### 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 $