Question

An iron of length 2 m and cross-section area of 50 mm${}^2$, stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young's modulus of the iron rod is

• A. $19.6\times {10}^{10}N/m^2$
• B. $19.6\times {10}^{15}N/m^2$
• C. $19.6\times {10}^{18}N/m^2$
• D. $19.6\times {10}^{20}N/m^2$

Answer 1

The correct option is A $19.6\times {10}^{10}N/m^2$

Given,

Length, $L=2m$

Area, $A=50m{m}^2=50\times {10}^{-6}{m}^2$

Extension,$\Delta L=0.5\times {10}^{-3}m$

Weight, $F=mg=250\times 9.8=2450N$

Young modulus = Stress/ Strain

$Y=\frac{FL}{A\Delta L}=\frac{2450\times 2}{(50\times {10}^{-6})\times (0.5\times {10}^{-3})}=19.6\times {10}^{10}N{m}^{-2}$

Hence, Young modulus is $19.6\times {10}^{10}N{m}^{-2}$