Question

Calculate the modulus of elasticity for an applied force of 200 N on 0.1 $m^2$ area resulting in change in length of a chain of 2 m by 0.004 m.

• A. ${10}^8$ N per $m^2$
• B. ${10}^4$ N per $m^2$
• C. ${10}^6$ N per $m^2$
• D. ${10}^{-6}$ N per $m^2$

Answer 1

The correct option is C

${10}^6$ N per $m^2$

We know that, Modulus of elasticity = $\frac{F/A}{e/L}$

Given that, F = 200 N, A = 0.1 $m^2$, e = 0.004 m, L = 2 m

So, Modulus of elasticity = $\frac{200/0.1}{0.004/2}$ = $\frac{2000}{0.002}$ = ${10}^6$ N per $m^2$