To find the ratio of the kinetic energies of rotation of the sphere and the cylinder, we need to calculate the kinetic energy for each of them and then compare their values.
The moment of inertia $I_{\text{sphere}}$ of a solid sphere rotating about its diameter is given by:
$$I_{\text{sphere}} = \frac{2}{5} m R^2$$where $m$ is the mass and $R$ is the radius of the sphere.
The rotational kinetic energy $E_{\text{sphere}}$ of the sphere is calculated using the formula:
$$E_{\text{sphere}} = \frac{1}{2} I_{\text{sphere}} \omega^2$$Substituting $I_{\text{sphere}}$:
$$E_{\text{sphere}} = \frac{1}{2} \left(\frac{2}{5} m R^2\right) \omega^2 = \frac{1}{5} m R^2 \omega^2$$The moment of inertia $I_{\text{cylinder}}$ of a solid cylinder rotating about its geometrical axis is given by:
$$I_{\text{cylinder}} = \frac{1}{2} m R^2$$Given that the angular speed of the cylinder $2\omega$ is twice that of the sphere, the rotational kinetic energy $E_{\text{cylinder}}$ is calculated as:
$$E_{\text{cylinder}} = \frac{1}{2} I_{\text{cylinder}} (2\omega)^2$$Substituting $I_{\text{cylinder}}$ and simplifying:
$$E_{\text{cylinder}} = \frac{1}{2} \left(\frac{1}{2} m R^2\right) (2\omega)^2 = \frac{1}{2} \times \frac{1}{2} m R^2 \times 4\omega^2 = \frac{1}{2} m R^2 \omega^2$$Now, calculate the ratio $(E_{\text{sphere}} / E_{\text{cylinder}})$:
$$\text{Ratio} = \frac{E_{\text{sphere}}}{E_{\text{cylinder}}} = \frac{\frac{1}{5} m R^2 \omega^2}{\frac{1}{2} m R^2 \omega^2} = \frac{1}{5} \div \frac{1}{2} = \frac{2}{5}$$Thus, the ratio is 1:5.
Therefore, the correct answer is 1 : 5.
The center of mass of a thin rectangular plate (fig - x) with sides of length \( a \) and \( b \), whose mass per unit area (\( \sigma \)) varies as \( \sigma = \sigma_0 \frac{x}{ab} \) (where \( \sigma_0 \) is a constant), would be 