Question

A steel rod $25cm$ long has a cross-sectional area of $0.8c{m}^2$. Force required to stretch this rod by the same amount as the expansion produced by heating it through ${10}^{\circ }C$ is
(Coefficient of linear expansion of steel is ${10}^{-5}{/}^{\circ }C$ and Young's modulus of steel is $2\times {10}^{10}N/m^2)$.

• A. $160N$
• B. $360N$
• C. $106N$
• D. $260N$

Answer 1

Answer: (A)

$160N$

Given : $l=25cm=0.25m,A=0.8c{m}^2=0.8\times {10}^{-4}{m}^2,\alpha ={10}^{-5}{/}^{o}C,Y=2\times {10}^{10}N,\Delta t={10}^{o}C$ .

The coefficient of thermal linear expansion is given by:

$\alpha =\frac{\Delta l}{l\times \Delta t}$

$\frac{\Delta l}{l}=\alpha \times \Delta t$

$\frac{\Delta l}{l}={10}^{-5}\times 10={10}^{-4}$

Now, Young's modulus of steel rod is given by:

$Y=\frac{F\times l}{A\times \Delta l}$

whee $F=$ force required to stretch the rod of cross section area $A$ length $l$ by $\Delta l$

$F=\frac{YA\Delta l}{l}$

$F=2\times {10}^{10}\times 0.8\times {10}^{-4}\times {10}^{-4}=160N$