A ring and a disc roll on horizontal surface without slipping with same linear velocity. If both have same…

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
  1. $\frac{3}{2} \mathrm{~J}$
  2. $\frac{5}{2} \mathrm{~J}$
  3. $\frac{7}{2} \mathrm{~J}$
  4. $\frac{9}{2} \mathrm{~J}$

Solution

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) $\begin{aligned} & \frac{(\mathrm{K} \cdot \mathrm{E})_{\text {disc }}}{(\mathrm{K} . \mathrm{E})_{\text {ring }}}=\frac{\frac{3}{4} \mathrm{Mv}^2}{\mathrm{Mv}^2} \\ \therefore \quad & (\mathrm{~K} . \mathrm{E})_{\text {disc }}=(\mathrm{K} . \mathrm{E})_{\text {ring }} \times \frac{3}{4}=6 \times \frac{3}{4}=\frac{9}{2} \mathrm{~J} \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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