MCQ
A long spring is stretched by $2\ cm$ and its potential energy is $V.$ If the spring is stretched by $10\ cm,$ its potential energy will be:
- A$\frac{\text{V}}{5}$
- B$\frac{\text{V}}{25}$
- C$5\text{V}$
- ✓$25\text{V}$
✓
Answer
Correct option: D.
$25\text{V}$
$P.E.$ of a stretched spring, $\text{V}=\frac{1}{2}\text{kx}^2,$
where $k$ is the spring constant,
$\therefore\text{V}=\frac{1}{2}\text{k}\times2^2$ or $\text{k}=\frac{\text{V}}{2}$
And Now, $P.E., V' =\frac{1}{2}\text{k}\times10^2$
$=\frac{1}{2}\Big(\frac{\text{V}}{2}\Big)\times100$
$=25\text{V}$
where $k$ is the spring constant,
$\therefore\text{V}=\frac{1}{2}\text{k}\times2^2$ or $\text{k}=\frac{\text{V}}{2}$
And Now, $P.E., V' =\frac{1}{2}\text{k}\times10^2$
$=\frac{1}{2}\Big(\frac{\text{V}}{2}\Big)\times100$
$=25\text{V}$
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