JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A circular disc reaches from top to bottom of an inclined plane of length \('L'.\) When it slips down the plane, it takes time \('t_{1}'\). when it rolls down the plane, it takes time \(t_{2}\). The value of \(\frac{t_{2}}{t_{1}}\) is \(\sqrt{\frac{3}{x}}\). The value of \(x\) will be .... .
- A \(8\)
- B \(6\)
- C \(4\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
If disk slips on inclined plane, then it's aceleration \(a_{1}=g \sin \theta\) \(L=\frac{1}{2} a_{1} t_{1}^{2}\) \(\Rightarrow t_{1}=\sqrt{\frac{2 L}{a_{1}}}\ldots \ldots({i} )\) If disk rols on inclined plane, its acceleration,…
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