JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively \(\left( k _{\text {sph }}: k _{ cyl }\right)\) is \(2: \sqrt{ x }\), then value of \(x\) is .............
- A \(5\)
- B \(10\)
- C \(15\)
- D \(20\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
For solid sphere \(\frac{2}{5} mR ^2= mk _{ mph }^2\) \(k _{ sph }=\sqrt{\frac{2}{5}} R\) For solid cylinder \(\frac{ mR ^2}{2}= mk _{ cyl }^2\) \(\Rightarrow k _{ cyl }=\frac{ R }{\sqrt{2}}\)…
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