Physics Q24 – NEET-UG 2025

Home Physics Mechanical Properties of FluidsQuestion #265

A balloon is made of a material of surface tension

S

and its inflation outlet (from where gas is filled in it) has small area

A

. It is filled with a gas of density

ρ

and takes a spherical shape of radius

R

. When the gas is allowed to flow freely out of it, its radius

r

changes from

R

to 0 (zero) in time

T

. If the speed

v (r)

of gas coming out of the balloon depends on

r

r^{a}

and

T \propto S^{α} A^{β} ρ^{γ} R^{δ}

then

Clue 1 of 3: Conceptual Principle

The problem involves the efflux of gas from a balloon due to internal pressure, which is governed by surface tension. This is a classic application of fluid dynamics and surface tension.

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● Video Walkthrough

Step 1: Determine the efflux speed

v (r)

and exponent

a

✦ Active

The excess pressure inside a spherical balloon of radius $r$ due to surface tension $S$ is given by $P_{e x c e s s} = \frac{2 S}{r}$ . According to Torricelli's law, the speed of gas efflux $v$ through a small opening is $v = \sqrt{\frac{2 P_{e x c e s s}}{ρ}}$ , where $ρ$ is the gas density.

v = \sqrt{\frac{2 (2 S / r)}{ρ}} = \sqrt{\frac{4 S}{ρ r}} = 2 \sqrt{\frac{S}{ρ r}} = 2 S^{1 / 2} ρ^{- 1 / 2} r^{- 1 / 2}

Comparing this with the given dependence $v \propto r^{a}$ , we find that $a = - \frac{1}{2}$ .

💡 Teacher's Secret Hint

Remember that for a liquid drop or a bubble, the pressure difference is $2 S / R$ or $4 S / R$ respectively. For a balloon, it's typically considered as a single surface, hence $2 S / r$ for the excess pressure.

Step 2: Calculate the total time

T

for deflation○ Expand

The volume of the spherical balloon is $V = \frac{4}{3} π r^{3}$ . The rate of change of volume is $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ . The rate of gas flowing out through the outlet of area $A$ is $A v$ . Since the volume is decreasing, we have $\frac{d V}{d t} = - A v$ .

4 π r^{2} \frac{d r}{d t} = - A v ⟹ d t = - \frac{4 π r^{2}}{A v} d r

Substitute the expression for $v$ from Step 1:

d t = - \frac{4 π r^{2}}{A (2 S^{1 / 2} ρ^{- 1 / 2} r^{- 1 / 2})} d r = - \frac{2 π}{A} S^{- 1 / 2} ρ^{1 / 2} r^{2 - (- 1 / 2)} d r = - \frac{2 π}{A} S^{- 1 / 2} ρ^{1 / 2} r^{5 / 2} d r

To find the total time $T$ for the radius to change from $R$ to $0$ , we integrate:

T = \int_{R}^{0} d t = \int_{R}^{0} - \frac{2 π}{A} S^{- 1 / 2} ρ^{1 / 2} r^{5 / 2} d r = \frac{2 π}{A} S^{- 1 / 2} ρ^{1 / 2} \int_{0}^{R} r^{5 / 2} d r

Evaluating the integral:

T = \frac{2 π}{A} S^{- 1 / 2} ρ^{1 / 2} {[\frac{r^{7 / 2}}{7 / 2}]}_{0}^{R} = \frac{2 π}{A} S^{- 1 / 2} ρ^{1 / 2} \frac{2}{7} R^{7 / 2} = \frac{4 π}{7} A^{- 1} S^{- 1 / 2} ρ^{1 / 2} R^{7 / 2}

💡 Teacher's Secret Hint

Pay close attention to the sign convention for $d V / d t$ and the limits of integration. The negative sign ensures that $T$ is positive as $r$ decreases.

Step 3: Determine the exponents

α, β, γ, δ

○ Expand

From the derived expression for $T$ , we have $T \propto A^{- 1} S^{- 1 / 2} ρ^{1 / 2} R^{7 / 2}$ . Comparing this with the given proportionality $T \propto S^{α} A^{β} ρ^{γ} R^{δ}$ , we can identify the exponents:

α = - \frac{1}{2}

β = - 1

γ = \frac{1}{2}

δ = \frac{7}{2}

Combining the results from Step 1 and Step 3, we have $a = - \frac{1}{2}, α = - \frac{1}{2}, β = - 1, γ = \frac{1}{2}, δ = \frac{7}{2}$ . This set of values matches option (3).

💡 Teacher's Secret Hint

Ensure all exponents are correctly matched to the corresponding physical quantities. A common mistake is to mix up the powers or signs.

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Physics Question 24 – NEET-UG 2025

Clue 1 of 3: Conceptual Principle

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