The bob of a simple pendulum executes simple harmonic motion in water with a period $$t$$, while the period of oscillation of the bob is $$t_0$$ in air. Neglecting frictional force of water and given that the density of the bob is $$\left(\frac{4}{3}\right) \times 1000 \text{ kg/m}^3$$. What relationship between $$t$$ and $$t_0$$ is true?

The time period of a simple pendulum in air is given by:

$$ t_0 = 2\pi \sqrt{\frac{l}{g}} $$

When the pendulum is immersed in water, it experiences an upward buoyant force, which reduces the effective acceleration due to gravity ($$g_{\text{eff}}$$).

$$ g_{\text{eff}} = g \left( 1 - \frac{\rho_{\text{water}}}{\rho_{\text{bob}}} \right) $$

Given the density of the bob is $$\rho_{\text{bob}} = \frac{4}{3} \times 1000 \text{ kg/m}^3$$ and knowing the density of water is $$\rho_{\text{water}} = 1000 \text{ kg/m}^3$$, we can find the ratio:

$$ \frac{\rho_{\text{water}}}{\rho_{\text{bob}}} = \frac{1000}{\frac{4}{3} \times 1000} = \frac{3}{4} $$

Substitute this into the $$g_{\text{eff}}$$ equation:

$$ g_{\text{eff}} = g \left( 1 - \frac{3}{4} \right) = \frac{g}{4} $$

Now, the new time period $$t$$ in water is:

$$ t = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{g/4}} $$

$$ t = 2\pi \sqrt{\frac{4l}{g}} = 2 \left( 2\pi \sqrt{\frac{l}{g}} \right) $$

Since $$t_0 = 2\pi \sqrt{\frac{l}{g}}$$, we get:

$$ t = 2t_0 $$

 Option (C) $$t = 2t_0$$