JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Rotational Mechanics

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A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg , kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

(A) $3.5 m / s^{2}$ ✓ Correct
(B) $0.35 m / s^{2}$
(C) $2.5 m / s^{2}$
(D) $0.25 m / s^{2}$

Answer: (A)

Solution

Taking torque about the bottom contact point (instantaneous axis of rotation):

τ = F \times 2 r = I_{b o tt o m} α

For a solid sphere,

I_{ce n t er} = \frac{2}{5} m r^{2}

. Using parallel axis theorem:

I_{b o tt o m} = I_{ce n t er} + m r^{2} = \frac{2}{5} m r^{2} + m r^{2} = \frac{7}{5} m r^{2}

Therefore:

49 \times 2 r = \frac{7}{5} m r^{2} α

98 r = \frac{7}{5} \times 20 \times r^{2} α

98 = 28 r α

For rolling without slipping,

a = r α

a = \frac{98}{28} = 3.5 m/s^{2}

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Rotational Mechanics

Solution

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