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# A simple pendulum has a time period $T = 2$ sec in air. If the whole arrangement is placed in a non viscous liquid whose density is $1/2$ times the density of bob. The time period of the simple pendulum in the liquid will be(A) $\dfrac{1}{{\sqrt 2 }}$ sec(B) $4$ sec(C) $2\sqrt 2$ sec(D) $4\sqrt 2$ sec

Last updated date: 21st Jun 2024
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Hint: To solve this question, we need to use the formula for the time period of a simple pendulum. Then, we have to find out the effective acceleration due to gravity in the liquid. Substituting it in the formula for the time period, we will get the final answer.

Formula used: The formula used for solving this question is given by
$T = 2\pi \sqrt {\dfrac{l}{{{g_{eff}}}}}$, here $T$ is the time period of a simple pendulum whose string has a length of $l$, and ${g_{eff}}$ is the effective acceleration due to gravity.

Complete step-by-step solution:
Let the density of the bob of the simple pendulum be $\rho$ and that of the given non-viscous liquid be $\sigma$. According to the question, the density of the liquid is half of the density of the bob. So we can write
$\sigma = \dfrac{\rho }{2}$.............(1)
Now, let the length of the string of the given pendulum be $l$. We know that the time period of a simple pendulum is given by
$T = 2\pi \sqrt {\dfrac{l}{{{g_{eff}}}}}$..........(2)
According to the question, $T = 2s$ in air. We know that the effective acceleration due to gravity in air is equal to $g$. Therefore, substituting these in (2) we get
$2 = 2\pi \sqrt {\dfrac{l}{g}}$
Or
$2\pi \sqrt {\dfrac{l}{g}} = 2$ …………. (3)
When the pendulum is dipped inside the liquid, then the liquid will apply an upward force on the bob. We know from the Archimedes’ Principle that the magnitude of this upward force is equal to the weight of the fluid displaced. Since the whole of the bob is inside the liquid, the volume displaced by the bob is equal to the volume of the bob. If the mass of the bob is equal to $m$, then its volume can be given by
$V = \dfrac{m}{\rho }$ ……….(4)
Since the density of the fluid is $\sigma$, so the mass of the displaced fluid becomes
$m' = V\sigma$
From (4)
$m' = \dfrac{{m\sigma }}{\rho }$
So the weight of the fluid displaced, or the upward force on the bob becomes
$U = m'g$
$\Rightarrow U = \dfrac{{m\sigma }}{\rho }g$...................(5)
The weight of the bob which will act vertically downwards on the bob is given by
$W = mg$...............(6)
So the net downward force on the bob becomes
$F = W - U$
Putting (5) and (6) in the above equation, we get
$F = mg - \dfrac{{m\sigma }}{\rho }g$
$F = mg\left( {1 - \dfrac{\sigma }{\rho }} \right)$..................(7)
The effective acceleration due to gravity is equal to the net downward force on the body per unit mass. So the effective acceleration due to gravity becomes
${g_{eff}} = \dfrac{F}{m}$
From (7)
${g_{eff}} = \dfrac{{mg}}{m}\left( {1 - \dfrac{\sigma }{\rho }} \right)$
$\Rightarrow {g_{eff}} = \left( {1 - \dfrac{\sigma }{\rho }} \right)g$
Putting (1) in the above equation, we get
${g_{eff}} = \left( {1 - \dfrac{1}{2}} \right)g$
$\Rightarrow {g_{eff}} = \dfrac{g}{2}$
Substituting this in (2) we get the required time period of the given pendulum as
$T' = 2\pi \sqrt {\dfrac{l}{{g/2}}}$
$T' = 2\pi \sqrt {\dfrac{l}{g}} \sqrt 2$
Substituting (3) in the above equation, we finally get
$T' = 2\sqrt 2 s$
Thus, the time period of the simple pendulum in the liquid will be equal to $2\sqrt 2$ sec.

Hence the correct answer is option C.

Note: Do not think that the upward force on the bob will make the string of the pendulum loose. This is because the net force on the bob is still downwards not upwards.