The force between two charges is given by Coulomb's law: \[ F = \frac{k q_1 q_2}{r^2}, \] where \( F \) is the force, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them. If the distance \( r \) is doubled, the new distance becomes \( 2r \). The new force \( F' \) is: \[ F' = \frac{k q_1 q_2}{(2r)^2} = \frac{k q_1 q_2}{4r^2} = \frac{F}{4}. \] Thus, the force decreases by a factor of 4 when the distance is doubled. Hence, the correct answer is \( \boxed{(1)} \).
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,

The correct relation between $\gamma=\frac{ c _{ p }}{ c _{ v }}$ and temperature $T$ is :
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 : \(A_2 \rightleftharpoons 2A\)