Q 11-08-025JEE MainJEE Main 2026 (4 Apr, Shift 2)Medium
A metal string $A$ is suspended from a rigid support and its free end is attached to a block of mass $M$. Second block having mass $2M$ is suspended at the bottom of the first block using a string $B$. The area of cross sections of strings $A$ and $B$ are same. The ratio of lengths of strings of $A$ to $B$ is $2$ and the ratio of their Young's moduli $(Y_A/Y_B)$ is $0.5$. The ratio of elongations in $A$ to $B$ is ______.
Answer: (D) $6$
String $A$ holds both blocks: $T_A = 3Mg$. String $B$ holds only the lower block: $T_B = 2Mg$.
$\Delta L = \dfrac{TL}{AY}$, so
$$\frac{\Delta L_A}{\Delta L_B} = \frac{T_A}{T_B}\cdot\frac{L_A}{L_B}\cdot\frac{Y_B}{Y_A} = \frac{3}{2} \times 2 \times 2 = 6$$
Solution by Sreeraj P, M.Sc Physics