Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height h above the earth surface. Clock-1 and Clock-2 operate at time periods 4 s and 6 s respectively. Then the value of h is (consider radius of earth $$R_E = 6400$$ km and $$g$$ on earth $$10 \ m s^{-2}$$)

We know that the time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{l}{g'}}$$, where $$g'$$ is the effective gravitational acceleration at that location. Since both clocks use identical pendulums (same length $$l$$), the ratio of their time periods depends only on the ratio of gravitational accelerations.

On the Earth's surface, $$g_{\text{earth}} = g$$, and at height $$h$$ above the surface, $$g_h = g\left(\frac{R_E}{R_E + h}\right)^2$$.

The ratio of time periods gives us $$\frac{T_2}{T_1} = \sqrt{\frac{g}{g_h}} = \frac{R_E + h}{R_E}$$.

Substituting $$T_1 = 4$$ s and $$T_2 = 6$$ s, we get $$\frac{6}{4} = \frac{R_E + h}{R_E}$$, which gives $$\frac{3}{2} = \frac{R_E + h}{R_E}$$.

So $$R_E + h = \frac{3}{2}R_E$$, which means $$h = \frac{1}{2}R_E = \frac{1}{2} \times 6400 = 3200$$ km.

Hence, the correct answer is Option C.