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Physics · Rotational Dynamics

JEE Main 2026 — 5 April, Evening Shift — Question 6

An object of uniform density rolls up the curved path with the initial velocity v0v_0 as shown in the figure. If the maximum height attained by an object is 7v0210g\frac{7v_0^2}{10g} (g = acceleration due to gravity), the object is a

Question figure
  1. Option A:

    solid cylinder

  2. Option B:

    ring

  3. Option C:

    disc

  4. Option D:

    solid sphere

    Correct

Answer: D

Step-by-step solution

Using work-energy theorem: mgh=12mv02+12Iω2mgh = \frac12 mv_0^2 + \frac12 I \omega^2 with ω=v0/R\omega = v_0/R. Substituting h=7v02/(10g)h = 7v_0^2/(10g) gives I=25mR2I = \frac{2}{5}mR^2, which is solid sphere.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Physics
Chapter
Rotational Dynamics
Topic
Rolling Motion

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