Question

A simple pendulum oscillating in the air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is \[\dfrac{1}{16}th\] of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is:
\[A.\,4T\sqrt{\dfrac{1}{15}}\]
\[B.\,2T\sqrt{\dfrac{1}{10}}\]
\[C.\,T\sqrt{\dfrac{1}{14}}\]
\[D.\,2T\sqrt{\dfrac{1}{14}}\]

Answer 1

Hint: The formula for calculating the time period of the oscillations of a simple pendulum should be used to solve this problem. There are 2 cases involved, in one case, the pendulum is immersed in the liquid and in the other case, the pendulum is not immersed. So, representing these cases in terms of the formula, we will find the period of an oscillation produced by the oscillator immersed in the liquid.
Formula used:
\[T=\dfrac{1}{2\pi }\sqrt{\dfrac{L}{g}}\]

Complete answer:
The formula for computing the time period of the oscillation of an oscillator is given by the formula as follows.
\[T=\dfrac{1}{2\pi }\sqrt{\dfrac{L}{g}}\]
Where L is the length of the string of the pendulum and g is the gravitational constant.
As we know, the gravitational force acts with different magnitude on the different materials. So, the gravitational pull on the oscillator when kept outside will be different from that of the gravitational pull on the oscillator when placed inside some liquid.
So, we have two equations.
The time period of the oscillations of the oscillator when kept outside is given by the formula as follows.
\[T=\dfrac{1}{2\pi }\sqrt{\dfrac{L}{g}}\]…… (1)
The time period of the oscillations of the oscillator when placed inside some liquid is given by the formula as follows.
\[T'=\dfrac{1}{2\pi }\sqrt{\dfrac{L}{g'}}\]
This gravitational pull g’ can be calculated as follows.
\[g'=g\left( 1-\dfrac{{{\rho }_{\text{liquid}}}}{{{\rho }_{\text{body}}}} \right)\]
Now substitute the given value, that is, density of the liquid is\[\dfrac{1}{16}th\]of material of the bob, in the above equation. So, we get,
\[\begin{align}
  & g'=g\left( 1-\dfrac{1}{16} \right) \\
 & \Rightarrow g'=g\dfrac{15}{16} \\
\end{align}\]
Substitute this value of the gravitational pull in the equation of the time period.
\[T'=\dfrac{1}{2\pi }\sqrt{\dfrac{L}{g\left( {}^{15}/{}_{16} \right)}}\] …… (2)
Now divide the equations (1) and (2).
\[\begin{align}
  & \dfrac{T'}{T}=\dfrac{2\pi \sqrt{\dfrac{L}{{}^{15g}/{}_{16}}}}{2\pi \sqrt{\dfrac{L}{g}}} \\
 & \Rightarrow T'=T\dfrac{4}{\sqrt{15}} \\
\end{align}\]
As, the value of the period of oscillation of a bob immersed in the liquid is obtained to be equal to \[\,4T\sqrt{\dfrac{1}{15}}\].

So, the correct answer is “Option A”.

Note:
The density of the liquid refers to change in the value of the gravitational pull. So, in order to confuse, the term density is used. The units of the parameters should be taken care of. Even by giving the values of the parameters, the further question can be extended.