JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 25)
Two identical spheres each of mass $$2 \mathrm{~kg}$$ and radius $$50 \mathrm{~cm}$$ are fixed at the ends of a light rod so that the separation between the centers is $$150 \mathrm{~cm}$$. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is $$\frac{x}{20} \mathrm{~kg} \mathrm{m^{2 }}$$, where the value of $$x$$ is ___________.
Answer
53
Explanation
_31st_January_Evening_Shift_en_25_1.png)
$$\begin{aligned} & \mathrm{I}=\left(\frac{2}{5} \mathrm{mR}^2+\mathrm{md}^2\right) \times 2 \\ & \mathrm{I}=2\left(\frac{2}{5} \times 2 \times\left(\frac{1}{2}\right)^2+2 \times\left(\frac{3}{4}\right)^2\right)=\frac{53}{20} \mathrm{~kg}-\mathrm{m}^2 \\ & \mathrm{X}=53 \end{aligned}$$
Comments (0)
