A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity 3 ~m / s (as shown in figure). Maximum height with respect to the initial position covered by it will be \underline\hspace2cm cm.

Rotational Motion PYQ — JEE Main 2023 · 8 Apr · Shift 2 · Q64 with Solution

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Correct answer: 75

Concept used: conservation of mechanical energy
As the hollow spherical ball rolls without slipping, its initial kinetic energy consists of:

At the maximum height, the ball momentarily comes to rest, so both translational and rotational kinetic energies become zero.

Thus,

\text{Initial total KE} = \text{Gain in potential energy}

Moment of inertia of a hollow spherical ball
For a hollow sphere (spherical shell),

I = \frac{2}{3}mR^2

v = \omega R

v = 3\,\text{m/s}

So translational kinetic energy is

K_t = \frac{1}{2}mv^2

Rotational kinetic energy is

K_r = \frac{1}{2}I\omega^2

Substitute $I = \frac{2}{3}mR^2$ and $\omega = \frac{v}{R}$ :

K_r = \frac{1}{2}\left(\frac{2}{3}mR^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{3}mv^2

Therefore total initial kinetic energy is

K = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2

mgh = \frac{5}{6}mv^2

Cancel $m$ :

gh = \frac{5}{6}v^2

So,

h = \frac{5v^2}{6g}

h = \frac{5\times 3^2}{6\times 10} = \frac{45}{60} = 0.75\,\text{m}

Convert to cm:

0.75\,\text{m} = 75\,\text{cm}

\boxed{75}