Q.32. A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body

Q: What is the correct answer for JEE Main Physics Q32?

The correct answer is B. Sol. Let the mass of the body be $m$ and the radius of the circular loop be $R$. At the highest point of the circular loop, the forces acting on the body are gravity $mg$ downwards and the normal force $N$ downwards. The net force provides the centripetal force: $N + mg = \frac{mv^2}{R}$, where $v$ is the speed at the highest point. Given that the body exerts a force of three times its weight on t

Q: Q.32. A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height as shown. If the body exerts a force of three times its weight on the…

The correct answer is B. Sol. Let the mass of the body be $m$ and the radius of the circular loop be $R$. At the highest point of the circular loop, the forces acting on the body are gravity $mg$ downwards and the normal force $N$ downwards. The net force provides the centripetal force: $N + mg = \frac{mv^2}{R}$, where $v$ is the speed at the highest point. Given that the body exerts a force of three times its weight on t

Question

Q.32. A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height  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 . The value of  is ________.

FirstInTest · Accepted Answer

Sol. Let the mass of the body be $m$ and the radius of the circular loop be $R$. At the highest point of the circular loop, the forces acting on the body are gravity $mg$ downwards and the normal force $N$ downwards. The net force provides the centripetal force: $N + mg = \frac{mv^2}{R}$, where $v$ is the speed at the highest point. Given that the body exerts a force of three times its weight on the plane, this means the normal force $N$ exerted by the plane on the body is $3mg$. So, $3mg + mg = \frac{mv^2}{R} \Rightarrow 4mg = \frac{mv^2}{R} \Rightarrow v^2 = 4gR$. Now, apply conservation of mechanical energy between the initial position (height $h$) and the highest point of the loop (height $2R$). Initial energy $E_i = mgh$. Final energy $E_f = mg(2R) + \frac{1}{2}mv^2$. By conservation of energy, $E_i = E_f$: $mgh = mg(2R) + \frac{1}{2}mv^2$. Substitute $v^2 = 4gR$: $mgh = mg(2R) + \frac{1}{2}m(4gR)$. $mgh = 2mgR + 2mgR = 4mgR$. Divide by $mg$: $h = 4R$. The question states $h = aR$, so $a=4$.

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2

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4

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3

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6

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