Q 1 :

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

 

Assertion (A): The property of body, by virtue of which it tends to regain its original shape when the external force is removed, is elasticity.

 

Reason (R): The restoring force depends upon the bonded inter atomic and inter molecular force of solid.

 

In the light of the above statements, choose the correct answer from the given below:                                          [2024]

  • (A) is false but (R) is true

     

  • (A) is true but (R) is false

     

  • Both (A) and (R) are true and (R) is the correct explanation of (A)

     

  • Both (A) and (R) are true but (R) is not the correct explanation of (A)

     

(3)

 



Q 2 :

The density and breaking stress of a wire are 6×104kg/m3 and 1.2×108N/m2 respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is 13rd of the value on the surface of Earth. The maximum length of the wire with breaking is _____ m (take, g = 10 m/s2).            [2024]



(600)

T=mg

σ=TA=mgA=(ρA)gA

=σρg=1.2×108×36×104×10=600

 



Q 3 :

A 100 m long wire having cross-sectional area 6.25×10-4 m2 and Young's modulus is 1010 Nm-2 is subjected to a load of 250 N. Then the elongation in the wire will be:     [2023]

  • 6.25×10-3 m

     

  • 4×10-3 m

     

  • 4×10-4 m

     

  • 6.25×10-6 m

     

(2)

Elongation in wire δ=FlAY

δ=250×1006.25×10-4×1010=4×10-3 m



Q 4 :

A force is applied to a steel wire ‘A’, rigidly clamped at one end. As a result, elongation in the wire is 0.2 mm. If the same force is applied to another steel wire ‘B’ of double the length and a diameter 2.4 times that of the wire ‘A’, the elongation in the wire ‘B’ will be (wires having uniform circular cross sections).         [2023]

  • 6.06×10-2 mm

     

  • 2.77×10-2 mm

     

  • 3.0×10-2 mm

     

  • 6.9×10-2 mm

     

(4)

Y=F/AΔll

F=YAlΔl

(AΔll)1=(AΔll)2

Δl2Δl1=A1A2×l2l1

Δl20.2=12.4×2.4×21

Δl2=6.9×10-2 mm



Q 5 :

A wire of length 'L' and radius 'r' is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by 'l'.

Another wire of the same material of length '2L' and radius '2r' is pulled by a force '2f'. Then the increase in its length will be:            [2023]

  • 2l

     

  • l

     

  • 4l

     

  • l2

     

(2)

[IMAGE 76]

fπr2=YL

[IMAGE 77]

2fπ(2r)2=Y'2L

21=2''=



Q 6 :

A steel rod has a radius of 20 mm and a length of 2.0 m. A force of 62.8 kN stretches it along its length. Young's modulus of steel is 2.0×1011N/m2. The longitudinal strain produced in the wire is __________ ×10-5.             [2023]



(25)

Strain=stressY=62.8×103π×(0.02)22×1011

              =62.8×1033.14×4×10-4×2×1011=2.5×10-4=25×10-5



Q 7 :

A metal block of mass m is suspended from a rigid support through a metal wire of diameter 14 mm. The tensile stress developed in the wire under equilibrium state is 7×105Nm-2. The value of mass m is ________ kg. (Take, g=9.8ms-2 and π=227)                [2023]



(11)

Tensile stress, σ=FA=4mgπD2

 m=πD2σ4g

           =227×(14×10-3)2×7×1054×9.8=11 kg



Q 8 :

A steel rod of length 1 m and cross-sectional area 10-4 m2 is heated from 0°C to 200°C without being allowed to extend or bend. The compressive tension produced in the rod is _______ ×104 N. (Given Young's modulus of steel =2×1011 Nm-2, coefficient of linear expansion =10-5 K-1)             [2023]



(4)

Stress=Y×strain

Stress=Y×Δll=Y×lαΔTl=YαΔT

Compressive Tension=Stress×Area of cross section

                                            =YAαΔT=4×104 N



Q 9 :

The length of a wire becomes l1 and l2 when 100 N and 120 N tensions are applied respectively. If 10l2=11l1, the natural length of the wire will be 1xl1. Here the value of x is __________ .           [2023]



(2)

Let the original length be 'l0'

When T1=100 N, Extension=l1-l0

When T2=120 N, Extension=l2-l0

Then 100=K(l1-l0)                               ...(1)

and 120=K(l2-l0)                                  ...(2)

56=l1-l0l2-l0

5l2-5l0=6l1-6l0

l0=6l1-5l2

l0=6l1-5(11l110)

l0=l12    x=2