A thin circular plate of mass |RIGIDBODYDYNAMICSJEEMain2019

JEE MAIN 2019

08-04-2019 S1

Question

A thin circular plate of mass $M$ and radius $R$ has its density varying as $\rho(r)=\rho_0 r$ with $\rho_0$ as constant and $r$ is the distance from its center. The moment of inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $\mathrm{I}=\mathrm{a} \mathrm{MR}^2$. The value of the coefficient a is

Select the correct option:

$\frac{3}{2}$

$\frac{1}{2}$

$\frac{3}{5}$

$\frac{8}{5}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & M=\int_0^R \rho_0 r \times 2 \pi r d r=\frac{2 \pi \rho_0 R^3}{3} \\ & I_c=\int_0^R \rho_0 r \times 2 \pi r d r^2=\frac{2 \pi \rho_0 R^5}{5} \\ & \therefore I=I_C+M R^2=2 \pi \rho_0 R^5\left(\frac{1}{3}+\frac{1}{5}\right)=\frac{16 \pi \rho_0 R^5}{15} \\ & =\frac{8}{5}\left[\frac{2}{3} \pi \rho_0 R^3\right] R^2=\frac{8}{5} M R^2\end{aligned}$

Question Tags

JEE Main

Physics

Hard

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