Q66 / 69 · JEE Advanced 2011 (Paper 1) · Physics
Steel wire of length $L$ at $40^{\circ} \mathrm{C}$ is suspended from the ceiling and then a mass $m$ is hung from its free end. The wire is cooled down from $40^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$ to regain its original length $L$. The coefficient of linear thermal expansion of the steel is $10^{-5} /{ }^{\circ} \mathrm{C}$, Young's modulus of steel is $10^{11} \mathrm{~N} / \mathrm{m}^2$ and radius of the wire is $1 \mathrm{~mm}$. Assume that $L>$ diameter of the wire. Then, the value of $m$ in $\mathrm{kg}$ is nearly
Correct answer: 3
Solution
$\Delta l_1=\frac{F L}{A Y}=\frac{m g L}{\pi r^2 Y}=$ Increase in length $\Delta l_2=L \alpha \Delta \theta=$ Decrease in length To regain its original length,
$$
\begin{aligned}
& \Delta l_1 & =\Delta l_2 \\
\therefore & \frac{m g L}{\pi r^2 Y} & =L \alpha \Delta \theta \\
\therefore & m & =\left(\frac{r^2 Y \alpha \Delta \theta}{g}\right)
\end{aligned}
$$
Substituting the values, we get
$$
m \simeq 3 \mathrm{~kg}
$$
$\therefore$ Answer is 3 .
Analysis of Question
Question is very simple. In my opinion any student who have gone through the syllabus once can solve this problem easily.
Hence, the correct answer is 3.