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: Formula-first approach
Since $T\propto\dfrac{1}{\sqrt{g_{eff}}}$, we get $\dfrac{T'^{2}}{T^{2}}=\dfrac{g}{g_{eff}}$.

Step 2: Substitute
$\dfrac{4}{3}=\dfrac{1}{1-1/x}$, so $1-\dfrac{1}{x}=\dfrac{3}{4}$.

Step 3: Solve
$\dfrac{1}{x}=\dfrac{1}{4}$, hence $x=4$.

Final Answer:
Solving $1-1/x=3/4$ gives $x=4$. \[ \boxed{\text{(B) }4} \]