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 \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) ).
- A \(1\)
- B \(5\)
- C \(7\)
- D \(10\)
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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