A block ‘A’ takes 2 s to slide down a frictionless incline of 30° and length ‘l’, kept inside a lift going up with uniform velocity ‘v’. If the incline is changed to 45°, the time taken by the block, to slide down the incline, will be approximately:

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2.66 s

0.83 s

0.70 s

1.68 s

answer is C.

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Detailed Solution

$\begin{array}{l} a = g \sin θ \\ ℓ = \frac{1}{2} g \sin 30^{\circ} (2)^{2} . . . (1) \\ ℓ = \frac{1}{2} g \sin 45^{\circ} t^{2} . . . (2) \\ (\frac{1}{2}) (4) = \frac{1}{\sqrt{2}} t^{2} \Rightarrow t = \sqrt{2 \sqrt{2}} ≃ 1 .68 \end{array}$

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