Final Test Xiii(Xyz) Solutions Paper-1

8/2/2019 Final Test Xiii(Xyz) Solutions Paper-1

1/22

MATHEMATICS

XIII (XYZ) Final Test [Paper-I] Page # 1

PART-A

[SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.7 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

Q.1

1

1

xxxxdx6321nx l has the value equal to

(A) 0 (B) ln 6 (C*)3

1ln 6 (D)

3

2ln 6

[Sol. I =

1

1

xxxxdx6321nx l

Using King

I =

1

1

xxxx dx6321nx l

2I =

1

1

xxx

xxx

dx6321

6321nx l =

1

1

xxx

xxxx

dx1236

63216nx l

(Multiplying Nr and Dr by 6x)

=

1

1

2dx6nx l

2I = 2 ln 6 1

0

2 dxx = 3

2ln 6 I = 3

1ln 6. Ans.]

Q.2 Let f : R R be a differentiable function such that f '(x) > f(x) x R and f (x0) = 0, then

(A) f(x) < 0 0xx (B) f(x) 0 0xx (C*) f(x) > 0 0xx (D) f(x) 0 0xx

[Sol.

Let g(x) = ex f(x)

g'(x) = ex (f '(x) f(x)) > 0 Rxxx0

gI

g(x) is increasing on R g(x) > g(x

0) 0xx e

x f(x) >

0

0

x)x(fe 0

f(x) > 0 0xx [given f (x0) = 0]


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MATHEMATICS


Q.3 Let f(x) =

0x,x

xx1

0x,3

0x,x

5xcos)xsinx1(

x

1

2

3

2

If f(x) is continuous at x = 0, then the value of (2 + 2 + 2 + e2), is(A) 8 (B) 10 (C) 12 (D*) 26

[Sol. Since f(x) is continuous at x = 0, so at x = 0, both left and right limits must exist and both must be

equal to 3.

Now,2x

5xcos)xsinx1( =

2

2

x

.....x2

)5(

(Using series expansion of sin x and cos x)

If )x(fLim0x

exist, then + + 5 = 0 and )x(fLim0x

= 3 2 = 3

= 1, = 4.

Since, 3x

xx1Lim

x

1

2

3

0x

0

x

xxLim

2

3

0x

Now, )1(x1Lim x1

0x

= e = 3 e2 = 9

Hence, 2

+ 2

+ 2

+ e2

= (

1)2

+ (

4)2

+ (0)2

+ (3)2

= 1 + 16 + 0 + 9 = 26. Ans.]

Q.4 Number of values of x satisfying simultaneously

sin1x = 2 tan1 x and tan1 )1x(x + cosec1 2xx1 =2

, is

(A) 0 (B) 1 (C*) 2 (D) 3

Sol. Domain is x [1, 1]Given, sin1x = 2tan1x

2

1

x1

xtan =

2

1

x1

x2tan

2x1

x2

= 2

x1

x2

x = 0 or (1 x2)2 = 4 (1 x2) (1 x2) (3 + x2) = 0 x = 1, 0, 1 .......(1)

tan1 )1x(x + cosec1 2xx1 = 2

x(x 1) 0 x x2 0 x(x 1) = 0


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MATHEMATICS


x = 0, 1 .......(2)Now, (1) (2) gives x = 0, 1Hence, number of common solution are 2. Ans.]

Q.5 Tangent drawn to an ellipse 11

y

4

x22

at the point with eccentric angle 30 cuts the director circle

of ellipse at P and Q. The area of triangle OPQ is (where O is origin)

(A*)7194 (B)

7192 (C)

7196 (D)

7198

[Sol. Equation of tangent at M

2

1,3 is 3 x + 2y = 4

Also, distance of PQ from (0, 0) =7

4O

x

y

P

QM

x + y = 52 211

y

4

x 22

2

1,3

Now, PQ = 27

165 = 2

7

19

\Area of (OPQ) =7

192

7

4

2

1 =

7

194. Ans.]

Q.6 The area bounded by the curve y = sin1x + cos1x + tan1

x

1+ tan1 x, y-axis and

the line 2y = (x + 1) is equal to

(A) sq. units (B*)2

sq. units (C)

3

sq. units (D)

4

sq. units

[Sol. f(x) = sin

1x + cos

1x + tan

1

x

1

+ tan

1(x)

x [1, 0) f(x) =22

= 0

y

(1, )

(1,0)(1,0)x

(0, /2)

x (0, 1] f(x) =

22

Required area (shaded region)

= 2 2

122

1

sq. units. Ans.]


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MATHEMATICS


Q.7 The smallest positive root of the equation x1sin = xcos is

(A)2

1

4

3

(B*)

2

1

4

7

(C)

2

1

4

11

(D)

2

1

4

[Sol. sin (1 x) = cos x (Also cos x 0 and sin (1 x) 0)

cos x =

)x1(2

cos

\x = 2n

x1

2(+) Reject

() 2x = 2n2

+ 1

x = n4

+

2

1

n = 1 (out of domain)

n = 2 x =2

1

4

7

B is correct.]

[PARAGRAPH TYPE]


Paragraph for Question no. 8 to 10

Let P1

be a plane containing two lines L1

:1

4z

0

y

3

2x

and L2

:1

2z

0

y

2

7x

P2

be another plane containing a triangle whose vertices are (3, 2, 0), (2, 0, 0) and (0, 5, 0).

P is the point of intersection of L1

= 0 and L2

= 0.

Q.8 If M is the projection of the point P on the line P1

= 0 = P2

then area of triangle OPM is

(where 'O' is the origin)

(A)2

345sq. units (B) 345 sq. units (C*)

2

15sq. units (D) 0 sq. units

Q.9 If acute angle between the lines L1

= 0 and P1

= 0 = P2

is cot1 () then is(A*) 3 (B) 5 (C) 7 (D) 10

Q.10 Image of the line L1

= 0 in the plane P2

is

(A)13z

0y

35x

(B)

13z

0y

35x

(C*)

1z

0y

314x (D)

1z

0y

314x

[Sol. L1

:1

4z

0

y

3

2x

= l

(3l + 2, 0, l + 4)

L2

:1

2z

0

y

2

7x

= r

(2r + 7, 0, + r + 2)


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MATHEMATICS


For point of intersection of L1

= 0 & L2

= 0

3l + 2 = 2r + 7 l + 4 = r + 2

3l + 2r = 5 (i) l + r = 2 (ii)

From (i) & (ii)

l = r = 1

P (5, 0, 3)Equation of plane P

1containing L

1= 0 & L

2= 0 is -

102103

4zy2x

= 0 y = 0

Plane P1

is y = 0 i.e. z-x plane

Equation of plane P2

containing points (3, 2, 0), (2, 0, 0) & (0, 5, 0) is z = 0 i.e. x-y plane.

Line of intersection of x-y plane & z-x plane is x-axis i.e.0

z

0

y

1

x

(i) Projection of the point P on the line P1

= 0 = P2

(x-axis) is M (5, 0, 0) Area ofOPM

35

2

1

z

x

y

M(5, 0, 0)

P

4

O

(5, 0, 3)

z2

=2

15Ans.

(ii) Acute angle between L1

= 0 and P1

= 0 = P2

(x-axis)

cos =001109

0)1(0013

00,1,:axis-xofsDr'

10,3,:0LofsDr' 1

cos =10

3 cot = 3 = cot1 3 = cot1

= 3 Ans.(ii) Image of the point (5, 0, 3) in the plane P

2(x-y plane) is (5, 0, 3)

Point of intersection of the line L1

= 0 with x-y plane is -

(3l + 2, 0 l + 4) l + 4 = 0 l = 4 (14, 0, 0)Now line passing through (5, 0, 3) & (14, 0, 0) is

3

z

0

y

9

14x

1

z

0

y

3

14x

Ans. ]


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MATHEMATICS


[MULTIPLE CORRECT CHOICE TYPE]

Q.11 to Q.15 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct.

Q.11 Locus of the centre of the circle which touches the two circles x2 + y2 + 8x 9 = 0

and x2 + y28x + 7 = 0 externally, is a conic whose

(A) eccentricity is 2 (B*) auxiliary circle is x2 + y2 = 1

(C*) length of latus rectum is 30 (D*) co-ordinates of focus are (4, 0) and (4, 0)

[Sol. As, | cc1 cc2 | = | (r + r1) (r + r2) | = constant

where | r1 r

2| < c

1c

2 locus of C is a hyperbola with foci c1

and c2

i.e., (4, 0) and (4, 0).

Also, 2a = | r1 r

2| = 2 a = 1

Now, e =a2

ae2=

2

8= 4

CC1

(4, 0)(4, 0)

C

So, b2 = 12 (42 1) = 15

Hence, locus of centre of circle is hyperbola, whose equation is15

y

1

x22

= 1.

Now, verify the options. Ans.]

Q.12 In the given figure AO (O being origin) is the median through the

vertex A of the triangle ABC. Also, given the two upward

parabolas P1

and P2.

y

xO

C

B

A

P1

: y = x2 + 2px + q (p, q R) is passing through A and has its vertex at B.P

2: y = ax2 + 2bx + 1 (a, b R) is passing through A and has its vertex at C.

Which of the following is/are correct?

(A*) the sum of all the roots of the equation (x2 + 2px + q) (ax2 + 2bx + 1) = 0 is zero.

(B) ab > 0

(C*) p2 1 = q a

b2

.

(D*) ap + b = 0.Sol. Coordinates of B (p, q p2)

Coordinates of C

a

b1,

a

b2

Since mid point of B and C is origin. Hence2

a

bp

= 0 ap + b = 0 (D)

Similarly q

p

2

+ a

b1

2

= 0

(C)

Since a > 0 and abscissa of point C is positive. Hencea

b> 0 b is negative

Hence ab < 0 (Option B is false).

Sum of the roots of equation (x2 + 2px + q) (ax2 + 2bx + 1) = 0 is

a

b2p2 = 2

a

bp =

a

2(ap + b) = 0 (A) is true.


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MATHEMATICS


Q.13 The value of the constant term in the trinomial

10

2

2 2x

1x

is also equal to

(A) number of different dissimiliar terms in (x1

+ x2

+ ......... + x10

)10.

(B*) (10C0)2 + (10C

1)2 + (10C

2)2 + ........ + (10C

10)2

(C*) coefficient of x10 in (1 x)20.

(D) number of linear arrangements of 20 things of which 10 alike of one kind and rest all are different.

[Sol.514/bin

10

2

2 2x1x

=

20

x1x

[11th, 02-01-2011,P-2, PQ]

Tr + 1

= 20Crx20 r(1)r

rx

1= 20C

rx20 2r(1)r

20 2r = 0 r = 10So, term independent of x = 20C

10. Now verify alternatives. Ans.]

Q.14 Let f(x) be a non-constant differentiable function satisfying f(x) = x2 1

0

2dtx)t(f .

Then which of the following is/are correct ?

(A*) f(x) is monotonic.

(B*) x3cosec

3

4x

)x(fLim

is equal to 1

e .

(C*) Derivative of f(x) w.r.t ln x at x = 1 is equal to 3.

(D) Area bounded by f(x) and co-ordinate axes is3

2.

[Sol. f(x) = x2

1

0

1

0

1

0

22

dtxdt)t(fx2dt)t(f

f(x) = x2 + A + 2Bx x2 = A + 2Bx

A = 1

0

1

0

22 dtBt2Adt)t(f

B = 1

0

1

0

dtBt2Adt)t(f

f(x) = 3x 3

Now verify all the option. Ans.]


8/22


9/22

MATHEMATICS


where w = ei and a, b, c {z : z41 = 0}. [Ans. 55]

[Sol. A =

ww0b11

ca1

| A | = w(1 + b a c) 0 (for non-singular matrices) 1 + b a + cNow total matrices = 4 4 4 = 64 and 1 + b = a + c

2

2

4

1

i

i

1i

1

i

i

1

i

i

i

1

1

i

i

1

1

1

111

cab

Total possibility = 9Total possibility of 1 + b = a + c are 9

Hence 64 9 = 55 Ans.]

Q.314

PQ, a variable chord of the parabola y2 = 4x subtends a right angle at the vertex. The tangents at P and

Q meet at T and the normals at those points meet at N. If the locus of themid point of TN is a parabola,

whose length of latus rectum is L, then find the value of 2L. [Ans. 25]

[Sol. 22

221

1

at

at2

at

at2 = 1

t1t2

= 4

M

T

90

Q(t )2

P(t )1

at t , a(t + t )1 2 1 2

O

N

x

y

(h,

k)

Now T a t1t2, a(t1 + t2)and N a )2tttt(

2122

21

, a t1t2(t

1+ t

2)

Hence 2h = a((t1

+ t2)2 + 2)

& 2k = a(t1

+ t2) (1 t

1t2)

2k = 5a(t1

+ t2)

t1

+ t2

=a5

k2

2h =

2

a5

k22

2h =

2

22

a25

a50k4

y2 =2

25(x 1)

Hence latus rectum =2

25 2L = 25 Ans.]

Q.4 Bowl A has 6 red balls and 4 blue balls. 5 balls are drawn simultaneously and kept in a bowl B whichwas empty. Now a ball is randomly taken out from the bowl B and found to be blue. If the probability

that 2 red and 3 blue balls were transferred from bowl A to B is nm where m and n arerelatively prime, find (m + n). [Ans. 0019]

[Sol. Bowl A

6R

4B

5 ballsPut inBowl B

Let A : One ball drawn from the bowl B found to be blue.


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MATHEMATICS


B1

: 1 R + 4 B from bowl A to B.

B2


B3


B4


B5


P (B1) =

510

44

16

C

CC=

510

C

6; P (B

2) =

510

34

26

C

CC=

510

C

60; P (B

3) =

510

24

36

C

CC=

510 C

120

P (B4) =

510

14

46

C

CC=

510 C

60; P (B

5) =

510

04

56

C

CC=

510C

6

1BAP = 54

; 2BAP = 53

; 3BAP = 52

; 4BAP = 51

; 5BAP = 0

ABP 2 =

5

1i

ii

22

BAPBP

BAPBP

ABP 2 =124836

5

2436

=

1435

23

=

5

4023 = 42

15 =145 n

m

Hence (m + n) = 19. Ans.]

Q.5 Let the complex number z satisfies the inequalities |3z|log|3z|log2

2

1i > 0,

4)1z(amp

i

and | z | 5. If area of common region in which complex number z lies isb

awhere a, b are

relatively prime numbers then find the value of (a + b). [Ans. 641][Sol. log

2| z 3 | > log

2| z 3 | > 0

log2| z 3i | > log

2| z 3 | (z 3, 3i)

put z = x + iy

| z 3i | > | z 3 |

| x + i (y 3) | > | x 3 + iy |

x2 + (y 3)2 > (x 3)2 + y2

x > y

| amp ( z ( 1 + i)) | 4

(3, 0)

(5, 0)

(0, 2)

(1, 1)

A

| z | = 5

O

B

(0, 5)

x

y

x = y

y > x

x > y

4

amp (z (1 + i))

4

Region enclosed by z satisfying all the three given inequalities is sector AOB in which point (3, 0) and the

points on the line segment OB are not included.

Area of the sector AOB = Area of quater circle =4

25=

16

625

a + b = 625 + 16 = 641. Ans.]


11/22

MATHEMATICS


Q.6 For a posit ive constant t, let and are the roots of the equation x2 + t2x 2t = 0. If the minimum

value of dx11

x1

x

2

1

22

isc

ba . (where b and c are coprime), then find the

value of (a + b + c). [Ans. 0020]

[Sol. If and are the roots of x2 + t2x 2t = 0, then we have + = t2 and = 2t

so that 2

2

2

22

)(

2)(

)(

=

t

1

4

t2 and

t2

1

t4

11

)(

122

.

Now, I = dx11

x1

x

2

122

= dxt2

1

t4

1x

t

1

4

tx

2

1

2

22

= 3

t4

3

8

t32

2

.

Now 2

2

t43

8t3 32

6 =4

23 (Using A.M. G.M.)

Hence, I 8

9+ 3

Hence a = 3, b = 9 and c = 8.

a + b + c = 20. Ans.]


12/22

PHYSICS


PART-A



3 Marks (1)

Q.1 A uniform disk, hoop, and solid sphere each of mass m and radius r are rolled up a ramp without

slipping. All starting with the same center of mass velocity vcm

. Rank the order of the maximum height h

that each object reaches.

(A) hdisk> hhoop > hsphere (B*) hhoop > hdisk> hsphere(C) hsphere

> hdisk

> hhoop

(D) hhoop

> hsphere

> hdisk

Q.2 A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the

block and the surface is 0.6. If the acceleration of the truck is 5 m/s2, the frictional force acting on the

block is :

(A) 0.5 N (B*) 5 N (C) 6 N (D) 10 N

Q.3 A long cylindrical wire of radius R carries a current I uniformly distributed throughout its interior.Which

plot best represents the magnitude of the magnetic field as a function of r, the distance from the center of

the wire?

(A*)

O R 2R

Field

r

(B)

O R 2R

Field

r

(C)

O R 2R

Field

r

(D)

O R 2R

Field

r

Q.4 Sudbury's Creighton mine is one of the deepest mine in the world (depth = 2.07 km). In this mine the

conditions as compared to those at the surface are,

(A) lower air pressure, higher acceleration due to gravity

(B*) higher air pressure, lower acceleration due to gravity

(C) higher air pressure, higher acceleration due to gravity

(D) lower air pressure, lower acceleration due to gravity

[Sol. P1A = P

2A + mg P

1> P

2P1 P2 ]

Q.5 A 5 106 coulomb electric point charge is placed midway between two parallel long metal plates

connected to a 9-volt battery. If the electric charge experiences a force of 1.5 104 newtons, what is

the separation of the metal plates?

(A) 2.7 104 m (B) 3.7 103 m (C*) 0.30 m (D) 3.3 m

[Sol. F = qE = 5 106d

9= 1.5 104 ; d = 4

6

105.1

1045

= 0.30 m ]

Q.6 Four very long straight thin wires carry equal electric currents in the +z direction. They intersect the x-y

plane at (x, y) = (a, 0), (0, a), (a, 0) and (0,a). The magnetic force exerted on the wire at position

(a, 0) is along

(A) +y (B)y (C*) +x (D)x


13/22

PHYSICS


[Sol.

y

x+ve x-direction. ]

Q.7 Two beams of light having intensities I and 4I interfere to produce a fringe pattern on the screen. Phase

difference between the beams is2

at point A and at point B. Then the difference between resultant

intensities at A and B is

(A) I (B) 2I (C) 3I (D*) 4I

[Sol. IA

= I + 4I + 2I42 cos 2

= 5I ; I

B= I + 4I + 2I42 cos = I ; IAIB = 4I ]

[PARAGRAPH TYPE]


3 Questions 3 Marks (1)

Paragraph for question nos. 8 to 10

The energy-level scheme for the hypothetical one electron element Bansalium is shown in figure. The

potential energy is taken to be zero for an electron at an infinite distance from the nucleus.

2eVn = 45eVn = 3

10eVn = 2

20eVn = 1

Q.8 A sample of atoms Bansalium are in all the 3 excited state shown above. What is the possible wavelength

that can be emitted by atom in visible range ?(A*) 414 nm (B) 620 nm (C) 124 nm (D) 920 nm

[Sol. n = 4 3 E = 3eV =3

1242= 414 nm ]

Q.9 If a Bansalium atom is in ground state, which of the following photons cannot excite the atom to a higher

state ?

(A) 10 eV (B) 15 eV (C) 18 eV (D*) 12 eV

[Sol. E = exactly the energy difference ]

Q.10 If photons emitted from Bansalium transitions n = 4 n = 2 and from n = 2 n = 1 will eject

photoelectrons from an unknown metal but the photon emitted from the transition n = 3n = 2 will not,what are the limits (maximum and minimum possible values) of the work function of the metal ?

(A) 8 eV < < 10 eV (B) 5 eV < < 10 eV(C*) 5 eV < < 8 eV (D) 5 eV < < 12 eV

[Sol. E4 2 = 8eV

E2 1 = 10 eV

< 8eVE

3 2 = 5eV

> 5eV ]


14/22

PHYSICS




5 Question 4 Marks

Q.11 In the given circuit,

~

R = 5 V =30 voltC

V =20 voltL1A

(A*) current leads the voltage, (emf of source) in phase

(B) voltage (emf of source) leads the current in phase.

(C*) power factor of the circuit is5

1

(D) Applied voltage (emf of source) is 15 V.

[Sol. Current leads the voltagei VR

V0

V VC L

V0

= 2R

2

CLV)VV(

= 25100 = 55 V P.F. =0

R

V

V=

55

5=

5

1]

Q.12 A satellite of mass m orbits a planet of mass M in a circular orbit of radius R. The time required for one

revolution is

(A) independent of M (B) proportional to m

(C*) proportional to R3/2 (D) proportional to R2

[Sol. T2 =GM

42

R3; T =GM

2R3/2 ]

Q.13 A ray of light from a denser medium strikes the plane boundary of a rarer medium at an angle of incidence

I. The angle of refraction is r. If the reflected and the refracted rays are mutually perpendicular, the critical

angle of the medium is

(A) sin1(cot I) (B*) sin1(tan I) (C*) sin1(cot r) (D) sin1(tan r)

[Sol. r = 90I

r

II sin I = 1 sin r = cos I

= cot I

sin QC

= 1 sin 90

sin QC

= 1

sin QC

=1

= tan I ; QC

= sin1 (tan I) = sin1 (cot r) ]

Q.14 A block connected to a spring executes simple harmonic motion. When the block is found at position

(from mean position) x =2

xmax

then

(A) speed= vmax

/ 2 (B*) speed= 3 vmax / 2

(C) acceleration = 3 amax / 2 (D*) acceleration = amax / 2


15/22

PHYSICS


[Sol. v = 2

2

2

AA

=

2

3 v

max; a =2x =

2

A2

=2

amax

]

Q.15 According to the principles of newtonian physics, which of the following depend on the frame of reference?

(A) Real Force (B) Time elapsed in an event (C*) Workdone by force

(D*) Kinetic energy

PART-C

[INTEGER TYPE]

Q.1 to Q.6 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits)

6 Quesion 5 Marks

Q.1 An air bubble of radius r = 3mm moves up in a viscous liquid with a constant velocity v = 0.7 cm/sec.

If the density of the liquid is 1.75 gm/cc. Find the viscosity of liquid (in poise). Neglect the density of air

in comparison to that of the liquid. [Ans. 50 poise]

[Sol. 6rv = B =3

4r3PL g ; = 9

2r2

v

gPL

=9

2

7.0

100075.1)9.0(2

= 50 poise ]

Q.2 A solid hemisphere rests in equilibrium on a rough ground and against a smooth wall. The curved surface

touches the wall and the ground. The angle of inclination of the circular base (flat surface) to the horizontal

is 30, Find minimum coefficient of friction required between ground and hemisphere. Fill the value of

32 in OMR sheet. [Ans. 0006]

[Sol. f = mg 8R3 cos 60

30

60

mg

3R/8

30

N1

N2

60

N2

= f

N1

= mg

f =16

mg3

fN1

16

3 32 = 6 ]


16/22

PHYSICS


Q.3 The figure shows two circuits with a charged capacitor that is to be discharged through a resistor as

shown. The initial charge on capacitors is1

2

q

q= 2. If both switches are closed at t = 0, the charges

become equal at 104ln2 sec. Find the resistance R (in ). [Ans.0006]

q1

q1 5F

5q2

q2 10/3F

R

[Sol. q = q1

11CR

t

e

= q2

22CR

t

e

;2

1

q

q=

2211 CR

1

CR

1t

e ; R = 6 ]

Q.4 A wooden block of mass 0.6 kg of size 10 cm 10 cm 10 cm is floating over an unknown liquid as

shown in the figure

4 cm

How much minimum mass (in gm) should be kept on the wooden block so that it completely submerges

into liquid ? [Ans. 0400 ]

[Sol. PL 6 102 g = 600 g; mg + 600 g = P

L 1000 g ; m = 1000600 = 400 gm

Q.5 A spherical black body of diameter 0.10 m is held at constant temperature. If the power radiated by the

body (emiissive power) is 5.67 102 Js1 , the temperature of the body (in K) is [Ans. 1000 ]

[Sol.dt

dE= ATT4; T4 =

)1067.5()05.0(4

)1067.5(82

2

= 1012 [A = 4r2]

T = 1000 K ]

Q.6 A uniformly wound solenoidal coil of self-inductance 1.8 104 henry and resistance 6 ohms is broken

up into two identical coils. These identical coils are then connected in parallel across a 12 volt battery of

negligible resistance. The time constant for the current in the circuit is x 105 seconds. Fill the value of

x in OMR. [Ans. 0003 ][Sol. L

1= L

2=

2

L= 9 105 H ; R

1= R

2=

2

R= 3

eqL

1=

1L

1+

2L

1 Leq = 2

L1

=2

9 105 H

eqR

1 =1

R1 +

2R1 R

eq=

2R1 = 1.5 =

RL =

2

3

102

9 5= 3 105 sec. ]


17/22

CHEMISTRY

XIII (XYZ) Final test [Paper-I] Page # 1

PART-A



Q.1 Which of the crystal system has none of the interfacial angle equal to 90?

(A) Hexagonal (B*) Trigonal (C) Monoclinic (D) Orthorhombic

Q.2 The minimum number of 90 angles between hybrid orbitals is observed in

(A) sp3d2 (B) d2sp3 (C*) dsp2 (D) sp3d

[Sol. (A) A

Hybridisation : sp3d

Number of angles at 90 = 12

2

x x

x x

x

x

90

90

(B) A

Hybridisation : sp3

d2

Number of angles at 90 = 12

x x

x x

x

x

(C) A

Hybridisation : sp2

dNumber of angles at 90 = 4

x x

x x90 (D) A

x

x

xx

Hybridisation : sp3d

Number of angles at 90=6

x 12090 ]

Q.3

OO||||

HCCCl

)excess(NaOH

(A) COONaCH||

O

(B*)COONa|COONa

+COONa|

OHCH2

(C)

O||

COONaCCl +

O||

OHCHCCl 2 (D)

OH|

COONaCH|Cl

[Sol. Cl CCH HO CCH

Na OC C ONa+ +

Na OCCH OH+

2

Cl CCH

O O

OO

O O

O

Omorereactive

OH

O

OH

NaOH Cannizaro

+

]


18/22

CHEMISTRY


Q.4 Which of the following ideal aqueous solution will show maximum boiling point?

(A) 0.5 M NaCl showing 50% dissociation.

(B*) 0.3 M K2Fe[Fe(CN)

6]

(C) 1 M Glucose solution

(D) 1 mole of AgCl is mixed with 0.5 lof H2O

Q.5equivalent1

/SOHHNO 423

(A)

ONaCH3

(B).

Product (B) is

(A) (B) (C*) (D)

[Sol.

423 SOHHNO

ArN

S,

ONaCH3

]

Q.6 Statement-1 : 7N15 is a emitter.

Statement-2 : emission decreases n/p ratio and increases stability if n/p is greater than 1 for lower

atomic number.

(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

(C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

Q.7 Incorrectstatement about given carbohydrate is

OH OH

H HH

OH H

CH OH2

O

OOH

H OH

H OH

HO

H

CH OH2

(A) Above compound is a reducing sugar. (B) Above compound undergo mutarotation.

(C*) Above compound is a non-reducing sugar.(D) Above compound has a glycosidic linkage.

[Sol.

OH OH

H HH

OH H

CHOH2

O

OOH

H OH

H OH

HO

H

CHOH2

Glycosidiclinkage

Due to this it will be reducing sugar the ring can open into aldehydic

form and it can undergo mutarotation.]


19/22

CHEMISTRY


[PARAGRAPH TYPE]

Q.8 to Q.10 has four choices (A), (B), (C), (D) out ofwhich ONLY ONE is correct.

Paragraph for question nos. 8 to 10

Covalent compounds undergo hydrolysis either via SN1 (unimolecular nucleophilic substitution) or via

SN2 (Bimolecular nucleophilic substitution) mechanism. For S

N2 mechanism, within the molecule atom

should have at least one vaccant orbital, if it is not there then hydrolysis takes place via SN1 mechanism

(dissociative step) in drastic conditions.

Q.8 What are the hydrolysis product(s) of BeCl2 ?(I) [Be(OH)

4]2 (II) Be(OH)

2(III) HCl (IV) BeH

2

(A) I, II, III (B*) II, III (C) I, III (D) II, III and IV

Sol.

ClBeCl ClBeOHH O2sp

+slow HCl

HCl

FastBe

B

Cl

Cl

Cl

OH

OH2

OH2

+

+

H O2

Be(OH)2[Be(OH) ]2 n or

white ppt

OH[Be(OH) ]42

]

Beryllate ion is formed in basic medium.

Q.9 Least probable product formed on hydrolysis of BCl3, would be

(A*) [B(OH)4] (B) HCl (C) B(OH)

3(D) None of these

[Sol. BCl3

+ 3H2O H

3BO

3+ 3HCl

H3BO

3+ 2H

2O H

3O+ +

4)OH(B

During the hydrolysis of BCl3the medium becomes acidic due to formation of HCl hence in the presence

HCl feasibility of second reaction i.e. formation of [Be(OH)4]is least.]

Q.10 CCl4

is inert towards hydrolysis under ordinary conditions because

(I) No vaccant orbital on attacking site of C-atom is present.

(II) H2O molecule can not approach carbon due to steric crowding.

(III) Bond dissociation energy of CCl bond is highest.

(IV) CCl4

is non-polar and does not react with polar H2O molecules.

Select correct code:

(A) I, II and IV (B) I and IV (C*) I and II (D) I, II, III and IV

[Sol. (i) Under ordinary condition, H2O molecule can not approach to the vaccant antibonding M.O. of

ClCl bond (yet having low energy) due to steric crowding over central carbon atom.

(ii) CCl bond dissociation energy is not very high as its magnitude is less than that of CF. ]


20/22

CHEMISTRY




Q.11 For 0.1 M ClNHHC 356 , 0.1 M KCl, 0.1 M Glucose and 0.1 M Na2C2O4 aqueous solution:

(A*) The solution with highest boiling point is 0.1 M Na2C

2O

4.

(B) The solution with highest freezing point is 0.1 M Na2C

2O

4.

(C*) 0.1 M C6H

5NH

3Cl and 0.1 M NaCl are isotonic at same temperature (assume complete dissociation).

(D*) 0.1 M glucose solution has highest vapour pressure at given temperature.

Q.12 Which of the following specie(s) is / are having only one corner shared per tetrahedron?

(A*) 672OSi (B*)

272OCr (C*)

272OS (D)

693OSi

[Sol. (A) O O O

O O

O O

Si Si(B) O O O

O O

O O

Cr Cr

(C) O O O

O O

O O

S S(D) or

Si

SiSi

O

O

OO

O

OO

O O

]

Q.13 H

H

Me

Me

Incorrect statement about this compound is:

(A) It shows geometrical isomerism. (B*) It posses centre of symmetry.

(C) It posses plane of symmetry. (D) It shows optical isomerism.

[Sol. H

H

Me

MeIt can show GI It do not posses COS

It posses POS It can show OI (it is meso isomer) ]

Q.14 Which of the following statement(s) is/are correct with respect to behaviour of real gas?

(A*) For every Vander Waal gas at critical condition, attractive forces will be dominant.

(B*) The liquid and gaseous state can be distinguished only if the temperature is below the gas's critical

temperature.(C*) At very high pressures, real gases occupy greater volume as compared to ideal gas having same

moles at same temperature and exerting same pressure.

(D*) For a real gas, the 'y' intercept ofT

PVmVs P curve where 'P' represents pressure (in atm) , V

m

represents molar volume and 'T' represents temperature in kelvin is equal to 0.0821 atm-litre/mol-

Kelvin.


21/22

CHEMISTRY


Q.15

H (A) 4NaBH (B)

H(C) (major).

Product (C) is

(A) (B) (C*) (D)

[Sol.OH

H

2

H

H

OH

H

2

]

PART-C[INTEGER TYPE]

Q.1 to Q.6 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits)

Q.1 1 l H2O

2solution having 1 M concentration was left open causing first order decay of H

2O

2with rate

constant equal to 6.93 102 sec1. If after 20 sec, the solution is reacted with excess of I(aq) forming

I2(g) with water, then calculate the amount of energy released in kJ. [Ans. 50]

[Assume no change in volume of H2O

2solution]

Given Data : mole/kJ200)aq(OHH 22of

; mole/kJ60)aq(IHof

mole/kJ60)g(IH 2of ; kJ290),OH(H 2of

Q.2 How many geometrical isomer(s) is / are formed by the complex of the type [Ma2bcde] where, M is the

central metal and a,b,c,d ande are unidentate ligands?

If your answer is 2, write the answer as 0002. [Ans. 9]

Q.3 (a) How many alkenes on catalytic reduction will produce the alkane, ?

(b)

OH

32OAl Total number of alkenes produced.

Write answer of part (a) and (b) in the same order and present the two digit number asanswer in OMR sheet. For example : If the answers (a) is 9 and (b) is 5, then fill as 0905 in

OMR sheet. [Ans. 0503]

Q.4 When a graph is plotted betweenm

xlog and log P, it is a straight line with an angle 45 and an intercept

of 0.3010 on 'y' axis. If initial pressure is 0.3 atm, then what will be the amount of gas (in gm) adsorbed

on 10 gm of absorbent . [Ans. 0006]


22/22

CHEMISTRY

Q.5 How many reactions among them will produce homodiatomic molecules ?

(i) NH4Cl

(ii) (NH4)

2Cr

2O

7

(iii) PtCl4

(iv) NaN3

(v) CaCO3

(vi) Microcosmic salt

(vii) Al2O

3

(viii) HgO

(ix) Pb(NO3)

2 [Ans.0005]

Q.63

HN

CH(R)CO2H has its Ka

1= 107. The isoelectric point of the -amino acid occur at

pH = 8.2. If pKa2

of the conjugate acid of-amino acid is equal to X, then what will be the value of

10X?[Hint: If pKa2

is 2.4 write 0024 in OMR sheet] [Ans. 0094]

[Sol. Isoelectric pH of a aminoacid =2

21

aa pKpK

8.2 =2

7 2apK 2

apK = 16.47 = 9.4 ]

Documents

Final Test Xiii(Xyz) Solutions Paper-1