JEE Main 2026 Apr 6 Shift 2, Physics Q3: Rolling Motion

JEE Main2026Apr 6, Shift 2Physics

Q.

A solid sphere (A) of mass $5 m$ and spherical shell (B) of mass $m$ , both having same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of A and B, they start rolling without slipping with an acceleration of $a_{A}$ and $a_{B}$ , respectively. The ratio of $a_{A}$ and $a_{B}$ is _____.

A
$5 : 21$
B
$6 : 10$
C
$21 : 25$
D
$1 : 5$

Solution

Solid sphere A ( $I = \frac{2}{5} (5 m) R^{2}$ , rolling: $a_{A} = R α$ ):

Newton's second law: $F + f = 5 m a_{A}$

Torque about centre: $(F - f) R = \frac{2}{5} (5 m) R^{2} \cdot \frac{a _{A}}{R}$ $\Rightarrow$ $F - f = 2 m a_{A}$

Adding: $2 F = 7 m a_{A}$ $\Rightarrow$ $a_{A} = \frac{2 F}{7 m}$ .

Spherical shell B ( $I = \frac{2}{3} m R^{2}$ ):

$F + f = m a_{B}$ and $(F - f) R = \frac{2}{3} m R^{2} \cdot \frac{a _{B}}{R}$ $\Rightarrow$ $F - f = \frac{2}{3} m a_{B}$

Adding: $2 F = \frac{5}{3} m a_{B}$ $\Rightarrow$ $a_{B} = \frac{6 F}{5 m}$ .

$\frac{a _{A}}{a _{B}} = \frac{2 F /7 m}{6 F /5 m} = \frac{2 \times 5}{7 \times 6} = \frac{10}{42} = \frac{5}{21}$ .

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