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\[\begin {array} {1 1} (a)\;7.2 \times 10^{-7} \\ (b)\;15.2 \times 10^{-7} \\ (c)\;8.4 \times 10^{-7} \\ (d)\;12.1 \times 10^{-7} \end {array}\]

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$Mass= 50000\;kg$

Area of cross section of 4 cylindrical column

$A=4 \pi (0.6^2-0.3^2)$

$Y= 2.0 \times 10^{11} Nm^{-2}$

$Y=\large\frac{MgL}{A \Delta L}$

$\large\frac{\Delta L}{L}=\frac{Mg}{AY}$

$\qquad= \large\frac{50,000 \times 9.8}{4 \times 3.14 \times (0.6^2 -0.3)^2) \times 2 \times 10^{11}}$

$\qquad= 7.2 \times 10^{-7}$

Hence a is the correct answer.

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