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Step-by-Step Solution
Step 1: Original Time Period of the Simple Pendulum
For a simple pendulum of length $l$ under gravity $g$, the time period $T$ is given by:
$T = 2\pi \sqrt{\frac{l}{g}}$.
Step 2: Effective Weight When Immersed in a Liquid
When the bob is immersed in a liquid of density $\sigma$, it experiences a buoyant force. If the density of the bob is $\rho$ and is given that $\sigma = \frac{1}{4}\rho$, then the effective weight of the bob becomes:
$\displaystyle W_{\text{eff}} = mg - \text{Buoyant force}.$
The buoyant force is
$\displaystyle \text{Buoyant force} = v \,\sigma\, g$,
where $v$ is the volume of the bob. Since the mass of the bob is $m=\rho\,v$, we have $v = \frac{m}{\rho}$. Substituting $\sigma = \frac{\rho}{4}$:
$\displaystyle \text{Buoyant force} = \left(\frac{m}{\rho}\right)\left(\frac{\rho}{4}\right)g = \frac{m}{4}g.$
Therefore,
$\displaystyle W_{\text{eff}} = mg - \frac{m}{4}g = \frac{3mg}{4}.$
Hence the effective acceleration due to gravity when the bob is immersed in the liquid is
$\displaystyle g_{\text{eff}} = \frac{3g}{4}.$
Step 3: New Length of the Thread
The length of the thread is increased by one-third of the original length. Thus, the new length $l_1$ is:
$\displaystyle l_1 = l + \frac{l}{3} = \frac{4l}{3}.$
Step 4: New Time Period of the Pendulum
With the new effective acceleration $g_{\text{eff}} = \frac{3g}{4}$ and the new length $l_1 = \frac{4l}{3}$, the new time period $T_1$ is given by the same pendulum formula but with these modified values:
$\displaystyle T_1 = 2\pi \sqrt{\frac{l_1}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{\frac{4l}{3}}{\frac{3g}{4}}}.$
Simplify inside the square root:
$\displaystyle T_1 = 2\pi \sqrt{\frac{\frac{4l}{3}}{\frac{3g}{4}}}
= 2\pi \sqrt{\frac{4l}{3} \cdot \frac{4}{3g}}
= 2\pi \sqrt{\frac{16l}{9g}}
= 2\pi \cdot \frac{4}{3} \sqrt{\frac{l}{g}}.$
Notice that $2\pi\,\sqrt{\frac{l}{g}}$ is the original time period $T$. Hence,
$\displaystyle T_1 = \frac{4}{3} T.$
Step 5: Conclusion
Therefore, when the bob is immersed in the liquid of density one-fourth of the bob's density and the thread length is increased by one-third of its original length, the new time period becomes:
$\displaystyle \boxed{\frac{4}{3} T}.$
