Question

A uniform steel wire of cross-sectional area $0.20m{m}^2$ is held fixed by clamping its two ends. Find the extra force exerted by each clamp on the wire if the wire is cooled from ${100}^{\circ }C$ to $0^{\circ }C$. Young's modulus of steel = $2.0\times {10}^{11}N/m^2$. Coefficient of linear expansion of steel = $1.2\times {10}^{-5}{/}^{\circ }C$.

• A. 500 N
• B. 480 N
• C. 4.8 N
• D. 48 N

Answer 1

The correct option is D 48 N

When the wire cools down the length of the wire should decrease. But as it is clamped, the length remains the same. This creates a strain in the wire, thereby stressing it.

Let, $l_{100}$ and $l_0$ be the lengths at ${100}^{\circ }C$ and $0^{\circ }C$ respectively. We know -

$L_{100}$ = $L_0(1+100\alpha )$.

At $0^{\circ }C$, the unstrained length would be $l_0$.

But, the strained length is $l_{100}$ (which is constant because of the clamping).

$\therefore$ Strain = $\left[\frac{l_{100}-l_0}{l_0}\right]$ = $100\alpha$. Also, stress = (Young's modulus) $\times$ strain.

$\therefore$ $T$ (the extra tension developed = force at the clamps) = stress $\times$ area,

$\Rightarrow$ $T$ = $2\times {10}^{11}\times 1.2\times {10}^{-5}\times 100\times 6.2\times {10}^{-6}N$

$\Rightarrow$ $T$ = $48N$.