Official Answer Key & Detailed Solution
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Correct Answer: Option (C) — \frac{1}{\sin Q} \sqrt{\frac{2h}{g}}
Concept & Option Breakdown
Standard Reference VerifiedLet the length of the incline be $l$ and height be $h$. From the right triangle, $\sin Q = \frac{h}{l} \implies l = \frac{h}{\sin Q}$. Acceleration along the frictionless inclined plane is $a = g\sin Q$. Using the equation of motion for a body starting from rest ($u=0$): $l = \frac{1}{2} a t^2 \implies \frac{h}{\sin Q} = \frac{1}{2} (g\sin Q) t^2$ $t^2 = \frac{2h}{g\sin^2 Q} \implies t = \frac{1}{\sin Q}\sqrt{\frac{2h}{g}}$.
As $Q \to 90^\circ$, $\sin Q \to 1$, time reduces to free fall $t = \sqrt{\frac{2h}{g}}$. Only option (c) satisfies this limit.
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