A solid sphere rolls down an inclined plane and its velocity at the bottom is v1. Then same sphere slides down the plane (without friction) and let its velocity at the bottom be v2. Which of the following relation is correct

The correct option is None of these.

To find the correct relation between the velocities, we can use the principle of conservation of mechanical energy for both cases. Let the vertical height of the inclined plane be $h$.

Case 1: Solid sphere rolling down the inclined plane (with friction, rolling without slipping)
When the sphere rolls without slipping, its potential energy at the top is converted into both translational kinetic energy and rotational kinetic energy at the bottom:
$mgh=\frac{1}{2}m{v}_1^2+\frac{1}{2}I{\omega }^2$
Here, $m$ is the mass of the sphere, $I$ is the moment of inertia of the solid sphere, and $\omega$ is the angular velocity.

For a solid sphere, the moment of inertia is:
$I=\frac{2}{5}m{R}^2$
Since the sphere rolls without slipping, we have $\omega =\frac{v_1}{R}$. Substituting these values into the energy equation:
$mgh=\frac{1}{2}m{v}_1^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\left(\frac{v_1}{R}\right)}^2$
$mgh=\frac{1}{2}m{v}_1^2+\frac{1}{5}m{v}_1^2$
$mgh=\frac{7}{10}m{v}_1^2$
Solving for $v_1^2$:
$v_1^2=\frac{10}{7}gh$

Case 2: Solid sphere sliding down the inclined plane (without friction)
When the sphere slides down without friction, it does not rotate. Therefore, its potential energy is converted entirely into translational kinetic energy:
$mgh=\frac{1}{2}m{v}_2^2$
Solving for $v_2^2$:
$v_2^2=2gh$

Finding the relation between $v_1$ and $v_2$
We divide the equation for $v_1^2$ by the equation for $v_2^2$:
$\frac{v_1^2}{v_2^2}=\frac{\frac{10}{7}gh}{2gh}=\frac{5}{7}$
Taking the square root on both sides:
$v_1=\sqrt{\frac{5}{7}}{v}_2$

Since the derived relation $v_1=\sqrt{\frac{5}{7}}{v}_2$ does not match any of the given options ($v_1=v_2$, $v_1=\frac{5v_2}{7}$, or $v_1=\frac{7v_2}{5}$), the correct choice is None of these.