Q95 / 125 · BITSAT 2015 · Physics
An elastic string of unstretched length $\mathrm{L}$ and force constant $\mathrm{k}$ is stretched by a small length $\mathrm{x}$. It is further stretched by another small length y. The work done in the second stretching is
- A$1 / 2 \mathrm{Ky}^{2}$
- Correct answerB$1 / 2 \mathrm{Ky}(2 \mathrm{x}+\mathrm{y})$
- C$1 / 2 \mathrm{~K}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)$
- D$1 / 2 \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$
Solution
In the string elastic force is conservative in nature. $\therefore \mathrm{W}=-\Delta \mathrm{U}$
Work done by elastic force of string,
$$
\begin{aligned}
\mathrm{W} &=-\left(\mathrm{U}_{\mathrm{F}}-\mathrm{U}_{\mathrm{i}}\right)=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{F}} \\
\mathrm{W} &=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{\mathrm{k}}{2}(\mathrm{x}+\mathrm{y})^{2} \\
&=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{1}{2} \mathrm{k}\left(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xy}\right) \\
&=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{1}{2} \mathrm{k} \mathrm{y}^{2}-\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}(2 \mathrm{xy}) \\
&=-\mathrm{k} \mathrm{x} \mathrm{y}-\frac{1}{2} \mathrm{k} \mathrm{y}^{2}
\end{aligned}
$$
Therefore, the work done against elastic force
$\mathrm{W}_{\text {external }}=-\mathrm{W}=\frac{\mathrm{K} y}{2}(2 \mathrm{x}+\mathrm{y})$
Hence, the correct answer is option: 'B'.