The moment of inertia \( I \) of each sphere about the central axis (using the parallel axis theorem) is:
\[ I_{\text{total}} = 2 \left( I_{\text{sphere}} + md^2 \right). \]For a solid sphere:
\[ I_{\text{sphere}} = \frac{2}{5}mR^2 = \frac{2}{5} \times 2 \times (0.5)^2 = 0.2 \, \text{kg m}^2. \]Distance \( d \) from the center of each sphere to the midpoint of the rod is \( 0.75 \, \text{m} \).
So,
\[ I_{\text{total}} = 2 \left( 0.2 + 2 \times (0.75)^2 \right) = 2 \left( 0.2 + 1.125 \right) = \frac{53}{20} \, \text{kg m}^2. \]Thus, \( x = 53 \).
A, B and C are disc, solid sphere and spherical shell respectively with the same radii and masses. These masses are placed as shown in the figure.
The moment of inertia of the given system about PQ is $ \frac{x}{15} I $, where $ I $ is the moment of inertia of the disc about its diameter. The value of $ x $ is:
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)