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NVAEasyJEE 2023Moment of Inertia & Radius of Gyration

JEE Physics 2023 Question with Solution

A uniform solid cylinder with radius RR and length LL has moment of inertia I1I_1, about the axis of the cylinder. A concentric solid cylinder of radius R′=R2R' = \frac{R}{2} and length L′=L2L' = \frac{L}{2} is carved out of the original cylinder. If I2I_2 is the moment of inertia of the carved-out portion of the cylinder, then I1I2\frac{I_1}{I_2} is:

Answer

Correct answer:32

Step-by-step solution

Standard Method

Given: A uniform solid cylinder of radius RR and length LL has moment of inertia I1I_1 about its axis. A concentric solid cylinder of radius R2\frac{R}{2} and length L2\frac{L}{2} is carved out, and its moment of inertia is I2I_2.

Find: The value of I1I2\frac{I_1}{I_2}.

The moment of inertia of a uniform solid cylinder about its own axis is

I=12mR2I = \frac{1}{2} mR^2

For the original cylinder,

m1=ρπR2Lm_1 = \rho \pi R^2 L

Therefore,

I1=12m1R2=12ρπR2L⋅R2=ρπR4L2I_1 = \frac{1}{2} m_1 R^2 = \frac{1}{2} \rho \pi R^2 L \cdot R^2 = \frac{\rho \pi R^4 L}{2}

For the carved-out cylinder, its mass is proportional to its volume:

m2=ρπ(R2)2(L2)=ρπR2L8m_2 = \rho \pi \left(\frac{R}{2}\right)^2 \left(\frac{L}{2}\right) = \frac{\rho \pi R^2 L}{8}

Its moment of inertia about its own axis is

I2=12m2(R2)2=12⋅ρπR2L8⋅R24=ρπR4L64I_2 = \frac{1}{2} m_2 \left(\frac{R}{2}\right)^2 = \frac{1}{2} \cdot \frac{\rho \pi R^2 L}{8} \cdot \frac{R^2}{4} = \frac{\rho \pi R^4 L}{64}

Now take the ratio:

I1I2=ρπR4L2ρπR4L64=642=32\frac{I_1}{I_2} = \frac{\frac{\rho \pi R^4 L}{2}}{\frac{\rho \pi R^4 L}{64}} = \frac{64}{2} = 32

Therefore, the required value is 3232.

Scaling Argument

Given: Both cylinders are uniform and concentric.

Find: I1I2\frac{I_1}{I_2} using proportionality.

Two horizontal solid cylinders are shown side by side. The first has radius R and length l, and the second has radius R/2 and length l/2, with arrows marking these dimensions.

Since for a solid cylinder about its axis,

I∝mR2I \propto mR^2

and mass is proportional to volume,

m∝R2Lm \propto R^2 L

so

I∝R2L⋅R2=R4LI \propto R^2 L \cdot R^2 = R^4 L

Hence,

I1I2=R4L(R2)4(L2)=R4LR416⋅L2=32\frac{I_1}{I_2} = \frac{R^4 L}{\left(\frac{R}{2}\right)^4 \left(\frac{L}{2}\right)} = \frac{R^4 L}{\frac{R^4}{16} \cdot \frac{L}{2}} = 32

Therefore, the correct answer is 3232.

Common mistakes

  • Using only the radius change and ignoring the change in length. The carved-out mass also changes because volume depends on both RR and LL. First scale the mass using volume, then apply the moment of inertia formula.

  • Assuming I∝R2I \propto R^2 only. For a solid cylinder about its axis, I=12mR2I = \frac{1}{2}mR^2, and since mm itself depends on R2LR^2L, the full dependence is I∝R4LI \propto R^4L.

  • Comparing the remaining cylinder with the original cylinder instead of comparing the carved-out portion with the original cylinder. Here I2I_2 is for the removed inner cylinder, not for the leftover body.

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