JEE Main 12 Jan 2019 Morning Question 29
Question: Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also J, is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 14 cm
B) 16 cm
C) 12 cm
D) 18 cm
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{mR_1^{2}+R_2^{2}}{2}=mK^{2} $
$ \frac{m(10^{2}+20^{2})}{2}=mK^{2}\Rightarrow \frac{100+400}{2}=K^{2} $
$ K^{2}=250 $
$ \Rightarrow K,\underline{\approx },16,cm $
