JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2023
- Medium
- Numerical
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 .............
Pick an option to check your answer.
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 \(\frac{ mR ^2}{2}= mk _{ cyl }^2\)
\(\Rightarrow k _{ cyl }=\frac{ R }{\sqrt{2}}\)
\(\frac{ k _{ rph }}{ k _{ cyl }}=\frac{\sqrt{\frac{2}{5}}}{\frac{1}{\sqrt{2}}}=\frac{2}{\sqrt{5}}=\frac{2}{\sqrt{ x }}\)
\(\therefore x =5\)
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