Free Objective Test 02 Practice Test - 11th and 12th 

Question 1

The Poisson’s ratio of the material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross sectional area by 4%. The percentage increase in its length is

A. 1%
B. 2%
C. 2.5%
D. 4%

SOLUTION

Solution : D

Given Poisson's ration= 0.5
It shows that the density of material is constant
Therefore the change in volume of the wire is zero
Thus
V=A×l = constant
log V=log A+log |
therefore  VV=0= AA+ ll
 ll= AA
or  % increase in length= ll×100
                                         =-(-4)
                                         =4%

Question 2

Two wires are made of the same material and have the same volume. However, wire 1 has cross sectional area 3A. If length of wire 1 increased by x on applying force F, how much force is needed to stretch wire 2 by the same amount?

A. 4F
B. 6F
C. 9F
D. F

SOLUTION

Solution : C

Y=FlA l
so,F=YA2 lAl=YA2  lV
where,Al=V= Volume of wire
hence F  A2
FF=(3A)2A2=9
or F'=9F

Question 3

When a rod is heated but prevented from expanding, the stress developed is independent of

A. material of the rod
B. rise in temperature
C. length of rod
D. none of these

SOLUTION

Solution : C

Lt=L0(1+α  θ)
 L=LtL0=L0 α  θ
If the same rod of lenghth L0 is subjected to stress along its length, then extension in length can be calculated by Hooke's law
Y=stressstrain=stress LLn
=L0× stress L
  Therefore  L=L0× stressY
If the rod is prevented from expanding, we have 
Lo α θ=Lo× stressY
 Therefore stress=Y α  θ= (independent of L0)

Question 4

The bulk modulus for an incompressible liquid is

A. zero
B. unity
C. infinity
D. between 0 and 1

SOLUTION

Solution : C

The bulk modulus is
K=pv v
If liquid is incompressible so  V=0
Hence, K=pv0
=infinity

Question 5

A wire is stretched under a force. If the wire suddenly snaps, the temperature of the wire

A. remains the same 
B. decreases
C. increases
D. first decreases then increases

SOLUTION

Solution : C

The work done on the wire to produce a strain in it will be stored as energy which is converted into heat, when wire snaps suddenly

Due to this the temperature increases

Question 6

A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the students measures the extension in the length of the wire to be 0.8 mm with an uncertainty of  +/- 0.05 mm at a load of exactly 1kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of +/-0.01 mm.The Young’s modulus obtained from the reading is (g= 9.8 ms2)

A. (2.0+/0.3)× 1011 Nm2
B. (2.0+/0.2) × 1011 Nm2
C. (2.0+/0.1)× 1011 Nm2
D. (2.0+/0.05)× 1011 Nm2

SOLUTION

Solution : B

Y=4Flπ D2 l
=4 ×(1X9.8)X2(227)(0.4×103)2X(0.8X103)
=2×1011N m2
Fractional error in y is
 yy=+/(2 DD+ ll)
or  y =+/(2 X0.010.4+0.050.8)×2×1011
=+/0.2 × 1011Nm2
Therefore y=(2× 1011+/0.2 × 1011)N m2

Question 7

A wire 3 m in length and 1 mm in diameter at 303K is kept at a low temperature of 103K and is stretched by hanging a weight of 10 kg at one end. The change in length of the wire is (Y = 2× 1011 Nm2, g = 10 m s2 and = 1.2× 105 K1)

A. 5.2 mm
B. 2.5 mm
C. 52 mm
D. 25 mm

SOLUTION

Solution : A

The contraction in the length of the wire due to change in the temperature
=α LΔT
=1.2×105× 3× (103303)
=7.2× 103 m
The expansion in the length of wire due to stretching force
=FLAY=(10 X 10)X 3(0.75×106)(2× 1011)
=2× 103 m
Resultant change in length
=7.2×103+2× 103
=5.2×103
=-5.2 mm
Negative sign shows contraction

Question 8

Two rods of different materials having coefficients of linear expansion α1  and α2 and Young’s modulus
Y1 and Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If
α1 : α2 = 2:3, the thermal stress developed in the two rods are equal provided Y1:Y2 equal to

A. 2 : 3
B. 4 : 9
C. 1 : 2
D. 3 : 2

SOLUTION

Solution : D

From the formula for thermal stress
(F/A)1=α1y1 T
(F/A)2=α2y2 T
Since,temperature  T is same
(FA)1(FA)2=α1y1α2y2
For thermal stress to be equal
α1y1=α2y2
or y1y2=α2α1=32 since α1α2=2:3)

Question 9

The pressure of a medium is changed from 1.01× 105 Pa to 1.165× 105 Pa and change in volume is 10% keeping temperature constant. The bulk modulus of the medium is

A. 204.8× 105Pa
B. 102.4× 105 Pa
C. 51.2× 105 Pa
D. 1.55× 105 Pa

SOLUTION

Solution : D

Here
 P=(1.165×1051.01×105)
=0.155× 105 Pa
× V/V=10/100=1/10
Bulk modulus k= p v/v
=0.155×105110
=1.55×105Pa

Question 10

If S is stress and Y is Young’s modulus of material of a wire, the energy stored in the wire per unit volume is

A. S2y
B. 2yS2
C. S22y
D. 2S2Y

SOLUTION

Solution : C

Energy stored per unit volume
U=12stress× strain

=12stress× strainy

=12S× Sy

=S22y

Question 11

When a certain weight is suspended from a long uniform wire, its length increases by one cm. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one then the increase in length will be -

A. 0.5 cm
B. 2 cm
C. 4 cm
D. 8 cm

SOLUTION

Solution : C

c): I=FLAY l1r2   (F,L and Y are constant)
l2l1=(r1r2)2=(2)2=4
   l2=4l1=4cm

Question 12

The relationship between Young's modulus Y, Bulk modulus K and modulus of rigidity  η is

A. y=9η kη+3K
B. y=9YKY+3K
C. y=9η k3+K
D. y=3η k9η+K

SOLUTION

Solution : A

Solution: a): Y =3K(12σ)and y =2η(1+σ)

Eliminating  σ We get y=9η kη+3K

Question 13

The Young's modulus of a rubber string 8 cm long and density  1.5 kgm3 is 5 ×108Nm2. It is suspended from the ceiling in a room. The increase in length due to its own weight will be

A. 9.6× 105 m
B. 9.6× 1011m
C. 9.6× 103m
D. 9.6 m

SOLUTION

Solution : B

b):l = L2dg2Y = (8×102)2×1.5×9.82×5×108 = 9.6×1011m

Question 14

A steel ring of radius r and cross-section area  'A' is fitted on to a wooden disc of radius R(R> r). If Young's modulus be E, then the force with which the steel ring is expanded is

A. AERr
B. AE(Rrr)
C. EA(RrA)
D. ErAR

SOLUTION

Solution : C

c): Initial length (circumference) of the ring =2π r

Final length (circumference) of the ring =2π R

Change in length=2π R2π r.

strain=change in lengthoriginal length=2π(Rr)2π r=Rrr

Now young's modulus  E=F/At/L=F/A(Rr)/r

F=AE(Rrr)

Question 15

A wire extends by 1 mm when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be

A. 8
B.
C. 2
D. 1

SOLUTION

Solution : A

a) I=FLπ r2r lFr2 (Y and L are constant)
l2l1=F2F1×(r1r2)2=2×(2)2=8
                                             l2=8l1=8×1=8

Question 16

The coefficient of linear expansion of brass and steel are
 α1 and α2. If we take a brass rod of length l1 and steel rod of length  l2 at 0C, their difference in length  (l2l1)will remain the same at a temperature if

A. α1l2=α2l1
B. α1l22=α2l21
C. α21l1=α22l2
D. α1l1=α2l2

SOLUTION

Solution : D

d) L2=l2(l+α2θ) and L1=l1(l+α1θ)
 (L2L1)=(l2l1)+θ(l2α2l1α1)
Now (L2L1)=(l2l1) so, l2α2l1α1=0

Question 17

The length of an elastic string is 'a' metre when the longitudinal tension is 4 N and 'b' metre when the longitudinal tension is 5 N. The length of the string in metre when the longitudinal tension is 9 N is-

A. a - b
B. 5b - 4a
C. 2b14a
D. 4a - 3b

SOLUTION

Solution : B

b) Let L is the original length of the wire and K is force constant of wire.

Final length=initial length+elongation

L' = L + FK
For first condition a = L + 4K.....(i)
For second condition b = L + 5K.....(ii)
By solving (i) and (ii) equation we get
L = 5a - 4b and K = 1ba
Now when the longitudinal tension is 9N, length of the string =L + 9K = 5a - 4b + 9(b-a) = 5b-4a.

Question 18

According to Hook's law of elasticity, if stress is increased, the ratio of stress to strain 

A. Increases
B. Decreases
C. Becomes zero
D. Remains constant

SOLUTION

Solution : D

(d)Y = StressStrain=Constant
It depends only on nature of material

Question 19

A pan with set of weights is attached with a light spring. When disturbed, the mass-spring system oscillates with a time period of 0.6s. When some additional weights are added then time period is 0.7s. The extension caused by the additional weights is approximately

A. 1.38 cm 
B. 3.5 cm
C. 1.75 cm
D. 2.45 cm

SOLUTION

Solution : B

b)2πmk=0.6....(i) and 2πm+mk=0.7   ...(ii)

Dividing (ii) by (i) we get (76)2=m+mm=4936
m+mm1=49361mm=1336 m13m36

Alsokm=4π2(0.6)2
Desired extension = mgk = 1336×mgk
1336×10×0.364π2 3.5 cm

Question 20

A rubber cord catapult has cross-sectional area 25mm2 and initial length of rubber cord is 10 cm It is stretched to 5cm and then released to project a missile of mass  5gm.Taking  Yrubber 5× 108 N/m2, velocity of projected missile is

A. 20 ms1
B. 100 ms1
C. 250 ms1
D. 200 ms1

SOLUTION

Solution : C

c): Potential energy stored in the rubber cord catapult will be converted into kinetic energy of mass.
12mv2=12YAl2L v=YAl2mL
5×108×25×106×(5×102)25×103×10×102=250 m/s