The formula for escape velocity is: \[ v_e = \sqrt{\frac{2GM}{R}}, \] where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet. From the formula, it is clear that \( v_e \propto \sqrt{\frac{M}{R}} \). - As the ratio \( \frac{M}{R} \) increases, the escape velocity \( v_e \) increases. Hence, Statement I is correct.
- However, \( v_e \) depends on \( R \) as seen from the formula, so escape velocity is not independent of the radius of the planet.
Hence, Statement II is incorrect. Thus, the correct answer is \( \boxed{(3)} \).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)