\(6 \times 10^6 km\)
\(3 \times 10^6 km\)
\(6 \times 10^7 km\)
\(3 \times 10^7 km\)
Using Kepler’s third law:
\( T^2 \propto R^3 \) or \( \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \)
Given:
\( T_1 = 1 \, \text{year}, \quad R_1 = 1.5 \times 10^6 \, \text{km}, \quad T_2 = 2.83 \, \text{years} \)
Substitute:
\[ \frac{1}{(2.83)^2} = \frac{(1.5 \times 10^6)^3}{R_2^3} \]
\[ R_2^3 = \left[ (2.83)^2 \cdot (1.5 \times 10^6)^3 \right] \]
Taking the cube root:
\[ R_2 = 3 \times 10^6 \, \text{km} \]
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$^{-1}$)