A disc and a ring of same mass are rolling and if their kinetic energies are equal, then the ratio of their velocities will be

The correct option is √4:√3.

To find the ratio of the velocities of the disc and the ring, we can write down the expressions for their total kinetic energy in rolling motion.

The total kinetic energy ($K$) of a rolling body of mass $M$ and radius $R$ moving with linear velocity $v$ is the sum of its translational kinetic energy and rotational kinetic energy:
$K=K_{\text{trans}}+K_{\text{rot}}$
$K=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$

For a body undergoing pure rolling, the angular velocity is given by:
$\omega =\frac{v}{R}$

Substituting this value into the kinetic energy equation gives:
$K=\frac{1}{2}M{v}^2+\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=\frac{1}{2}M{v}^2\left(1+\frac{I}{M{R}^2}\right)$

1. Kinetic Energy of the Disc ($K_{\text{disc}}$):
The moment of inertia of a disc is:
$I_{\text{disc}}=\frac{1}{2}M{R}^2$
Substituting this in the total kinetic energy equation for the disc moving with velocity $v_{\text{disc}}$:
$K_{\text{disc}}=\frac{1}{2}M{v}_{\text{disc}}^2\left(1+\frac{1}{2}\right)=\frac{3}{4}M{v}_{\text{disc}}^2$

2. Kinetic Energy of the Ring ($K_{\text{ring}}$):
The moment of inertia of a ring is:
$I_{\text{ring}}=M{R}^2$
Substituting this in the total kinetic energy equation for the ring moving with velocity $v_{\text{ring}}$:
$K_{\text{ring}}=\frac{1}{2}M{v}_{\text{ring}}^2\left(1+1\right)=M{v}_{\text{ring}}^2$

3. Finding the Ratio of Velocities:
Since their kinetic energies are equal ($K_{\text{disc}}=K_{\text{ring}}$):
$\frac{3}{4}M{v}_{\text{disc}}^2=M{v}_{\text{ring}}^2$

Canceling the mass $M$ from both sides gives:
$\frac{3}{4}{v}_{\text{disc}}^2=v_{\text{ring}}^2$
Rearranging the terms to get the ratio of velocities:
$\frac{v_{\text{disc}}^2}{v_{\text{ring}}^2}=\frac{4}{3}$

Taking the square root on both sides:
$\frac{v_{\text{disc}}}{v_{\text{ring}}}=\frac{\sqrt{4}}{\sqrt{3}}$

Therefore, the ratio of their velocities is √4:√3.