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JEE Main 2026 · 5 Apr · Shift 2, Physics question 31

An object of uniform density rolls up the curved path with the initial velocity vov_o as shown in the figure. If the maximum height attained by an object is 7vo210 g\dfrac{7v_o^2}{10\,g} (gg = acceleration due to gravity), the object is a ________.

A round object with speed v₀ at the bottom left of a curved ramp that rises ever more steeply to a vertical end at the right.
Choose one option
Answer and worked solution

Official answer: D

Energy for rolling without slipping:
12mvo2(1+k2R2)=mgh⇒h=vo22g(1+k2R2).\frac12mv_o^2\left(1+\frac{k^2}{R^2}\right)=mgh\quad\Rightarrow\quad h=\frac{v_o^2}{2g}\left(1+\frac{k^2}{R^2}\right).

Match. h=7vo210gh=\dfrac{7v_o^2}{10g} needs k2R2=25\dfrac{k^2}{R^2}=\dfrac25: a solid sphere.

(D)

Topics:Rolling motionMoment of inertia
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