Q118 / 160 · JIPMER 2004 · Physics
The spring extends by $x$ on loading, then energy stored by the spring is (if $T$ is the tension in spring and $k$ is spring constant)
- Correct answerA$\frac{T^2}{2 k}$
- B$\frac{T^2}{2 k^2}$
- C$\frac{2 k}{T^2}$
- D$\frac{2 T^2}{k}$
Solution
The energy stored in the spring is
$E=\frac{1}{2} k x^2 \quad \ldots$ (i)
and expansion in the spring is
$x=\frac{T}{k} \quad \ldots$ (ii)
From equations (i) and (ii), we get
$E=\frac{1}{2} k \frac{T^2}{k^2}=\frac{1}{2} \frac{T^2}{k}=\frac{T^2}{2 k}$
Hence, the correct answer is option: 'A'.