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A long spring, when stretched by a distance $x$, has potential energy $U$. On increasing the stretching to $n x$, the potential energy of the spring will be
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Verified Answer
The correct answer is:
$n^2 U$
Potential energy of the spring,
and
$U^{\prime}=\frac{1}{2} k(n x)^2 \Rightarrow U^{\prime}=n^2 \frac{1}{2} k x^2$
$\Rightarrow \quad U^{\prime}=n^2 U \quad \text{[From Eq. (i)]}$
and
$U^{\prime}=\frac{1}{2} k(n x)^2 \Rightarrow U^{\prime}=n^2 \frac{1}{2} k x^2$
$\Rightarrow \quad U^{\prime}=n^2 U \quad \text{[From Eq. (i)]}$
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