Q172 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Physics
The spring is initially in an equilibrium condition. Initially it is stretched by length $l$ and the work done is $U_1$. Again, it is stretched by length $\dfrac{7l}{2}$, when the work done is $U_2$, the value of $U_2$ is
- A$\dfrac{7}{2}U_1$
- Correct answerB$\dfrac{49}{4}U_1$
- C$\dfrac{2}{7}U_1$
- D$\dfrac{4}{49}U_1$
Solution
The work done in stretching a spring by a distance $x$ from its equilibrium position is given by $U = \dfrac{1}{2}kx^2$.
For the first case, the spring is stretched by a length $l$.
$U_1 = \dfrac{1}{2}kl^2$
For the second case, the spring is stretched by a length $\dfrac{7l}{2}$ from its equilibrium position.
$U_2 = \dfrac{1}{2}k\left(\dfrac{7l}{2}\right)^2$
$U_2 = \dfrac{49}{4} \left(\dfrac{1}{2}kl^2\right)$
$U_2 = \dfrac{49}{4}U_1$
Answer: $\dfrac{49}{4}U_1$
For the first case, the spring is stretched by a length $l$.
$U_1 = \dfrac{1}{2}kl^2$
For the second case, the spring is stretched by a length $\dfrac{7l}{2}$ from its equilibrium position.
$U_2 = \dfrac{1}{2}k\left(\dfrac{7l}{2}\right)^2$
$U_2 = \dfrac{49}{4} \left(\dfrac{1}{2}kl^2\right)$
$U_2 = \dfrac{49}{4}U_1$
Answer: $\dfrac{49}{4}U_1$
Hence, the correct answer is option: 'B'.