Question

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

$\therefore Y=\frac{F/A}{\Delta L/L}$

As $\Delta L=L\alpha \Delta T$

$\therefore Y=\frac{F}{A\alpha \Delta T}orF=YA\alpha \Delta T$

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

$A=\pi {(1\times {10}^{-3})}^2{m}^2,$

$\Delta T=66$

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

$=3.7\times {10}^{2}N$