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A weight of 200 kg is suspended by verti...

A weight of 200 kg is suspended by vertical wire of length `600.5 cm`. The area of cross-section of wire is `1mm^(2)`. When the load is removed, the wire contracts by `0.5 cm`. The Young's modulus of the material of wire will be

A

`2.35 xx 10^(12) N//m^(2)`

B

`1.35 xx 10^(10) N//m^(2)`

C

`13.5 xx 10^(11) N//m^(2)`

D

`23.5 xx 10^(9) N//m^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
l= 600.5cm = 6.005m, `A= 1 xx 10^(-6) m^(2)`
m= 200kg, F= mg= `200 xx 10= 2000N`
`Delta l= 0.5cm= 0.5 xx 10^(-2)m`
`Y= (Fl)/(A Delta l) = (2 xx 10^(3) xx 6.005)/(10^(-6) xx 0.5 xx 10^(-2))`
`=2.35 xx 10^(12) N//m^(2)`
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