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Step-by-Step Solution
Step 1: Period of the pendulum in air
The period of a simple pendulum of length $l$ in air (or vacuum) is given by the standard formula:
$ T_0 = 2\pi \sqrt{\frac{l}{g}} $
Step 2: Determining the effective acceleration in water
When the pendulum bob is fully immersed in water, there is an upward buoyant force. If the mass of the bob is $m$ and its volume is $V$, the net effective force on the bob becomes:
$ \text{Net force} = mg - \rho_{\text{water}}\,V\,g
$
where $\rho_{\text{water}} = 1000\ \text{kg/m}^3$ is the density of water. The effective acceleration $g_{\text{eff}}$ is then:
$ m\,g_{\text{eff}} = mg - \rho_{\text{water}}\,V\,g \quad\Longrightarrow\quad g_{\text{eff}} = g - \frac{\rho_{\text{water}}\,V\,g}{m}.
$
Step 3: Using the density of the bob to simplify
The problem states that the density of the bob is $\left(\frac{4}{3}\right)\times 1000\ \text{kg/m}^3 = \frac{4}{3}\times 1000 = 1333.3\ \text{kg/m}^3$. Thus,
$ \rho_{\text{bob}} = 1333.3\ \text{kg/m}^3.
$
The mass of the bob is $m = \rho_{\text{bob}}\,V$ and the volume is $V$. Hence,
$ \frac{m}{V} = \rho_{\text{bob}} = 1333.3\ \text{kg/m}^3.
$
Substituting into $g_{\text{eff}} = g - \frac{\rho_{\text{water}}\,V\,g}{m}$:
$ g_{\text{eff}} = g - \frac{1000\,V\,g}{\rho_{\text{bob}}\,V}
= g - \frac{1000\,g}{1333.3} = g - \frac{3}{4}g = \frac{g}{4}.
$
Step 4: Period of the pendulum in water
With this effective acceleration, the period $t$ of the pendulum in water is:
$ t = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{\frac{g}{4}}} = 2\pi \sqrt{\frac{4l}{g}}.
$
Step 5: Relating $t$ and $T_0$
We already know $ T_0 = 2\pi \sqrt{\frac{l}{g}}$. Therefore,
$ t = 2\pi \sqrt{\frac{4l}{g}} = 2\pi \times 2 \sqrt{\frac{l}{g}} = 2 \bigl(2\pi \sqrt{\tfrac{l}{g}}\bigr) = 2\,T_0.
$
Hence, the correct relationship is:
$ t = 2\,T_0.
