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MHT CET 2024 (2 May, Shift 2) · Rolling Motion

Rolling Motion question from MHT CET 2024 (2 May, Shift 2)

Q26MHT CET 2024 · 2 MAY · SHIFT 2Easy

Rolling Motion

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

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Answer: (b)

  1. a=gsin⁡θ1+k2/R2a = \frac{g\sin\theta}{1 + k^2/R^2}, with k2R2=25\frac{k^2}{R^2} = \frac{2}{5} for a solid sphere.
  2. a=57×g×0.5=5g14a = \frac{5}{7}\times g\times 0.5 = \frac{5g}{14}
  3. So the answer is (b) 5g14\frac{5g}{14}.
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