Question

A brass wire 1.8 m long at ${27}^{o}C$ is held taut with negligible tension between two rigid supports Diameter of the wire is 2mm, its coefficient of linear expansion,${\alpha }_{Brass}=2\times {10}^{-5}{}^o{C}^{-1}$ and its Young's modulus, $Y_{Brass}=9\times {10}^{10}N{m}^{-2}$,If the wire is cooled to temperature $-{39}^{o}C$, tension developed in the wire is :

• A. $2.7\times {10}^{2}N$
• B. $3.7\times {10}^{2}N$
• C. $4.7\times {10}^{2}N$
• D. $5.7\times {10}^{2}N$

Answer 1

The correct option is B $3.7\times {10}^{2}N$

Let $F$ be tension developed in the wire.
$\therefore Y=\frac{F/A}{△L/L}$

As $△L=L\alpha △T$
$\therefore Y=\frac{F}{A\alpha △T}$ or $F=YA\alpha △T$

Here , $\alpha =2.0\times {10}^{-5}{}^o{C}^{-1},Y=9\times {10}^{10}N{m}^{-2}$

$A=\pi (1\times {10}^{-3}{)}^2{M}^2$
$△T={66}^{o}C$

$\therefore F=9\times {10}^{10}\times \pi (1\times {10}^{-3}{)}^2\times 2.0\times {10}^{-5}\times 66$
$\Rightarrow 3.7\times {10}^{2}N$