A circular disc of mass M and radius R is… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Mechanics

2020integerhard

A circular disc of mass $\text{M}$ and radius $\text{R}$ is rotating about its axis with angular speed ${\omega }_{1}$ . If another stationary disc having radius $\frac{\text{R}}{2}$ and same mass $\text{M}$ is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed ${\omega }_{2}$ . The energy lost in the process is $\text{p%}$ of the initial energy. Value of $p$ is ______

Official previous-year question

Held on 4 Sept 2020 · Verified 6 Jul 2026.

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Solution

JEE Main 2020 Physics, Rotational Motion — solution figure

${I}_{1}{\omega }_{1}+{I}_{2}{\omega }_{2}=({I}_{1}+{I}_{2})\times {\omega }_{f}$

$\frac{{\mathrm{MR}}^{2}}{2}\times \omega +0(\frac{{\mathrm{MR}}^{2}}{2}+\frac{{\mathrm{MR}}^{2}}{8})\frac{16}{25}$

${\omega }_{f}=\frac{4}{5}\omega$

Final $K.E$ ., ${K}_{f}=\frac{1}{2}(\frac{{\mathrm{MR}}^{2}}{2}+\frac{{\mathrm{MR}}^{2}}{8})\frac{16}{25}{\omega }^{2}$

${K}_{f}=\frac{{\mathrm{MR}}^{2}{\omega }^{2}}{5}$ ; ${K}_{i}=\frac{1}{2}(\frac{{\mathrm{MR}}^{2}}{2}){\omega }^{2}=\frac{{\mathrm{MR}}^{2}{\omega }^{2}}{4}$

Percentage loss in kinetic energy $% \mathrm{loss}=\frac{\frac{{\mathrm{MR}}^{2}{\omega }^{2}}{4}-\frac{{\mathrm{MR}}^{2}{\omega }^{2}}{5}}{\frac{{\mathrm{MR}}^{2}{\omega }^{2}}{4}}\times 100=20%$

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