Two identical discs are moving with the same kinetic energy. One rolls and the other slides. The ratio of their speeds is

• A

  1 : 2
• B

  1 : 1
• C

  2 : 3
• Correct answer
  D

  $\sqrt{2}: \sqrt{3}$

Solution

According to the question, $\therefore$ Kinetic energy for the rolling disc, or $$ \begin{aligned} \mathrm{KE}_r= & \frac{1}{2} m v_1^2+\frac{1}{2} I \omega^2 \\ & \left(\therefore \text { Moment of inertia, } I=\frac{m R^2}{2}\right) \\ & =\frac{1}{2} m v_1^2+\frac{1}{2} \frac{m R^2}{2}\left(\frac{v_1}{R}\right)^2 \end{aligned} $$
Given, $\mathrm{KE}$ of rolling disc $=\mathrm{KE}$ of sliding disc or, $$ \begin{aligned} \frac{3}{4} m v_1^2=\frac{1}{2} m v_2^2 & \text { or } \frac{v_1^2}{v_2^2}=\frac{2}{3} \\ \frac{v_1}{v_2} & =\sqrt{\frac{2}{3}} \\ or \quad v_1: v_2 & =\sqrt{2}: \sqrt{3} \end{aligned} $$

Hence, the correct answer is option: 'D'.