Two discs of same mass and different radii are made of different materials such that their thicknesses are $1 \mathrm{~cm}$ and $0.5 \mathrm{~cm}$ respectively. The densities of materials are in the ratio $3: 5$. The moment of inertia of these discs respectively about their diameters will be in the ratio of $\frac{x}{6}$. The value of $x$ is ________.
Answer (integer)
5
Solution
$m=\rho \pi R^{2} t$
<br/><br/>$$
\begin{aligned}
& \text { so } R^{2}=\frac{m}{\rho \pi t} \\\\
& I=\frac{m R^{2}}{4}=\frac{m^{2}}{4 \rho \pi t}
\end{aligned}
$$
<br/><br/>So $\frac{I_{1}}{I_{2}}=\frac{\rho_{2} t_{2}}{\rho_{1} t_{1}}=\frac{5}{3} \times \frac{0.5}{1}=\frac{5}{6}$
<br/><br/>So $x=5$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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