ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is \(\sqrt{n}\) cm. The value of n is

  1. A 265
  2. B 100
  3. C 325
  4. D 125
Verified Solution

Answer & Solution

Correct Answer

(A) 265

Step-by-step Solution

Detailed explanation

Let radius of gyration is k \( \Rightarrow mk^{2} = \frac{2}{5} mR^{2} + md^{2}\) \(k^{2} = \frac{2}{3} \times 10^{2} + 15^{2} = 265\) \((\sqrt{n})^{2} = 265 \Rightarrow n = 265\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app