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MHT CET 2025 (20 Apr, Shift 1) · Rolling Motion

Rolling Motion question from MHT CET 2025 (20 Apr, Shift 1)

Q13MHT CET 2025 · 20 APR · SHIFT 1Easy

Rolling Motion

An inclined plane makes an angle 30∘30^\circ with the horizontal. A solid sphere rolling down an inclined plane from rest without slipping has linear acceleration ( g=g = acceleration due gravity) ( sin⁡30∘=0.5\sin 30^\circ = 0.5)

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Answer: (b)

  1. Rolling down an incline: a=gsin⁡θ1+k2/R2a = \frac{g\sin\theta}{1 + k^2/R^2}.
  2. Solid sphere: a=gsin⁡θ7/5=57g×0.5a = \frac{g\sin\theta}{7/5} = \frac{5}{7}g\times 0.5
  3. a=5g14a = \frac{5g}{14}
  4. So the answer is (b) 5g14\frac{5g}{14}.
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