CGS unit of viscosity) is equal to 1 poiseuille (SI unit of viscosity).


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Sol. = M 1 L –1 T –1

1 CGS units = g cm –1 s –1

1 SI units = kg m –1 s –1

= 1000 g (100 cm) –1 s –1

= 10 g cm –1 s –1

Thus, 1 Poiseuilli = 10 poise

Illustration :

. A calorie is a unit of heat or energy and it equals about 4.2 J, where 1 J = 1 kg m 2 /s2. Suppose weemploy a system of units in which the unit of mass equals kg, the unit of length equals metre,the unit of time is second. Show that a calorie has a magnitude 4.2 –1 –2 2 in terms of the newunits.

Sol. 1 cal = 4.2 kg m 2 s –2

SI New systemn1 = 4.2 n 2 = ? M 1 = 1 kg M 2 = kg L1 = 1 m L2 = metreT 1 = 1 s T 2 = second

Dimensional formula of energy is [ML 2T –2 ]Comparing with [M a LbT c ], we find that a = 1, b = 2, c = –2

Now,c

2

1

b

2

1

a

2

112

T

T

L

L

M

Mnn

= 221

221

2.4ss1

mm1

kgkg1

2.4

(II) TO CHECK THE DIMENSIONALCORRECTNESS OFAGIVEN PHYSICALRELATION:

It is based onprincipleof homogeneity, whichstates that a given physical relationis dimensionallycor-rect if the dimensions of the various terms oneitherside of the relationare the same.

(i) Powersaredimensionless

(ii) sin , e ,,cos , log gives dimensionlessvalue andin above expression is dimensionless

(iii)We canadd orsubtract quantityhavingsame dimensions.

Illustration :

Let us check the dimensional correctness of the relation v = u + at.

Here ‘u’represents the initial velocity, ‘v’represents the final velocity, ‘a’the uniform accelerationand ‘t’ the time.

Dimensional formula of ‘u’ is [M 0 LT –1 ]

Dimensional formula of ‘v’ is [M 0 LT –1 ]

Dimensional formula of ‘at’ is [M 0 LT –2 ][T] = [M 0 LT –1 ]


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Here dimensions of every term in the given physical relation are the same, hence the given physi-cal relation is dimensionally correct.

Illustration :

Let us check the dimensional correctness of the relation

x = ut + 21 at 2

Here ‘u’ represents the initial velocity, ‘a’ the uniform acceleration, 'x' the displacement and ‘t’ the time.

Sol. [x] = L

[ut] = velocity × time =time

length× time = L

2at 21

= [at 2 ] = accelecration × (time) 2

21

is a number hence dimentionless)

=time

velocity×(time) 2 =

timeelength/tim

× (time) 2 = L

Thus, the equation is correct as far as the dimensions are concerned.

(III) TO ESTABLISHA RELATION BETWEEN DIFFERENTPHYSICALQUANTITIES :

Ifweknowthe various factors onwhicha physical quantitydepends, then wecanfind a relationamongdifferent factors byusing principleofhomogeneity.

Illustration :

Let us find an expression for the time period t of a simple pendulum. The time period t maydepend upon (i) mass m of the bob of the pendulum, (ii) length of pendulum, (iii) accelerationdue to gravity g at the place where the pendulum is suspended.

Sol. Let (i) amt (ii) bt (iii) cgt Combining all the three factors, we get

c ba gmt or c ba gKmt

where K is a dimensionless constant of proportionality.

Writing down the dimensions on either side of equation (i), we get

[T] = [M a ][L b ][LT –2 ] c = [M a Lb+c T –2c ]

Comparing dimensions, a = 0, b + c = 0 , – 2c = 1

a = 0, c = – 1/2, b = 1/2

From equation (i) t = Km 0 1/2 g –1/2 or g

K g

K t2/1

Illustration :

When a solid sphere moves through a liquid, the liquid opposes the motion with a force F. Themagnitude of F depends on the coefficient of viscosity of the liquid, the radius r of the sphereand the speedv of the sphere.Assuming that F is proportional to different powers of these quantities, guess a formula for F using the method of dimensions.


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Sol. Suppose the formula is F = k a r b vc

Then, MLT –2 = [ML –1 T –1 ] a Lbc

TL

= M a L –a + b + c T –a – c

Equating the exponents of M, L and T from both sides,a = 1

–a + b + c = 1 –a – c = –2

Solving these, a = 1, b = 1 and c = 1Thus, the formula for F is F = k rv.

Illustration :

If P is the pressure of a gas and is its density, then find the dimension of velocity in terms of P

and .(A) P 1/2 –1/2 (B) P 1/2 1/2 (C) P –1/2 1/2 (D) P –1/2 –1/2

[Sol. v P a b

v = kP a b

[LT –1 ] = [ML –1T –2 ] a [ML –3 ]b (Comparing dimensions)

a =21

, b = – 21

[V] = [P 1/2 –1/2 ]

UNITS AND DIMENSIONS OF SOME PHYSICAL QUANTITIESQuantity SI Unit Dimensional FormulaDensity kg/m 3 M/L 3

Force Newton (N) ML/T 2

Work Joule (J)(=N-m) ML 2/T2

Energy Joule(J) ML 2/T2

Power Watt (W) (=J/s) ML 2/T3

Momentum kg-m/s ML/T

Gravitational constant N-m 2/kg2 L3/MT 2

Angularvelocity radian/s T –1

Angular acceleration radian/s 2 T –2

Angularmomentum kg-m 2/s ML 2/T

Moment of inertia kg-m 2 ML 2

Torque N-m ML 2/T2

Angular frequency radian/s T –1

Frequency Hertz (Hz) T –1

Period s TSurfaceTension N/m M/T 2

Coefficient of viscosity N-s/m2

M/LTWavelength m L

Intensity of wave W/m 2 M/T 3


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Temperature kelvin (K) K

Specific heat capacity J/(kg-K) L 2/T2K

Stefan’s constant W/(m 2 –K 4) M/T 3K 4

Heat J ML 2/T2

Thermal conductivity W/(m-K) ML/T 3K

Current density A/m 2 I/L2

Electrical conductivity 1/ -m(=mho/m) I 2T3/ML 3

Electric dipole moment C-m LIT

Electric field V/m (=N/C) ML/IT 3

Potential (voltage) volt (V) (=J/C) ML 2/IT3

Electric flux V-m ML 3/IT3

Capacitance farad (F) I 2T4/ML 2

Electromotive force volt (V) ML 2/IT3Resistance ohm ( ) ML2/I2T3

Permittivityofspace C 2/N-m 2 (=F/m) I 2T4/ML 3

Permeabilityof space N/A 2 ML/I 2T2Magnetic field Tesla (T) (= Wb/m 2) M/IT 2

Magnetic flux Weber (Wb) ML 2/IT2

Magnetic dipole moment N-m/T IL 2

Inductance Henry (H) ML 2/I2T2

LIMITATIONS OF DIMENSIONAL ANALYSIS(i) Dimension does not depend on the magnitude. Due to this reasonthe equation x = ut + at 2 is alsodimensionallycorrect. Thus, a dimensionallycorrect equation need notbe actuallycorrect.(ii) The numericalconstants having nodimensionsconnot bededucedbythe method of dimensions.(iii) This method is applicable onlyif relation is ofproduct type. It fails in the case of exponential andtrigonometricrelations.

SI Prefixes : The magnitudes of physical quantites varyover a wide range. The mass of an electron is9.1 × 10 –31 kg and that of our earth is about 6 × 10 24 kg. Standard prefixes for certain power of 10.Table shows theseprefixes :

Power of 10 Prefix Symbol

12 tera T

9 giga G

6 mega M

3 kilo k

2 hecto h

1 deka da

–1 deci d


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–2 centi c

–3 milli m

–6 micro µ

–9 nano n

–12 pico p

–15 femto f

ORDER-OF MAGNITUDE CALCULATIONS

If value of phycal quantity P satisfy

0.5×10 x < P 5 × 10x

x is an integer

x is called order of magnitude Illustration :

The diameter of the sun is expressed as 13.9 × 10 9 m. Find the order of magnitude of the diameter ?Sol. Diameter = 13.9 × 10 9 m

Diameter = 1.39 × 10 10 morder of magnitude is 10.

SYMBOLS AND THERE USUAL MEANINGSThescientific groupinGreeceused followingsymbols.

Thetaα Alpha

Beta Gamma Delta Delta Mu

Lambda, Omega

Pi, Phi

epsilon

Psi Roh Nu Eta

Sigma

Tau Kappa chi

Approximatelyequal to


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Solved ExamplesQ.1 Find the dimensional formulaeof follwoing quantities:

(a) The surface tension S,(b) The thermalconductivityk and(c) The coefficient of vescosity .Someequationinvolving these quntitiesare

S =2

hr g Q =

d )t – ( A

k 12

and F = – A12

12

x – xv – v

;

where thesymbols have their usual meanings. ( - density, g - accelerationdue to gravity, r - radius, h -height,A - area, 1& 2 - temperatures, t - time, d - thickness, v 1 & v 2 - velocities, x 1 & x 2 - positions.)

Sol. (a) S =2

hr g

or [S] = [ ] [g]L 2 = 2 L M

· 2T L

·L2 = MT –2.

(b) Q = k d

)t – A( 12

or k = )t – A( Qd

12 .

Here,Q is theheatenergyhaving dimensionML 2T –2, 2 – 1 is temperature,Aisarea,d is thickness andt is time. Thus,

[k] = KT L

T ML2

–22

= MLTT –3 K –1 .

(d) F = –hA12

12

x – xv – v

or MLT –2 = [ ]L2 L L/T

= [ ] T L2

or, [ ] = M L –1T –1.

Q.2 SupposeA = B nCm, whereAhas dimensions LT, B has dimensions L 2T –1, and C has dimensions LT 2.Then the exponents n and m have the values:(A) 2/3; 1/3 (B) 2; 3 (C) 4/5; -1/5 (D*) 1/5; 3/5(E) 1/2; 1/2

Sol. LT = [L 2T –1]n [LT 2]m

LT = L2n+m

T2m–n

2n + m = 1 ....(i) –n + 2m = 1 ....(ii)


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8 ACC PH VECTOR

Solved Example

Q.1 Given that 0CBA , but of three two are equal inmagnitude and the magnitude of third vector is

2 timesthatofeitherof the vectors two having equal magnitude.Then the anglesbetween vectors are

given by-(A) 30°, 60°, 90° (B) 45°, 45°, 90° (C) 45°, 60°, 90° (D) 90°, 135°, 135°

Sol. (D)From polygon law, therevectors having summationzero, shouldfrom a closedpolygon (triangle).

Since the two vectors are havingsame magnitude and the third vector is 2 times that ofeitherof two

havingequal magnitude. i.e. thetriangle shouldberightangledtriangle.

BC

A

Angle betweenAand B is 90°

Angle between B and C is 135°

Angle betweenAand C is 135°

Q.2 If a particlemoves 5min + x-direction. Showthe displacement of the particle-

(A) 5 j (B) 5 i (C) - 5 j (D) 5 k Sol. Magnitudeofvector= 5

Unit vector in + x direction is i

displacement = 5 i

5x

y

0

Hencecorrect answer is (B).

Q.3 A car travels 6 km towards north at an angle of45° to the east then travels distnace of 4 km towardsnorth atanangle of135° to the east. How far is its final position due east and due north? How far is the pointfromthe stratingpoint? What angle does thestraight line joiningits initial and finalpositionmakeswith the east ? What is the total distnace travelled bythe car ?


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ACC PH VECTOR 9

Sol. Net movementalong X - direction

= (6–4) cos 45°

i

= 2× 2

1

= 2 km

45°

6 km4 km

W E

N

S

(X)

(Y)

4 sin 45°( ) j (6 sin 45°) j

4 cos 45°(– )i 6 cos 45° i

+

Netmovement alongY– direction

= (6 + 4) sin 45°

j

= 10 ×2

1= 5 2 km

Netmovement form starting point (Total distance travelled)= 6 + 4 = 10 km

Angle which makes with the east direction

tan = component – Xcomponent – Y

=2

25

= tan –1 (5)

Q.4 Abodyis movingwithuniform speedv ona horizontal circle inanticlockwise direction fromAasshownin figure. What is the changeinvelocityin (a) half revolution(b) first quarter revolution.

W Evv

vB

C

A (x)

(y)

(S)

N

Sol. Changeinvelocityin halfrevolution

v = v C – v A

= v (–

j ) – v ( j )


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6 2 A C C P H U N I T D I M EN S I O N


EXERCISE-1 (Exercise for JEE Main)

[SINGLE CORRECT CHOICE TYPE]

Q.1 In the S.I. system, the unit of temperature is-(A) degree centigrade (B) Kelvin(C) degree Celsius (D) degree Fahrenheit

[1010110687]Q.2 In the S.I. systemthe unit of energy is-

(A) erg (B) calorie (C) joule (D) electron volt[1010111385]

Q.3 The dimensions of the ratio of angular momentum to linear momentum is(A) [M 0LT 0] (B) [MLT –1] (C) [ML 2T –1] (D) [M –1L –1T –1]

[1010111480]Q.4 IfForce = (x/density) + C is dimensionally correct, the dimensionofx are -

(A) MLT –2 (B) MLT –3 (C) ML 2T –3 (D) M 2L –2T –2[1010113500]

Q.5 The dimensional formula for angular momentumis-(A) ML 2T –2 (B) ML 2T –1 (C) MLT –1 (D) M 0L2T –2

[1010113444]Q.6 For 10 (at+3) , the dimension of a is-

(A) M 0 L0 T0 (B) M 0 L0 T1 (C) M 0 L0 T –1 (D) None of these[1010113287]

Q.7 The velocityofa moving particle depends upontime t asv = at +ct

b . Thendimensional formula for b

is -[1010112939]

(A) [M 0 L0 T0] (B) [M 0 L1 T0] (C) [M 0 L1 T –1 ] (D) [M 0 L1 T –2]

Q.8 The pairs havingsamedimensional formula -(A)Angular momentum, torque(B)Torque, work (C) Plank'sconstant, boltzman's constant(D) Gasconstant, pressure

[1010112800]

Q.9 If F = ax + bt 2 + c where F is force, x is distance and t is time. Then what is dimension of 2bt axc

?

(A) [M L 2 T –2] (B) [M L T –2] (C) [M 0 L0 T0] (D) [M L T –1][1010112784]

Q.10 If force, time and velocity are treated as fundamental quantities then dimensional formula of energywill be(A) [FTV] (B) [FT 2V] (C) [FTV 2] (D) [FT 2V2]

[1010112441]Q.11 Which ofthe followingphysicalquantities donot havethe samedimensions

(A) Pressure,Yongsmodulus, stress (B) Electromotive force,voltage,potential(C) Heat,Work, Energy (D)Electricdipole,electricf ield,flux

[1010112085]


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EXERCISE-2 (Miscellaneous Exercise)

Q.1 Taking force, length and time to be the fundamental quantities find the dimensions of

(I) Density (II) Pressure (III) Momentum and (IV) Energy

[1010112351]

Q.2 The frequency of vibration of a string depends on the length L between the nodes, the tension F inthe string and its mass per unit length m. Guess the expression for its frequency from dimensionalanalysis.

[1010111605]

Q.3 The intensity of X-rays decreases exponentially according to the law x0I I e , where 0I is the initialintensityofX-rays and I is theintensityafter it penetratesa distancex throughlead. If be theabsorption

coefficient, thenfindthedimensional formulafor .[1010110571]

Q.4 Findthe dimensionsofPlanck's constant h fromthe equationE = h where E is the energyand is thefrequency.

[1010110859]

Q.5 If the velocityof light (c), gravitational constant (G) and the Planck’s constant (h) are selected as thefundamental units, find the dimensionalformulaeformass, length andtime inthis newsystemofunits.

[1010110105]

Q.6 The distancemoved bya particle intime fromcentreof ring under the influence ofits gravityis given byx=asin t wherea and areconstants. If is found todependon the radiusof the ring(r), itsmass (m)anduniversalgravitation constant (G),findusingdimensional analysisanexpressionfor intermsofr,mand G.

[1010111222]

Q.7 Find the dimensions of (a) the specific heat capacity c,(b) the coefficient of linearexpansion and(c) the gas constant R.Some of the equations involving these quantities are Q = mc (T 2 –T 1) l t = l 0 [1 + (T2 – T1)] andPV= nRT.(Where Q = heat enegry, m= mass , T 1 & T 2 = temperatures, l t = length at temperature t ºC,l

0 = length at temperature 0 ºC, P = pressure, v = volume, n = mole )

[1010111672]

Q.9 A particle is ina uni-directional potential field where the potential energy(U)ofa particledepends onthex-coordinate given byU x = k(1 – cosax) and k and a are constants. Find the physical dimensions of aand k.

[1010113021]

Q.10 Considera planet ofmass (m), revolving round the sun. The timeperiod(T) of revolutionofthe planetdepends uponthe radius of the orbit (r), mass of the sun (M) and the gravitational constant (G). Usingdimensional analysis,verifyKepler’s third lawofplanetarymotion.

[1010112624]


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EXERCISE-3

SECTION-A(IIT JEE Previous Year's Questions)

PASSAGE (1 to 5)Thevan-derWaals equationis

,RT) bV(Va

P 2

where P ispressure, V is molar volumeand T is the temperature of the givensample of gas. R is calledmolargas constant, a and b are called van-der Wall constants.

[1010110877]

Q.1 The dimensional formula for b issameas that for (A) P (B) V (C) PV 2 (D) RT

Q.2 The dimensional formula for a is sameas that for (A) V 2 (B) P (C) PV 2 (D) RT

Q.3 Which of the following does not possess the samedimensional formula as that for RT(A) PV (B) Pb (C) a/V 2 (D)ab/V 2

Q.4 The dimensional formula for ab/RTis(A) ML 5T –2 (B) M 0L3T0 (C) ML –1T –2 (D) M 0L6T0

Q5 The dimensional formulaof RT is sameas that of

(A) energy (B) force (C) specific heat (D) laten heat

Q.6 Match the physical quantities incolumnAwith their dimensional formulaeexpressed incolumnB.

[1010110434]

ColumnA Column B(1)Angular Momentum (a) ML 2T-2

(2) Latent Heat (b) ML 2T-2A -2

(3) Torque (c) ML 2T-1

(4) Capacitance (d) ML 3T-3A -2

(5) Inductance (e) M -1L-2T4A2

(6) Resistivity (f) ML 2T-2A -1(7) Magnetic Flux (g) ML -1T-2

(8) Magnetic EnergyDensity (h) L 2T-2


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EXERCISE

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

Q.6 C Q.7 B Q.8 B Q.9 B Q.10 AQ.11 D Q.12 C Q.13 B Q.14 D Q.15 BQ.16 C Q.17 B Q.18 B Q.19 C Q.20 B

EXERCISE 2

Q.1 (I) FL –4T2 (II) FL –2 (III) FT (IV) FL Q.2k L

Fm

Q.3 L –1

Q.4 M L2

T –1

Q.5

1 1 1 3 1 1 5 1 12 2 2 2 2 2 2 2 2

M c G h L c G h T c G h, , Q.6 3r

GM

Q.7 (a) L 2 T –2 K –1 (b) K –1 (c) ML 2 T –2 K –1 (mol) –1 Q.9 L –1, M L2 T –2 Q.10 T 2 =GMkr 3

Q.11 Equationisdimensionallycorrect Q.13 F µ v 2 Q.14 T = k r

S

3

EXERCISE 3

SECTION AQ.1 B Q.2 C Q.3 C Q.4 D Q5 A

Q.6 [1 (c ), 2 (h), 3 (a), 4 (e), 5 (b), 6 (d), 7 ( f ), 8 (g)]Q.7 (A) P, Q ; (B) R, S ; (C) R, S ; (D) R, S ]

SECTION B

Q.1 B


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EXERCISE-1 (Exercise for JEE Main)

[SINGLE CORRECT CHOICE TYPE]

3 pl

= mvmvr

= [r] = L

4 [x] = [force × density] = MLT –2 3LM

[x] = M 2L –2T –2

5 [l ] = [mvr] = (MLT –1.L) = ML 2T –1

6 [a] = T –1

7 ]v[t b

[b] = LT –1 . T = L

9 2 btcax

= 2

22

MLT

MLTMLT= MLTT –2

10 [E] = ML 2T –2 = (MLT –2) . (L . T –1) (T) = [FTV]

11 Flux = AEso [ ] [E]& dq pso [p] [E]

12 (13) I = Mr 2

[I] = ML 2 moment of inertia[ ] = [r . F] = [L . MLT –2] = ML 2T –2[ ] moment of force

13 f = Cm x . k y

[ f ] = [ m x] [k y] T –1 = M x My T –2y

x + y = 0

y = 21

x = 21

HINTS AND SOLUTION


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7 inch

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2.5 inch2.5 inch

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