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When a mass is suspended from the end of a wire the top end of which is attached to the roof of the lift, the extension is ‘e’ when the lift is stationary. If the lift moves up with a constant acceleration g/2, the extension of the wire would be

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$\frac{2 e}{3}$

$\frac{3 e}{2}$

answer is B.

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Detailed Solution

$Y = \frac{F / A}{Δℓ / ℓ}$

$\frac{{Δℓ}^{'}}{Δℓ} = \frac{F^{'}}{F} = \frac{g + a}{g} = \frac{3}{2}$

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