Q 11-08-042JEE MainJEE Main 2024 (29 Jan, Shift 2)Medium
Two metallic wires $P$ and $Q$ have the same volume and are made up of the same material. If their areas of cross section are in the ratio $4:1$ and force $F_1$ is applied to $P$, an extension of $\Delta l$ is produced. The force which is required to produce the same extension in $Q$ is $F_2$. The value of $\dfrac{F_1}{F_2}$ is ______.
Numerical value type. Enter your answer.
Answer: 16
Equal volumes: $A_1L_1 = A_2L_2 \Rightarrow \dfrac{L_1}{L_2} = \dfrac14$.
$F = \dfrac{YA\,\Delta l}{L}$, so for the same extension
$$\frac{F_1}{F_2} = \frac{A_1}{A_2}\times\frac{L_2}{L_1} = 4\times4 = 16$$
Solution by Sreeraj P, M.Sc Physics