JEE Physics 2025 Question with Solution

Given: A tangential force $F = 49 \, \text{N}$ acts at the highest point of a solid sphere of mass $m = 20 \, \text{kg}$ on a rough horizontal plane. The sphere rolls without slipping.

Find: The acceleration of the center of the sphere.

For linear motion along the horizontal direction,

$$F + f_s = ma$$

where $f_s$ is the static friction.

For rotational motion about the center of mass,

$$Fr - f_s r = I\alpha$$

For a solid sphere,

$$I = \frac{2}{5}mr^2$$

and for rolling without slipping,

$$a = \alpha r$$

So,

$$r(F - f_s) = \left(\frac{2}{5}mr^2\right)\left(\frac{a}{r}\right)$$

which gives

$$F - f_s = \frac{2}{5}ma$$

Now add the two equations:

$$(F + f_s) + (F - f_s) = ma + \frac{2}{5}ma$$ $$2F = \frac{7}{5}ma$$

Hence,

$$a = \frac{10F}{7m}$$

Substituting the given values,

$$a = \frac{10 \times 49}{7 \times 20}$$ $$a = \frac{490}{140} = 3.5 \, \text{m/s}^2$$

Therefore, the acceleration of the center of the sphere is $3.5 \, \text{m/s}^2$. The correct option is A.

Given: The sphere rolls without slipping under a force $49 \, \text{N}$ applied at the top.

Find: The acceleration of its center.

Take torque about the point of contact with the ground. Then friction gives no torque about that point.

For a solid sphere, moment of inertia about the contact point is

$$I_{contact} = I_{cm} + mr^2 = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2$$

The applied force has lever arm $2r$, so

$$\tau = F \cdot 2r = I_{contact}\alpha$$

Thus,

$$49 \cdot 2r = \frac{7}{5}mr^2\alpha$$

With $m = 20 \, \text{kg}$ and $a = r\alpha$,

$$98r = \frac{7}{5}\cdot 20 \cdot r^2 \alpha = 28r^2\alpha$$ $$98 = 28a$$ $$a = 3.5 \, \text{m/s}^2$$

Therefore, the acceleration of the center is $3.5 \, \text{m/s}^2$. This shortcut works because taking torque about the contact point removes the unknown friction term immediately.