Answer :

It is given that AB and AC are two frictionless planes inclined to the horizontal at angles θ_{1} and θ_{2} respectively.

The kinetic energy at B will be equal to the kinetic energy at C which will be the potential energy at A due to law of conservation of energy since there are no losses.

Let v_{1} and v_{2} be the final velocities of the stones on the planes AB and AC respectively and m be the mass of each stone.

Then, mgh = (1/2)mv_{1}^{2} = (1/2)mv_{2}^{2}

⇒ v_{1} = v_{2} = v (say)

If a_{1} and a_{2} are the accelerations of the two stones along the planes AB and AC respectively, then

a_{1} = gsinθ_{1}

a_{2} = gsinθ_{2}

As θ_{2}>θ_{1}, a_{2}>a_{1}.

Initial velocities of the two stones are zero.

From Newton’s first equation of motion,

v = u + at

⇒ v = 0 + at

⇒ v = at

⇒ t = v/a

Since v is constant for both the stones,

t ∝ 1/a

Since a_{2}>a_{1}, t_{2}<t_{1}.

So, the stone on plane AC will take less time than the stone on plane AB to reach the bottom.

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