The moment of inertia of a hollow sphere about its diameter axis is given by:
\[ I_{\text{sphere}} = \frac{2}{3} MR^2 = M k_1^2 \]
where \(k_1\) is the radius of gyration of the hollow sphere.
The moment of inertia of a solid cylinder about its diameter axis is:
\[ I_{\text{cylinder}} = \frac{1}{12} M(4R^2) + \frac{1}{4} MR^2 + M(2R)^2 = \frac{67}{12} MR^2 = M k_2^2 \]
Calculating the ratio of the radii of gyration:
\[ \frac{k_1}{k_2} = \sqrt{\frac{\frac{2}{3}}{\frac{67}{12}}} = \sqrt{\frac{8}{67}} \]
Hence, the value of \(x\) is 67.
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).