A circular disc of mass M and… - 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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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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Solution

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Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics