Question

A uniform steel of wire of density $7800kg/m^3$ is 2.5 m long and weighs $15.6\times {10}^{-3}kg$. It exceeds by 1.25 mm when loaded by 8kg. Calculate the value of young's modulus of elasticity of steel.

• A. $1.96\times {10}^{11}N/m^2$
• B. $19.6\times {10}^{11}N/m^2$
• C. $196\times {10}^{11}N/m^2$
• D. None of these

Answer 1

The correct option is C $196\times {10}^{11}N/m^2$

Young's modulus, $Y=\frac{FL}{A\Delta L}$

$A=\frac{Volume}{Area}=\frac{Mass}{Density\times Length}$

$=\frac{15.6\times {10}^{-3}}{7800\times 2.5}=8\times {10}^{-7}{m}^2$

Then,

$Y=\frac{8\times 9.8\times 2.5}{(8\times {10}^{-7})\times (1.25\times {10}^{-3})}=1.96\times {10}^{11}M/m^2$