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Physics · Work, Power & Energy

JEE Main 2026 — 6 April, Morning Shift — Question 6

A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height h 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

Question figure
  1. Option A:

    2

  2. Option B:

    4

    Correct
  3. Option C:

    3

  4. Option D:

    5

Answer: B

Step-by-step solution

At highest point, normal force N = 3mg (exerts force 3 times weight on plane? Actually "exerts a force of three times its weight on the plane" means N = 3mg. Then centripetal force: mg+N=mv2/R⇒4mg=mv2/R⇒v2=4gRmg + N = mv^2/R \Rightarrow 4mg = mv^2/R \Rightarrow v^2 = 4gR. Energy conservation: mg(h−2R)=12m(4gR)⇒h−2R=2R⇒h=4Rmg(h-2R) = \frac12 m(4gR) \Rightarrow h-2R = 2R \Rightarrow h=4R. So α=4\alpha = 4.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Physics
Chapter
Work, Power & Energy
Topic
Vertical Circular Motion

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