Q56 / 75 · JEE Main 2026 (04 April Shift 2) · Physics

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 ______.

  • A
    $1$
  • B
    $4$
  • C
    $8$
  • Correct answer
    D
    $6$

Solution

Tension in string $B$ is due to the block of mass $2M$:

$T_B = 2Mg$

Tension in string $A$ is due to both blocks of mass $M$ and $2M$:

$T_A = Mg + 2Mg = 3Mg$

The elongation in a string is given by Hooke's law as $\Delta L = \dfrac{TL}{AY}$.

The ratio of elongations in string $A$ to string $B$ is:

$\dfrac{\Delta L_A}{\Delta L_B} = \left(\dfrac{T_A}{T_B}\right) \left(\dfrac{L_A}{L_B}\right) \left(\dfrac{A_B}{A_A}\right) \left(\dfrac{Y_B}{Y_A}\right)$

Given that $\dfrac{L_A}{L_B} = 2$, $\dfrac{A_B}{A_A} = 1$, and $\dfrac{Y_A}{Y_B} = 0.5 \Rightarrow \dfrac{Y_B}{Y_A} = 2$.

Substituting the values into the ratio equation:

$\dfrac{\Delta L_A}{\Delta L_B} = \left(\dfrac{3Mg}{2Mg}\right) \times (2) \times (1) \times (2)$

$\dfrac{\Delta L_A}{\Delta L_B} = \dfrac{3}{2} \times 4 = 6$

Answer: $6$

Hence, the correct answer is option: 'D'.

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