Start free

Physics · Electrostatics

NEET (UG) 2025 — Question 22

A balloon is made of a material of surface tension SS and its inflation outlet (from where gas is filled in it) has small area AA. It is filled with a gas of density ρ\rho and takes a spherical shape of radius RR. When the gas is allowed to flow freely out of it, its radius changes from RR to 0 (zero) in time TT. If the speed v(r)v(r) of gas coming out of the balloon depends on rr as rar^{a} and T∝SαAβργRδT \propto S^{\alpha} A^{\beta} \rho^{\gamma} R^{\delta} then

  1. Option A:

    a=12,α=12,β=−1,γ=+1,δ=32a=\frac{1}{2}, \alpha=\frac{1}{2}, \beta=-1, \gamma=+1, \delta=\frac{3}{2}

  2. Option B:

    a=−12,α=−12,β=−1,γ=−12,δ=52a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=-\frac{1}{2}, \delta=\frac{5}{2}

  3. Option C:

    a=−12,α=−12,β=−1,γ=12,δ=72a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=\frac{1}{2}, \delta=\frac{7}{2}

    Correct
  4. Option D:

    a=12,α=12,β=−12,γ=12,δ=72a=\frac{1}{2}, \alpha=\frac{1}{2}, \beta=-\frac{1}{2}, \gamma=\frac{1}{2}, \delta=\frac{7}{2}

Answer: C

Step-by-step solution

T →\to Time \quad [T1][\text{T}^1]

S →\to Surface Tension S=Fℓ=[M1L1T−2][L1]=[M1T−2]\text{S} = \frac{\text{F}}{\ell} = \frac{[\text{M}^1\text{L}^1\text{T}^{-2}]}{[\text{L}^1]} = [\text{M}^1\text{T}^{-2}]

A →\to Area \quad [L2][\text{L}^2] ρ→\rho \to density \quad [M1L−3][\text{M}^1\text{L}^{-3}] R →\to Radius \quad [L1][\text{L}^1]

With Dimensional Analysis →\to T ∝SαAβργRδ\propto \text{S}^{\alpha} \text{A}^{\beta} \rho^{\gamma} \text{R}^{\delta}

[M0L0T1]=[MT−2]α[L2]β[ML−3]γ[L]δ[\text{M}^0 \text{L}^0 \text{T}^1] = [\text{MT}^{-2}]^{\alpha} [\text{L}^2]^{\beta} [\text{ML}^{-3}]^{\gamma} [\text{L}]^{\delta} [M0L0T1]=[Mα+γ,L2β+δ−3γ,T−2α][\text{M}^0 \text{L}^0 \text{T}^1] = [\text{M}^{\alpha+\gamma}, \text{L}^{2\beta+\delta-3\gamma}, \text{T}^{-2\alpha}]

  ⟹  −2α=1α+γ=0\implies -2\alpha = 1 \quad \alpha + \gamma = 0

  ⟹  α=−12γ=−α=12\implies \alpha = -\frac{1}{2} \quad \gamma = -\alpha = \frac{1}{2}

2β+γ−3γ=02\beta + \gamma - 3\gamma = 0

2β+δ−3(12)=02\beta + \delta - 3\left(\frac{1}{2}\right) = 0

2β+δ=322\beta + \delta = \frac{3}{2}

Answer key and solution verified before publishing.

Practise Electrostatics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
NEET (UG) 2025
Paper
2025 paper
Subject
Physics
Chapter
Electrostatics
Topic
Electric Dipole

More Electrostatics questions from this paper