A solid sphere rolling without friction on a horizontal surface with a constant speed of 2 m/s, rolls up on an inclined ramp which is inclined at 30°. The maximum distance travelled by the sphere on the inclined ramp is (acceleration due to gravity g = 10 m/s², sin 30° = 1/2) (a) 56 cm (b) 25 cm (c) 47 cm (d) 30 cm

Answer: (a). Rolling sphere KE = 7/10mv² becomes mgh at the highest point, with h = dsin 30°. 7/10× 2² = 10 × d × 1/2 d = 2.8/5 = 0.56 m = 56 cm. So the answer is (a).

A solid sphere rolling without friction on a horizontal surface with… | MHT CET 2025 (20 Apr, Shift 2)

Q19MHT CET 2025 · 20 APR · SHIFT 2Medium

Rolling Motion

A solid sphere rolling without friction on a horizontal surface with a constant speed of $2\text{ m/s}$ , rolls up on an inclined ramp which is inclined at $30^\circ$ . The maximum distance travelled by the sphere on the inclined ramp is (acceleration due to gravity $g = 10\text{ m/s}^{2}, \sin 30^\circ = \frac{1}{2}$ )

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Answer: (a)

Rolling sphere KE $= \frac{7}{10}mv^2$ becomes $mgh$ at the highest point, with $h = d\sin 30^\circ$ .
$\frac{7}{10}\times 2^2 = 10 \times d \times \frac{1}{2}$
$d = \dfrac{2.8}{5} = 0.56$ m $= 56$ cm.
So the answer is (a).

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Rolling Motion question from MHT CET 2025 (20 Apr, Shift 2)

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