Q164 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
A simple pendulum oscillating in air has a period of $\sqrt{3}$ second. If it is immersed in a non-viscous liquid, having density $\rho = \left(\dfrac{1}{x}\right)\sigma$, where $\sigma$ is the density of the material of the bob, the new period is $2$ second. The value of $x$ is
- A$6$
- Correct answerB$4$
- C$2$
- D$1$
Solution
The time period of a simple pendulum in air is given by:
$T_1 = 2\pi \sqrt{\dfrac{L}{g}} = \sqrt{3}$
When immersed in a liquid, the effective acceleration due to gravity becomes:
$g_{eff} = g\left(1 - \dfrac{\rho}{\sigma}\right)$
The new time period is:
$T_2 = 2\pi \sqrt{\dfrac{L}{g_{eff}}} = 2$
Taking the ratio of the two time periods:
$\dfrac{T_2}{T_1} = \sqrt{\dfrac{g}{g_{eff}}} = \dfrac{1}{\sqrt{1 - \dfrac{\rho}{\sigma}}}$
Substituting the given values $T_1 = \sqrt{3}$, $T_2 = 2$, and $\rho = \dfrac{\sigma}{x}$:
$\dfrac{2}{\sqrt{3}} = \dfrac{1}{\sqrt{1 - \dfrac{1}{x}}}$
Squaring both sides:
$\dfrac{4}{3} = \dfrac{1}{1 - \dfrac{1}{x}}$
$1 - \dfrac{1}{x} = \dfrac{3}{4}$
$\dfrac{1}{x} = 1 - \dfrac{3}{4} = \dfrac{1}{4}$
$x = 4$
Answer: $4$
$T_1 = 2\pi \sqrt{\dfrac{L}{g}} = \sqrt{3}$
When immersed in a liquid, the effective acceleration due to gravity becomes:
$g_{eff} = g\left(1 - \dfrac{\rho}{\sigma}\right)$
The new time period is:
$T_2 = 2\pi \sqrt{\dfrac{L}{g_{eff}}} = 2$
Taking the ratio of the two time periods:
$\dfrac{T_2}{T_1} = \sqrt{\dfrac{g}{g_{eff}}} = \dfrac{1}{\sqrt{1 - \dfrac{\rho}{\sigma}}}$
Substituting the given values $T_1 = \sqrt{3}$, $T_2 = 2$, and $\rho = \dfrac{\sigma}{x}$:
$\dfrac{2}{\sqrt{3}} = \dfrac{1}{\sqrt{1 - \dfrac{1}{x}}}$
Squaring both sides:
$\dfrac{4}{3} = \dfrac{1}{1 - \dfrac{1}{x}}$
$1 - \dfrac{1}{x} = \dfrac{3}{4}$
$\dfrac{1}{x} = 1 - \dfrac{3}{4} = \dfrac{1}{4}$
$x = 4$
Answer: $4$
Hence, the correct answer is option: 'B'.