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 as r^a and T ∝ S^α A^β ρ^γ R^δ

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NEET 2025 Official Paper (Held On: 04 May, 2025)

\(a = \frac{1}{2}, \alpha = \frac{1}{2}, \beta = -1, \gamma = +1, \delta = \frac{3}{2}\)
\(a = -\frac{1}{2}, \alpha = -\frac{1}{2}, \beta = -1, \gamma = -\frac{1}{2}, \delta = \frac{5}{2}\)
\(a = -\frac{1}{2}, \alpha = -\frac{1}{2}, \beta = -1, \gamma = \frac{1}{2}, \delta = \frac{7}{2}\)
\(a = \frac{1}{2}, \alpha = \frac{1}{2}, \beta = -\frac{1}{2}, \gamma = \frac{1}{2}, \delta = \frac{7}{2}\)

Answer 1

Calculation:

T ∝ S^a A^β ρ^γ R^δ

M⁰ L⁰ T¹ = K (M T⁻²)^α (L²)^β (M L⁻³)^γ L^δ

M⁰ L⁰ T¹ = K [M^α+γ L^2β−3γ+δ T^−2α]

On comparing

−2α = 1

⇒ α = −1/2

α + γ = 0

⇒ γ = 1/2

2β − 3γ + δ = 0

2β − 3(1/2) + δ = 0

By hit and trial

Put β = −1

2(−1) − 3/2 + δ = 0 ⇒ δ = 7/2

Correct option is: (3) a = −1/2, α = −1/2, β = −1, γ = 1/2, δ = 7/2