MHT CET · Physics · Rotational Motion
A ring and a disc roll on horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 6 J then total kinetic energy of the disc is
- A \(\frac{3}{2} \mathrm{~J}\)
- B \(\frac{5}{2} \mathrm{~J}\)
- C \(\frac{7}{2} \mathrm{~J}\)
- D \(\frac{9}{2} \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(\frac{9}{2} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Total \((K . E)_{\text {ring }}=M v^2\)...(i)
Total (K.E) disc \(=\frac{3}{4} \mathrm{Mv}^2\)...(ii)
Dividing equation (ii) by equation (i)
\(\begin{aligned}
& \frac{(\mathrm{K} \cdot \mathrm{E})_{\text {disc }}}{(\mathrm{K} . \mathrm{E})_{\text {ring }}}=\frac{\frac{3}{4} \mathrm{Mv}^2}{\mathrm{Mv}^2} \\
\therefore \quad & (\mathrm{~K} . \mathrm{E})_{\text {disc }}=(\mathrm{K} . \mathrm{E})_{\text {ring }} \times \frac{3}{4}=6 \times \frac{3}{4}=\frac{9}{2} \mathrm{~J}
\end{aligned}\)
Total (K.E) disc \(=\frac{3}{4} \mathrm{Mv}^2\)...(ii)
Dividing equation (ii) by equation (i)
\(\begin{aligned}
& \frac{(\mathrm{K} \cdot \mathrm{E})_{\text {disc }}}{(\mathrm{K} . \mathrm{E})_{\text {ring }}}=\frac{\frac{3}{4} \mathrm{Mv}^2}{\mathrm{Mv}^2} \\
\therefore \quad & (\mathrm{~K} . \mathrm{E})_{\text {disc }}=(\mathrm{K} . \mathrm{E})_{\text {ring }} \times \frac{3}{4}=6 \times \frac{3}{4}=\frac{9}{2} \mathrm{~J}
\end{aligned}\)
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