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MHT CET 2024 (11 May, Shift 1) · Rolling Motion

Rolling Motion question from MHT CET 2024 (11 May, Shift 1)

Q24MHT CET 2024 · 11 MAY · SHIFT 1Easy

Rolling Motion

An inclined plane makes an angle 30∘30^\circ with horizontal. A solid sphere rolls down from the top of the inclined plane from rest without slipping has a linear acceleration along the plane equal to (where gg is acceleration due to gravity) (given sin⁡30∘=0.5\sin 30^\circ = 0.5)

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Answer: (a)

  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=57gsin⁡30∘=57×g2=5g14a = \frac{5}{7}g\sin 30^\circ = \frac{5}{7}\times\frac{g}{2} = \frac{5g}{14}
  3. So the answer is (a) 5g14\frac{5g}{14}.
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