A simple pendulum oscillating in air has a period of 3 second. If it is immersed in a non-viscous liquid, having density ρ = ( x 1 ) σ , where σ is the density of the material of the bob, the new

Question

A simple pendulum oscillating in air has a period of 3 ​ second. If it is immersed in a non-viscous liquid, having density ρ = ( x 1 ​ ) σ , where σ is the density of the material of the bob, the new period is 2 second. The value of x is

TestHub · Accepted Answer

The time period of a simple pendulum in air is given by:

$T_{1} = 2 π \frac{L}{g} = 3$

When immersed in a liquid, the effective acceleration due to gravity becomes:

$g_{e ff} = g (1 - \frac{ρ}{σ})$

The new time period is:

$T_{2} = 2 π \frac{L}{g _{e ff}} = 2$

Taking the ratio of the two time periods:

$\frac{T _{2}}{T _{1}} = \frac{g}{g _{e ff}} = \frac{1}{1 - \frac{ρ}{σ}}$

Substituting the given values $T_{1} = 3$ , $T_{2} = 2$ , and $ρ = \frac{σ}{x}$ :

$\frac{2}{3} = \frac{1}{1 - \frac{1}{x}}$

Squaring both sides:

$\frac{4}{3} = \frac{1}{1 - \frac{1}{x}}$

$1 - \frac{1}{x} = \frac{3}{4}$

$\frac{1}{x} = 1 - \frac{3}{4} = \frac{1}{4}$

$x = 4$

Answer: $4$

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