The time period of a simple pendulum is $T$ in air. When the bob is completely immersed in a non-viscous liquid of density $\rho / 10$ (where $\rho$ is the density of the bob), the new time period of oscillation is:
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Upthrust always opposes gravity, meaning $g_{\text{eff}}$ decreases and the pendulum swings slower (time period $T$ increases). This immediately eliminates options where the factor is less than 1.
Step 1: Recall the pendulum period. A simple pendulum has period \[ T = 2\pi\sqrt{\frac{L}{g}} \] It depends on the length $L$ and the effective gravity $g$ pulling the bob down.
Step 2: See what changes in liquid. When the bob is dipped in a liquid, the liquid pushes up on it with a buoyant force. This upward push makes the effective gravity smaller. A smaller effective gravity makes the swing slower.
Step 3: Write the buoyant force. Let the bob have volume $V$ and density $\rho$, so its mass is $V\rho$. The liquid has density $\rho/10$. The upthrust is \[ U = V\left(\frac{\rho}{10}\right)g = \frac{V\rho g}{10} = \frac{mg}{10} \]
Step 4: Find the effective gravity. The net downward force is the weight minus the upthrust. \[ mg_{eff} = mg - \frac{mg}{10} = \frac{9mg}{10} \] So \[ g_{eff} = \frac{9g}{10} \]
Step 5: Compare the two periods. The length stays the same, so \[ \frac{T'}{T} = \sqrt{\frac{g}{g_{eff}}} = \sqrt{\frac{g}{\tfrac{9g}{10}}} = \sqrt{\frac{10}{9}} \]
Step 6: State the new period. Weaker pull means a longer time, which matches the answer being more than $T$. \[ \boxed{T' = T\sqrt{\frac{10}{9}}} \]
