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JEE MAIN 2024
31.01.24 S2
Question
Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 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 $\_\_\_\_$ .
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Solution
$\begin{aligned} & \mathrm{V}=\frac{\mathrm{K} \overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{9 \times 10^9(6 \mathrm{q} \ell)}{\mathrm{r}^2} \cos \left(120^{\circ}\right) \\ & =-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^2}\right) \times 10^9 \mathrm{Nm}^2 \mathrm{c}^{-2} \\ & \Rightarrow \alpha=27\end{aligned}$
Question Tags
JEE Main
Physics
Easy
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