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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

A cylinder is rolling down on an inclined plane of inclination \(60^{\circ}\). It's acceleration during rolling down will be \(\frac{\mathrm{x}}{\sqrt{3}} \mathrm{~m} / \mathrm{s}^2\), where \(\mathrm{x}=\)___________. (use \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) ).

  1. A \(1\)
  2. B \(5\)
  3. C \(7\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(10\)

Step-by-step Solution

Detailed explanation

For rolling motion, \(\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}_{\mathrm{cm}}}{\mathrm{MR}^2}}\) \( a=\frac{g \sin \theta}{1+\frac{1}{2}} \) \( =\frac{2 \times 10 \times \frac{\sqrt{3}}{2}}{3} \) \( =\frac{10}{\sqrt{3}}\) Therefore \(\mathrm{x}=10\)
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