The diameter of a brass rod is 4 mm and Young's modulus of brass is `9 xx 10^(10) N//m^(2)` '. The force required to stretch by 0.1% of its length is

Options

A. 30 `pi ` N
B. 36 N
C. `144 pi xx 10^(3)` N
D. `36 pi xx 10^(5)` N

Answer

A

Explanation

A `r =2 xx 10^(-3) m , Y = 9xx10^(10) N//m^(2) , l=0.1 % L implies (l)/(L ) =0.001`
As `Y = ( F)/(A) (L)/(l) :. F = YA (l)/(L) = 9xx 10^(10) xx pi (2xx 10^(-3))^(2) xx 0.001 =360 pi N`