QJEEM23Phb10f: JEE Main 2023 Physics - Class 11 JEEM23Phb10f Solution: For a rolling spherical shell, the ratio of rotational kinetic…

For a rolling spherical shell, we must consider the fact that it has both translational and rotational kinetic energy. The total kinetic energy ( $K_{total}$ ) can be expressed as the sum of the translational kinetic energy ( $K_{trans}$ ) and the rotational kinetic energy ( $K_{rot}$ ):

$K_{total} = K_{trans} + K_{rot}$

The translational kinetic energy of an object with mass (m) and linear velocity (v) is given by:

$K_{trans} = \frac{1}{2}mv^2$

The rotational kinetic energy of a rolling spherical shell with moment of inertia (I) and angular velocity (ω) is given by:

$K_{rot} = \frac{1}{2}Iω^2$

For a rolling object without slipping, the relationship between linear velocity (v) and angular velocity (ω) is:

$v = Rω$

Where R is the radius of the spherical shell.

The moment of inertia for a spherical shell is given by:

$I = \frac{2}{3}mR^2$

Now, we can substitute the moment of inertia and the relationship between linear and angular velocity into the equation for rotational kinetic energy:

$K_{rot} = \frac{1}{2}\left(\frac{2}{3}mR^2\right)\left(\frac{v}{R}\right)^2$

Simplifying the equation:

$K_{rot} = \frac{1}{2}\left(\frac{2}{3}mR^2\right)\frac{v^2}{R^2}$

$K_{rot} = \frac{1}{3}mv^2$

Now, we can find the ratio of rotational kinetic energy to total kinetic energy:

$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3}mv^2}{\frac{1}{2}mv^2 + \frac{1}{3}mv^2}$

Simplifying the equation:

$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3}}{\frac{1}{2} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{5}{6}}$

Multiplying both the numerator and the denominator by 6:

$\frac{K_{rot}}{K_{total}} = \frac{2}{5}$

Comparing this to the given ratio of $\frac{x}{5}$ , we can determine that the value of $x$ is 2.