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Steel wire of length 'L' at 40^@C is sus...

Steel wire of length 'L' at `40^@C` is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from `40^@C to 30^@C` to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is `10^-5//^@C`, Young's modulus of steel is `10^11 N//m^2` and radius of the wire is 1mm. Assume that `L gt gt` diameter of the wire. Then the value of 'm' in kg is nearly

A

2 kg

B

3 kg

C

4 kg

D

5 kg

Text Solution

Verified by Experts

The correct Answer is:
B
`Y = (MgL)/(pier^(2)(DeltaL)) " " DeltaL = (MgL)/(pier^(2)Y)` ….(1)
Similarly if `alpha` is the coefficient of linear expansion,
then `alpha = (DeltaL)/(LDeltaT)` ……(2)
`:. From (1) and (2), (MgL)/(pier^(2)Y) = alphaLDeltaT`
`:. M = (alphaDeltaT (pier^(2)Y))/(g)`
`:. M = (10^(-5) xx (40 - 30) xx 3.14 xx (10^(-3))^(2) xx 10^(11))/(10)`
`= 3.14 div 3 kg`
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