JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Rotational Mechanics

An object of uniform density rolls up the curved path with the initial velocity

v_{0}

as shown in the figure. If the maximum height attained by an object is

\frac{7 v _{0}^{2}}{10 g}

(

g =

acceleration due to gravity), the object is a ______.

Answer: (D)

Solution

For a body of mass

m

rolling without slipping at speed

v_{0}

, the total kinetic energy is

\frac{1}{2} m v_{0}^{2} + \frac{1}{2} I ω^{2}

, and with

ω = \frac{v _{0}}{R}

this becomes

\frac{1}{2} m v_{0}^{2} (1 + \frac{I}{m R ^{2}})

. Equating it to the potential energy

m g h

at the greatest height gives

h = \frac{v _{0}^{2}}{2 g} (1 + \frac{I}{m R ^{2}})

. Setting

h = \frac{7 v _{0}^{2}}{10 g}

yields

1 + \frac{I}{m R ^{2}} = \frac{7}{5}

, so

\frac{I}{m R ^{2}} = \frac{2}{5}

, i.e.

I = \frac{2}{5} m R^{2}

. A moment of inertia of

\frac{2}{5} m R^{2}

identifies the body as a solid sphere.

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option