Question

At ${40}^{\circ }\text{C}$, a brass wire of $1\text{mm}$ radius is hung from the ceiling. A small mass, $M$ is hung from the free end of the wire. When the wire is cooled down from ${40}^{\circ }\text{C}\text{to}{20}^{\circ }\text{C}$ it regains its original length of $0.2\text{m}$. The value of $M$ is close to:
(Coefficient of linear expansion and Young's modulus of brass are ${10}^{-5}{/}^{\circ }\text{C}$ and ${10}^{11}{\text{Nm}}^{-2}$ respectively; $g=10{\text{ms}}^{-2})$

• A. $0.5\text{kg}$
• B. $6.3\text{kg}$
• C. $9\text{kg}$
• D. $1.5\text{kg}$

Answer 1

The correct option is B $6.3\text{kg}$

$\text{Given that,}A=\pi r^2=\pi ({10}^{-3}{)}^2$

$A=\pi \times {10}^{-6}{\text{m}}^2$

Here, the elongation due to the load and the contraction due to decrease in temperature are the same,

$\Delta L_{temp}=\Delta L_{load}$

$L\alpha \Delta T=\frac{FL}{AY}$

$\Rightarrow {10}^{-5}\times 20=\frac{Mg}{(\pi \times {10}^{-6})\times {10}^{11}}$

$\Rightarrow Mg=20\pi$

$\Rightarrow M=2\pi =6.28\text{kg}$

Hence, option $\left(D\right)$ is correct.