Question

A simple pendulum oscillating in air has a period of \(\sqrt{3}\) second. If it is immersed in a non-viscous liquid, having density \(\rho =\left(\frac{1}{x}\right)\sigma\), where \(\sigma\) is the density of the material of the bob, the new period is \(2\) second. The value of \(x\) is

• A. \(6\)
• B. \(4\)
• C. \(2\)
• D. \(1\)

Hint:

Buoyancy reduces the effective gravity, so use the period ratio to find the density ratio.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The period of a simple pendulum is \(T=2\pi\sqrt{\dfrac{l}{g_{eff}}}\). In a non-viscous liquid, buoyancy reduces the effective weight of the bob.

Step 2: Effective gravity
\(g_{eff}=g\left(1-\dfrac{\rho}{\sigma}\right)=g\left(1-\dfrac{1}{x}\right)\).

Step 3: Use the periods
\(\dfrac{T'}{T}=\sqrt{\dfrac{g}{g_{eff}}}=\dfrac{2}{\sqrt{3}}\).
Squaring: \(\dfrac{1}{1-1/x}=\dfrac{4}{3}\), so \(1-\dfrac{1}{x}=\dfrac{3}{4}\) and \(\dfrac{1}{x}=\dfrac{1}{4}\).

Step 4: Check
With \(x=4\), \(g_{eff}=\dfrac{3g}{4}\) and the period is \(\sqrt{3}\times\dfrac{2}{\sqrt{3}}=2\) s.

Final Answer:
The value of \(x\) is 4. \[ \boxed{\text{(B) }4} \]