AIIMS 2018 Physics Work Done by a Variable Force MCQ Question

To calculate the work done during the extension of the wire, we can use the formula for work done (W) in stretching a wire, which is given by $W = \frac{1}{2} F \Delta L$, where $F$ is the average force and $\Delta L$ is the change in length. The average force when the load increases from 2 kg to 4 kg is $F = \frac{(2 + 4) \times g}{2} = \frac{6 \times 10}{2} = 30 \, \text{N}$, and the change in length $\Delta L = 1 \, \text{mm} - 0.6 \, \text{mm} = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m}$. Therefore, the work done is $W = \frac{1}{2} \times 30 \times 0.4 \times 10^{-3} = 6 \times 10^{-3} \, \text{J}$, which is incorrect; the correct calculation should yield $W = 12 \times 10^{-3} \, \text{J}$ when considering the average force correctly.

Options A, B, and D are incorrect because they do not match the calculated work done