Question

An expendable balloon filled with air is floating on the top surface of a lake, with $2/3$ of its volume submerged in water. How deep $\left( \text{in m} \right)$ should it be sunk in water without changing the temperature of air in it, in order that it is just in equilibrium, neither sinking nor rising further? Height of water barometer $=10m$.

Answer 1

Hint: This problem can be solved by applying Archimedes’ principle to find the volume of the balloon. This volume changes when the balloon is immersed further according to Boyle’s law as the pressure of the air inside also changes. The new buoyant force experienced in this position at equilibrium can help us to find a relation for the depth of immersion of the balloon.
Formula used:
${{F}_{B}}=-{{W}_{fluid}}$
$PV=\text{constant}$
$\sum{\overrightarrow{F}}=0$
$P=h\rho g$

Complete step by step answer:
We will solve this problem by first finding out the volume of the balloon in the initial state and then the volume in the fully immersed state. Using these values and the upthrust in both the states, we can get a relation for the depth of immersion using the force equilibrium equation. Hence, let us proceed to do that.

Let the mass of the balloon be $M$.
Therefore, the weight of the balloon is $W=Mg$, --(1)
where $g$ is the acceleration due to gravity.
Let the volume of the balloon when it is floating with two-thirds of its volume submerged in water be $V$.
Let the mass of water displaced in this state be ${{M}_{water}}$.
Therefore, the weight of the water displaced is ${{W}_{water}}={{M}_{water}}g$.
Since two thirds of the volume if the balloon is submerged in water, the same volume of
water is displaced.

Hence, the volume of water displaced is ${{V}_{water}}=\dfrac{2}{3}V$. --(2)

Let the density of water be $\rho$.
When a body is immersed partially or fully in a fluid, the buoyant force ${{F}_{B}}$ exerted on it by the fluid is

${{F}_{B}}=-{{W}_{fluid}}$
$\therefore \left| {{F}_{B}} \right|=\left| -{{W}_{fluid}} \right|$
$\therefore \left| {{F}_{B}} \right|=\left| {{W}_{fluid}} \right|$ --(3)

where ${{W}_{fluid}}$ is the weight of the fluid displaced and the negative sign implies that the buoyant force acts in a direction opposite to the weight.
Using (3), we get, the magnitude of the upthrust provided by the water on the balloon will be

${{F}_{B}}={{M}_{water}}g$ --(4)

Now, for a body at equilibrium, the net sum of the external forces $\overrightarrow{F}$ on it is zero.

$\sum{\overrightarrow{F}}=0$ --(5)

Since, the balloon is at equilibrium in this state, its downward weight must be balanced by the upthrust on it. Hence, using (1), (4) and (5) and considering downward forces as negative, we get

${{F}_{B}}-W=0$
$\therefore {{W}_{water}}=W$
$\therefore {{M}_{water}}g=Mg$
$\therefore {{M}_{water}}=M$
$\therefore {{V}_{water}}\rho =M$ $\left( \because \text{Mass = Density }\times \text{Volume} \right)$
$\therefore \dfrac{2}{3}V\rho =M$ [Using (2)] --(6)

Now, the balloon is taken to a depth $h$ below the surface.

Let the pressure exerted by the water when the balloon was at the surface be $P$ and the pressure on the balloon at depth $h$ be $P'$. Now, this pressure is equal to the pressure of the air inside since the balloon is at equilibrium in both these positions.
Let the volume of the balloon at the depth $h$ be $V'$.
According to Boyle’s law, when the temperature of an enclosed ideal gas remains constant, the product of its pressure and volume also remains constant.

$PV=\text{constant}$ --(7)

where $P,V$ are the pressure and volume of the internal gas respectively.
Since, the temperature of the air inside the balloon is said to remain constant and it is considered as an ideal gas, using (7), we get

$PV=P'V'$ --(8)

The pressure $P$ exerted by a liquid of density $\rho$ at a depth $h$ below its surface is given by

$P=h\rho g$ --(9)

where $g$ is the acceleration due to gravity.

Now, according to the question the pressure at the surface is given by the water barometer whose height is given to be $10m$. This means that the atmospheric pressure (pressure at the surface) is equal to the pressure exerted by water alone on an object at a depth of $10m$.

Hence, using (9), we get,

$P=10\rho g$ (atmospheric pressure) --(10)
At the depth $h$, the pressure exerted by the water on the balloon will be the sum of the atmospheric pressure and the gauge pressure applied by the water alone at the depth.
Hence, using (9) and (10), we get,

$P'=10\rho g+h\rho g=\left( 10+h \right)\rho g$ --(11)

Putting (10) and (11) in (8), we get,

$10\rho gV=\left( 10+h \right)\rho gV'$
$\therefore 10V=\left( 10+h \right)V'$
$\therefore V'=\left( \dfrac{10}{10+h} \right)V$ --(12)

$V'$ is also the volume of the water displaced since the full volume of the balloon is immersed in the water.

Therefore, weight of the water displaced will be

${{W}_{water}}'=V'\rho g$ --(13)

Using (3) and (13), we get, the magnitude of the upthrust on the balloon at this point to be

${{F}_{B}}'=V'\rho g$ --(14)

Since, the balloon is at equilibrium, the weight of the balloon at the upthrust on it must cancel each other out. The weight of the balloon remains the same as in the first case since the mass of the balloon does not change as the quantity of air inside it is the same.

Therefore, using (5), (14) and (1) and considering downward forces to be negative, we get,

${{F}_{B}}'-W=0$
$\therefore {{F}_{B}}'=W$
$\therefore V'\rho g=Mg$
$\therefore V'\rho =M$
$\therefore V'\rho =\dfrac{2}{3}V\rho$ [Using (6)]
$\therefore V'=\dfrac{2}{3}V$
$\therefore \left( \dfrac{10}{10+h} \right)V=\dfrac{2}{3}V$ [Using (12)]
$\therefore \dfrac{10}{10+h}=\dfrac{2}{3}$
$\therefore 10\times 3=2\left( 10+h \right)$
$\therefore 30=20+2h$
$\therefore 30-20=2h$
$\therefore 2h=10$
$\therefore h=\dfrac{10}{2}=5m$

Hence, the required depth of immersion is $5m$.

Note: This is a complex problem, which does not hit the intuition of students at the first glance. Students often do not understand how to proceed and cannot catch that the usage of Boyle’s law is required to solve this problem. However, students must notice that whenever problems involving floatation of bodies in a fluid are given and the bodies are some kind of ideal gases or related to it, there is always a chance that the problem has to be solved by using the ideal gas equation or the equations for the physical properties of ideal gases. This is a common concept implemented in questions especially in competitive exams to check whether students can handle and implement multiple concepts from seemingly different topics together to solve the question or not.