Question

The periodic time of a simple pendulum, (the simple pendulum with its bob is in air) is T. But if the bob of the simple pendulum is completely immersed in a non viscous liquid, whose density is \(\frac{1}{20}\) times the density of material of the bob, then its new period will be

• A. \(T\)
• B. \(\frac{19T}{20}\)
• C. \(\frac{20T}{19}\)
• D. \(\sqrt{\frac{20}{19}}T\)

Hint:

Buoyancy reduces the effective weight of the bob, so the effective g is smaller and the period is longer.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The liquid is non-viscous, so only buoyancy changes the effective gravity.

Step 2: Calculation
$\frac{T'}{T}=\sqrt{\frac{g}{g_{eff}}}=\sqrt{\frac{g}{19g/20}}=\sqrt{\frac{20}{19}}$.

Step 3: Why the other options fail
$T$ ignores buoyancy. $\frac{19T}{20}$ is smaller than $T$, which is impossible for a reduced effective gravity. $\frac{20T}{19}$ forgets the square root.

Final Answer:
The new period is $\sqrt{\frac{20}{19}}\,T$. \[ \boxed{\text{(D) }\sqrt{\dfrac{20}{19}}\,T} \]