Home
Class 11
PHYSICS
Longitudinal stress of 1 kg//mm^(2) is a...

Longitudinal stress of `1 kg//mm^(2)` is applied on a wire. The percentage increase in length is `(Y = 10^(11) N//m^(2))`

A

0.002

B

0.001

C

0.003

D

0.01

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • ELASTICITY

    ERRORLESS |Exercise Bulk Modulus|21 Videos

  • ELASTICITY

    ERRORLESS |Exercise Rigidity Modulus|18 Videos

  • FLUID MECHANICS

    ERRORLESS |Exercise Fluid Mechanics|19 Videos

Similar Questions

Explore conceptually related problems

Longitudinal stress of 1" kg/mm"^(2) is applied on a wire. The percentage increase in length is (Y=10''" N/m"^(2))

A stress of 1kg//mm^(2) is applied on a wire. If the modulus of elasticity of the wire is 10^(10) "dyne"//cm^(2) , then the percentage increase in the length of the wire will be

A stress of 1 kg mm^(-2) is applied to a wire of which Young's modulus is 10^(11) Nm^(-2) and 1.1xx10^(11) Nm^(-2) . Find the percentage increase in length.

A uniform cylindrical wire is subjected to a longitudinal tensile stress of 5 xx 10^(7) N//m^(2) . Young's modulus of the material of the wire is 2 xx 10^(11) N//m^(2) . The volume change in the wire is 0.02% . The factional change in the radius is

A stress of 1 kg wt//mm^(2) is applied to a wire whose Young's modulus is 10^(12)"dyne"//cm^(2) . The percentage increase in its length is

What should be the weight suspended from the end of a steel wire 2 m in length and w mm in diameter to increase the length by 1 mm ? (Y=19xx10^(11)N//m^(2))

A brass wire of radius 0.25 mm and length 2 m is streteched by applying a force of 2 kg wt. Calculate (i) the increase in length of the wire (ii) the strain. [ Y = 9 xx 10^(10) N//m^(2)]

A stell wire of 1m long 1mm^(2) cross section area is hang from rigid end. The length of 1kg is hung from it then change in length will be (given Y = 2 xx 10^(11) N//m^(2) )

The energy stored per unit volume in copper wire, which produces longitudinal strain of 0.1% is (Y = 1.1 xx 10^(11) N//m^(2))

If a copper wire of length 0.9 m and cross section 1mm^(2) is stretched by a load of 1 kg then the extension in the wire will be (Y_(c )=1.2xx10^(11)N//m^(2) and g=10m//s^(2))