Question:easy
The ratio of moment of inertia with respect to their diameters of a circular disc and a solid sphere having same radii and same masses is:
The ratio of moment of inertia with respect to their diameters of a circular disc and a solid sphere having same radii and same masses is:
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Double check the axis specified in the question.
For a disc, the moment of inertia about its perpendicular axis is $\frac{1}{2}MR^2$, but about its diameter it is $\frac{1}{4}MR^2$.
A common mistake is using the perpendicular axis formula by mistake.
For a disc, the moment of inertia about its perpendicular axis is $\frac{1}{2}MR^2$, but about its diameter it is $\frac{1}{4}MR^2$.
A common mistake is using the perpendicular axis formula by mistake.
Updated On: Jul 22, 2026
- $4:5$
- $5:4$
- $8:5$
- $5:8$
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Recall the moment of inertia of a disc about a diameter.
For a uniform disc of mass $M$ and radius $R$, the moment of inertia about a diameter is $I_{\text{disc}} = \frac{1}{4}MR^2$.
Step 2: Recall the moment of inertia of a solid sphere about a diameter.
For a uniform solid sphere of the same mass and radius, the moment of inertia about a diameter is $I_{\text{sphere}} = \frac{2}{5}MR^2$.
Step 3: Form the ratio and simplify.
Writing both fractions over a common denominator of 20 gives $I_{\text{disc}} = \frac{5}{20}MR^2$ and $I_{\text{sphere}} = \frac{8}{20}MR^2$, so the ratio of the numerators gives the answer directly.
\[ I_{\text{disc}} : I_{\text{sphere}} = 5:8 \]
\[ \boxed{5:8} \]
For a uniform disc of mass $M$ and radius $R$, the moment of inertia about a diameter is $I_{\text{disc}} = \frac{1}{4}MR^2$.
Step 2: Recall the moment of inertia of a solid sphere about a diameter.
For a uniform solid sphere of the same mass and radius, the moment of inertia about a diameter is $I_{\text{sphere}} = \frac{2}{5}MR^2$.
Step 3: Form the ratio and simplify.
Writing both fractions over a common denominator of 20 gives $I_{\text{disc}} = \frac{5}{20}MR^2$ and $I_{\text{sphere}} = \frac{8}{20}MR^2$, so the ratio of the numerators gives the answer directly.
\[ I_{\text{disc}} : I_{\text{sphere}} = 5:8 \]
\[ \boxed{5:8} \]
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