Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis A B as shown in figure is \sqrt8 / x . The value of x is :

Rotational Motion PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q64 with Solution

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Correct answer: C

Radius of gyration relation

For any body about a given axis, $I = Mk^2$ where $I$ is moment of inertia, $M$ is mass, and $k$ is radius of gyration.

We need $\frac{k_{\text{hollow sphere}}}{k_{\text{solid cylinder}}}$ for the axis along the diameter $AB$ shown.

Since masses are equal, $\frac{k_1}{k_2} = \sqrt{\frac{I_1/M}{I_2/M}} = \sqrt{\frac{I_1}{I_2}}$ provided the same mass $M$ is used.

Moment of inertia of hollow sphere about a diameter

For a thin hollow sphere (spherical shell), moment of inertia about any diameter is $I_{\text{sphere}} = \frac{2}{3}MR^2$

So, $k_{\text{sphere}} = \sqrt{\frac{I}{M}} = \sqrt{\frac{2}{3}}R$

Moment of inertia of solid cylinder about its own diameter

For a solid cylinder of radius $R$ and length $L$ , moment of inertia about a central transverse axis (a diameter through the center, perpendicular to cylinder axis) is $I_{\text{cyl}} = \frac{1}{12}M(3R^2 + L^2)$

From the figure/context, the cylinder is the standard case where its length equals diameter, i.e. $L = 2R$

Therefore,

= \frac{1}{12}M(3R^2 + 4R^2) = \frac{7}{12}MR^2$$ Hence, $$k_{\text{cyl}} = \sqrt{\frac{7}{12}}R$$ --- 4. **Take the ratio** $$\frac{k_{\text{sphere}}}{k_{\text{cyl}}} = \sqrt{\frac{\frac{2}{3}}{\frac{7}{12}}} = \sqrt{\frac{2}{3}\cdot \frac{12}{7}} = \sqrt{\frac{8}{7}}$$ Given this is of the form $$\sqrt{\frac{8}{x}}$$ we get $$x = 7$$ --- 5. **Compare with options and stored answer** The derived value is $$x=7$$ which is **not present** in the listed options $34, 51, 67, 17$. So the question/options appear inconsistent. The stored correct answer is **C: 67**, but that does not match the physics result. Thus, I **disagree** with the stored answer.

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View written solutionFree

Correct answer: C

Radius of gyration relation

For any body about a given axis, $I = Mk^2$ where $I$ is moment of inertia, $M$ is mass, and $k$ is radius of gyration.

We need $\frac{k_{\text{hollow sphere}}}{k_{\text{solid cylinder}}}$ for the axis along the diameter $AB$ shown.

Since masses are equal, $\frac{k_1}{k_2} = \sqrt{\frac{I_1/M}{I_2/M}} = \sqrt{\frac{I_1}{I_2}}$ provided the same mass $M$ is used.

Moment of inertia of hollow sphere about a diameter

For a thin hollow sphere (spherical shell), moment of inertia about any diameter is $I_{\text{sphere}} = \frac{2}{3}MR^2$

So, $k_{\text{sphere}} = \sqrt{\frac{I}{M}} = \sqrt{\frac{2}{3}}R$

Moment of inertia of solid cylinder about its own diameter

From the figure/context, the cylinder is the standard case where its length equals diameter, i.e. $L = 2R$

Therefore,

= \frac{1}{12}M(3R^2 + 4R^2) = \frac{7}{12}MR^2$$ Hence, $$k_{\text{cyl}} = \sqrt{\frac{7}{12}}R$$ --- 4. **Take the ratio** $$\frac{k_{\text{sphere}}}{k_{\text{cyl}}} = \sqrt{\frac{\frac{2}{3}}{\frac{7}{12}}} = \sqrt{\frac{2}{3}\cdot \frac{12}{7}} = \sqrt{\frac{8}{7}}$$ Given this is of the form $$\sqrt{\frac{8}{x}}$$ we get $$x = 7$$ --- 5. **Compare with options and stored answer** The derived value is $$x=7$$ which is **not present** in the listed options $34, 51, 67, 17$. So the question/options appear inconsistent. The stored correct answer is **C: 67**, but that does not match the physics result. Thus, I **disagree** with the stored answer.

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