JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 25)
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
5
Explanation
$m=\rho \pi R^{2} t$
$$ \begin{aligned} & \text { so } R^{2}=\frac{m}{\rho \pi t} \\\\ & I=\frac{m R^{2}}{4}=\frac{m^{2}}{4 \rho \pi t} \end{aligned} $$
So $\frac{I_{1}}{I_{2}}=\frac{\rho_{2} t_{2}}{\rho_{1} t_{1}}=\frac{5}{3} \times \frac{0.5}{1}=\frac{5}{6}$
So $x=5$
$$ \begin{aligned} & \text { so } R^{2}=\frac{m}{\rho \pi t} \\\\ & I=\frac{m R^{2}}{4}=\frac{m^{2}}{4 \rho \pi t} \end{aligned} $$
So $\frac{I_{1}}{I_{2}}=\frac{\rho_{2} t_{2}}{\rho_{1} t_{1}}=\frac{5}{3} \times \frac{0.5}{1}=\frac{5}{6}$
So $x=5$
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