Question

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 \(\frac{7l}{2}\), when the work done is \(U_{2}\), the value of \(U_{2}\) is

• A. \(\frac{7}{2}U_{1}\)
• B. \(\frac{49}{4}U_{1}\)
• C. \(\frac{2}{7}U_{1}\)
• D. \(\frac{4}{49}U_{1}\)

Hint:

Work done in stretching a spring is proportional to the square of the extension.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The work done in stretching a spring by \(x\) from its natural length is \(U=\dfrac{1}{2}kx^2\), so \(U\propto x^2\).

Step 2: First stretch
For extension \(l\): \(U_1=\dfrac{1}{2}kl^2\).

Step 3: Second stretch
For extension \(\dfrac{7l}{2}\): \(U_2=\dfrac{1}{2}k\left(\dfrac{7l}{2}\right)^2=\dfrac{49}{4}\cdot\dfrac{1}{2}kl^2\).

Step 4: Result
Hence \(U_2=\dfrac{49}{4}U_1\).

Final Answer:
U2 equals 49/4 times U1. \[ \boxed{\text{(B) }\dfrac{49}{4}U_{1}} \]