WBJEE 2019 Physics Solved Paper

Given, Decrement in the length =1%
Poisson's ratio for material of the rod, σ=0.2
As we know that,
Volume, v=πr2l,[ where ,l is the length of the rod]
‌

∆V

V

=‌

2∆r

r

+‌

∆l

l

...(i)
since, Poisson's ratio, σ=‌

−∆D/D

∆L/L

=‌

−∆r/r

∆L/L

So, Eq. (i) can be written as,
‌

∆V

V

=‌

∆l

l

(1−2σ)
=−1×[1−2×(0.2)]
=−1×[1−0.4]
=−0.6%
Here, negative sign shows the decrement in the volume.