A solid sphere of radius 10 cm is rotating about an axis which is… — Solution | JEE Physics

NVAEasyJEE 2026Moment of Inertia & Radius of Gyration

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. Find the value of $n$ .

Answer

Correct answer:265

Step-by-step solution

Standard Method

Given: A solid sphere has radius $R = 10 \, \text{cm}$ and the axis is at a distance $d = 15 \, \text{cm}$ from its centre.

Find: The value of $n$ if the radius of gyration is $k = \sqrt{n} \, \text{cm}$ .

Concept: The radius of gyration is defined by

I = Mk^2

For a solid sphere about its centre,

I_{\text{cm}} = \frac{2}{5}MR^2

For an axis not passing through the centre, use the parallel axis theorem:

I = I_{\text{cm}} + Md^2

Substitute the given values:

I = \frac{2}{5}M(10)^2 + M(15)^2

I = M\left(\frac{2}{5}\times 100 + 225\right)

I = M(40 + 225)

I = 265M

Now use

Mk^2 = 265M

k^2 = 265

Since $k = \sqrt{n} \, \text{cm}$ , we get

\sqrt{n} = \sqrt{265}

Hence,

n = 265

Therefore, the value of $n$ is $265$ .

Common mistakes

Using the moment of inertia of the solid sphere only about its centre is incorrect because the given axis is $15 \, \text{cm}$ away from the centre. Use the parallel axis theorem and add $Md^2$ .
Confusing radius of gyration with the actual radius of the sphere is incorrect. The radius of gyration is defined through $I = Mk^2$ , so first find $I$ about the given axis and then compare with $Mk^2$ .
Substituting the sphere radius incorrectly into $I_{\text{cm}} = \frac{2}{5}MR^2$ can cause error. Since $R = 10 \, \text{cm}$ , use $R^2 = 100$ , not $10$ .

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