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Q. When the arrangement consisting of a horizontal thick copper wire of length $2L$ and radius $2R$ , one end of which is welded to an end of another horizontal thin copper wire of length $L$ and radius $R$ is stretched by applying forces at two ends. Ratio of the elongation in the thick wire to that in the thin wire at equilibrium is

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
$\Delta \ell =\frac{F L}{A Y}$
$\frac{\Delta \ell_1}{\Delta \ell_2}=\frac{L_1}{A_1} \frac{A_2}{L_2}=\frac{2 L}{\pi(2 R)^2} \times \frac{\pi R^2}{L}=\frac{1}{2}$