Work Energy Power

JEE Main level, single correct, medium

37 of 183

Question 37 of 183

A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height hh as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height h=αRh = \alpha R. The value of α\alpha is _______.

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Answer: (B) 44

Let vv be the velocity of the body at the highest point of the circular loop.

At the highest point, the forces acting on the body are its weight mgmg (downwards) and the normal reaction NN (downwards). The net downward force provides the necessary centripetal acceleration:

N+mg=mv2RN + mg = \dfrac{mv^2}{R}

Given that the body exerts a force equal to three times its weight on the track at the highest point, we have N=3mgN = 3mg. Substituting this into the equation:

3mg+mg=mv2R3mg + mg = \dfrac{mv^2}{R}

4mg=mv2R⇒v2=4gR4mg = \dfrac{mv^2}{R} \Rightarrow v^2 = 4gR

Applying the principle of conservation of mechanical energy between the point of release and the highest point of the loop (which is at a height of 2R2R from the bottom):

mgh=mg(2R)+12mv2mgh = mg(2R) + \dfrac{1}{2}mv^2

Substituting v2=4gRv^2 = 4gR into the energy equation:

mgh=2mgR+12m(4gR)mgh = 2mgR + \dfrac{1}{2}m(4gR)

mgh=2mgR+2mgR=4mgRmgh = 2mgR + 2mgR = 4mgR

Therefore, h=4Rh = 4R.

Comparing this with the given relation h=αRh = \alpha R, we get α=4\alpha = 4.

Answer: 44

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