MHT CET · Physics · Work Power Energy
When a long spring of \(4 \mathrm{~cm}\) is stretched by \(1 \mathrm{~cm}\), the potential energy stored in the spring is \(U\). If it is stretched by \(4 \mathrm{~cm}\), the potential energy stored in it is
- A \(4 U\)
- B \(16 U\)
- C \(9 U\)
- D \(25 U\)
Answer & Solution
Correct Answer
(B) \(16 U\)
Step-by-step Solution
Detailed explanation
The potential energy \(U\) relates to an extension \(x\) via,
\(\begin{aligned} & U=\frac{1}{2} K x^2 \\ & \therefore \frac{U^{\prime}}{U}=\frac{x^{\prime 2}}{x^2}=\frac{4^2}{1^2}=16\end{aligned}\)
\(\begin{aligned} & U=\frac{1}{2} K x^2 \\ & \therefore \frac{U^{\prime}}{U}=\frac{x^{\prime 2}}{x^2}=\frac{4^2}{1^2}=16\end{aligned}\)
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