Physics Q28 – JEE-MAIN 2026

Home Physics Rotational MotionQuestion #435

A solid sphere (A) of mass

5 m

and a 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 _______.

Clue 1 of 3: Rolling without slipping

For an object rolling without slipping, relate linear and angular acceleration, and consider all forces and torques acting on the object.

🧠 Full Solution Path

Step 1: Identify Moments of Inertia and General Acceleration Formula✦ Active

For a solid sphere (A) of mass $M_{A} = 5 m$ and radius $R$ , its moment of inertia is $I_{A} = \frac{2}{5} M_{A} R^{2} = \frac{2}{5} (5 m) R^{2} = 2 m R^{2}$ . For a spherical shell (B) of mass $M_{B} = m$ and radius $R$ , its moment of inertia is $I_{B} = \frac{2}{3} M_{B} R^{2} = \frac{2}{3} m R^{2}$ . For rolling without slipping, the linear acceleration $a$ and angular acceleration $α$ are related by $a = R α$ . When a force $F$ is applied tangentially at the highest point, the equations of motion are:

F - f = M a (Translational)

F R + f R = I α = I (a / R) (Rotational about CM)

F + f = \frac{I a}{R^{2}}

Adding the translational and rotational equations gives:

2 F = (M + \frac{I}{R^{2}}) a

Thus, the acceleration is given by:

a = \frac{2 F}{M + \frac{I}{R^{2}}}

Step 2: Calculate Acceleration for Solid Sphere (A)○ Expand

Using the derived formula for acceleration with $M_{A} = 5 m$ and $I_{A} = 2 m R^{2}$ :

a_{A} = \frac{2 F}{M_{A} + \frac{I_{A}}{R^{2}}} = \frac{2 F}{5 m + \frac{2 m R^{2}}{R^{2}}} = \frac{2 F}{5 m + 2 m} = \frac{2 F}{7 m}

💡 Teacher's Secret Hint

Ensure correct substitution of mass and moment of inertia for the solid sphere.

Step 3: Calculate Acceleration for Spherical Shell (B) and Find the Ratio○ Expand

Using the derived formula for acceleration with $M_{B} = m$ and $I_{B} = \frac{2}{3} m R^{2}$ :

a_{B} = \frac{2 F}{M_{B} + \frac{I_{B}}{R^{2}}} = \frac{2 F}{m + \frac{\frac{2}{3} m R^{2}}{R^{2}}} = \frac{2 F}{m + \frac{2}{3} m} = \frac{2 F}{\frac{5}{3} m} = \frac{6 F}{5 m}

Now, find the ratio of $a_{A}$ to $a_{B}$ :

\frac{a_{A}}{a_{B}} = \frac{\frac{2 F}{7 m}}{\frac{6 F}{5 m}} = \frac{2 F}{7 m} \times \frac{5 m}{6 F} = \frac{10}{42} = \frac{5}{21}

💡 Teacher's Secret Hint

Double-check the algebraic simplification when calculating the ratio.

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Physics Question 28 – JEE-MAIN 2026

Clue 1 of 3: Rolling without slipping

🧠 Full Solution Path