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NVAEasyJEE 2025Moment of Inertia & Radius of Gyration

JEE Physics 2025 Question with Solution

The moment of inertia of a solid disc rotating along its diameter is 2.52.5 times higher than the moment of inertia of a ring rotating in a similar way. The moment of inertia of a solid sphere which has the same radius as the disc and rotating in similar way, is nn times higher than the moment of inertia of the given ring. Here, n=n = :

Answer

Correct answer:0.8

Step-by-step solution

Standard Method

Given: A ring, a solid disc, and a solid sphere rotate about their diameters. The sphere has the same radius as the disc.

Find: The value of nn such that the moment of inertia of the solid sphere is nn times the moment of inertia of the ring.

For a ring rotating about its diameter,

Iring=MR22I_{\text{ring}} = \frac{MR^2}{2}

For a solid sphere rotating about its diameter,

Isphere=2MR25I_{\text{sphere}} = \frac{2MR^2}{5}

Now,

Isphere=nIringI_{\text{sphere}} = n I_{\text{ring}}

So,

n=IsphereIring=2MR25MR22=25×21=45=0.8n = \frac{I_{\text{sphere}}}{I_{\text{ring}}} = \frac{\frac{2MR^2}{5}}{\frac{MR^2}{2}} = \frac{2}{5} \times \frac{2}{1} = \frac{4}{5} = 0.8

Therefore, the value of nn is 0.80.8.

Using the listed formulas

Given:

  • For a ring about diameter,
Iring=MR22I_{\text{ring}} = \frac{MR^2}{2}
  • For a solid disc about diameter,
Idisc=MR24I_{\text{disc}} = \frac{MR^2}{4}
  • For a solid sphere about diameter,
Isphere=2MR25I_{\text{sphere}} = \frac{2MR^2}{5}

Find: nn in

Isphere=nIringI_{\text{sphere}} = n I_{\text{ring}}

the solution states the disc is compared with the ring, but the required value of nn is obtained from the ratio of sphere to ring:

n=IsphereIringn = \frac{I_{\text{sphere}}}{I_{\text{ring}}}

Substituting,

n=25MR212MR2n = \frac{\frac{2}{5}MR^2}{\frac{1}{2}MR^2}

Cancel MR2MR^2:

n=25÷12n = \frac{2}{5} \div \frac{1}{2} n=25×2=45n = \frac{2}{5} \times 2 = \frac{4}{5} n=0.8n = 0.8

Therefore, the solid sphere has moment of inertia 0.80.8 times that of the ring, so the answer is 0.80.8.

Common mistakes

  • Using the moment of inertia of a ring about its central axis, MR2MR^2, instead of about its diameter, MR22\frac{MR^2}{2}. This is wrong because the axis in the question is along a diameter. Use the formula for the specified axis.

  • Using the moment of inertia of a solid disc about its central axis, 12MR2\frac{1}{2}MR^2, instead of about its diameter, 14MR2\frac{1}{4}MR^2. The axis changes the value, so always match the formula to the given axis of rotation.

  • Assuming the statement involving the disc must be used algebraically even though the required quantity is the ratio of sphere to ring. This can create unnecessary confusion. Focus on the expression n=IsphereIringn = \frac{I_{\text{sphere}}}{I_{\text{ring}}} from the solution.

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