JEE MAIN - Physics (2025 - 7th April Morning Shift - No. 22)
$\mathrm{A}, \mathrm{B}$ and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure.
_7th_April_Morning_Shift_en_22_1.png)
The moment of inertia of the given system about PQ axis is $\frac{x}{15} \mathrm{I}$, where I is the moment of inertia of the disc about its diameter. The value of $x$ is ____________.
Explanation
_7th_April_Morning_Shift_en_22_2.png)
All bodies have same mass and same radius.
A $\rightarrow$ Disc
B $\rightarrow$ Solid sphere
$\mathrm{C} \rightarrow$ Spherical shell
$$\begin{aligned} & \text { and, } \mathrm{I}=\frac{\mathrm{MR}^2}{4} \\ \mathrm{I}_{\mathrm{PQ}} & =\frac{\mathrm{MR}^2}{4}+\left(\frac{2}{5} \mathrm{MR}^2+\mathrm{MR}^2\right)+\left(\frac{2}{3} \mathrm{MR}^2+\mathrm{MR}^2\right) \\ \mathrm{I}_{\mathrm{PQ}} & =\frac{15 \mathrm{MR}^2+24 \mathrm{MR}^2+60 \mathrm{MR}^2+40 \mathrm{MR}^2+60 \mathrm{MR}^2}{60} \\ \mathrm{I}_{\mathrm{PQ}} & =\frac{199}{60} \mathrm{MR}^2=\frac{199}{15}\left(\frac{\mathrm{MR}^2}{4}\right) \\ & =\frac{199}{15} \mathrm{I} \end{aligned}$$
Comments (0)
