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) $6$
Concept. Each string carries the weight of everything below it, then $\Delta l=\dfrac{TL}{AY}$. Take the ratio so that the common quantities cancel.
Step 1. Tensions. String $A$ holds both blocks: $T_A=3Mg$. String $B$ holds only the lower block: $T_B=2Mg$.
Step 2. Ratio (same area):
$$\frac{\Delta l_A}{\Delta l_B}=\frac{T_A}{T_B}\cdot\frac{L_A}{L_B}\cdot\frac{Y_B}{Y_A}=\frac32\times2\times\frac1{0.5}=6$$
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
