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 ______.

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: \(6\)