An iron rod of length 2 m and cross-sectional area of `50 mm^(2)` is stretched by 0.5 mm, when a mass of 250 Kg is hung from its lower end. Young's modulus of iron rod is

Options

A. `19.6 xx 10^(20) Nm^(-2)`
B. `19.6 xx 10^(18) Nm^(-2)`
C. `19.6 xx 10^(10) Nm^(-2)`
D. `19.6 xx 10^(15) Nm^(-2)`

Answer

c

Explanation

Young's modulus of material of the body is
`Y =("Longitudinal stress")/("Longitudinal strain") = (MgL)/(AI)`
Putting the numerical values, we have
`L = 2m A = 50 mm^(2)`
`= 50 xx 10^(-6) m^(2)`
`I = 0.5 mm = 0.5 xx 10^(-3) m`
`M = 250 kg`
`:.Y = (250 xx 9.8 xx 2)/(50 xx 10^(-6) xx 0.5 xx 10^(-3))`
`= 19.6 xx 10^(10) Nm^(-2)`