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NEET 2025
Exam
NEET
Year
2025
Subject
Physics
Chapter
Units and Dimensions

NEET 2025 · Physics · Units and Dimensions

A balloon is made of a material of surface tension S and its...

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 η(r)\eta(r) of gas coming out of the balloon depends on rr as rαr^\alpha and T∝SαAβργRδT \propto S^\alpha A^\beta \rho^\gamma R^\delta then

Options

  1. Aa=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. Ba=−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. Ca=−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}
  4. Da=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}

Correct answer

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}

Step-by-step solution

T→ Time [T1]S→ Surface Tension S=Fℓ=[M1L1T−2][L1]=[M1T−2]A→ Area [L2]ρ→ density [M1L−3]R→ Radius [L1] With  Dimensional Analysis →T∝SαAβργRδ[M0L0T1]=[MT−2]α[L2]β[ML−3Vγ[L]δ[M0L0T1]=[Mα+γ,L2β+δ−3γ,T−2α]⇒−2α=1⇒α=−12⇒α+γ=02β+γ−3γ=02β+δ−3(12)=02β+δ=32\begin{aligned} & \mathrm{T} \rightarrow \text{ Time } \quad\left[\mathrm{T}^1\right] \\ & \mathrm{S} \rightarrow \text{ Surface Tension } \mathrm{S}=\frac{\mathrm{F}}{\ell}=\frac{\left[\mathrm{M}^1 \mathrm{ L}^1 \mathrm{ T}^{-2}\right]}{\left[\mathrm{L}^1\right]}=\left[\mathrm{M}^1 \mathrm{ T}^{-2}\right] \\ & \mathrm{A} \rightarrow \text{ Area } \quad\left[\mathrm{L}^2\right] \\ & \rho \rightarrow \text{ density }\left[\mathrm{M}^1 \mathrm{ L}^{-3}\right] \\ & \mathrm{R} \rightarrow \text{ Radius }\left[\mathrm{L}^1\right] \\ & \text{\text{ With } Dimensional Analysis } \rightarrow \\ & \mathrm{T} \propto \mathrm{S}^\alpha \mathrm{A}^\beta \rho^\gamma \mathrm{R}^\delta \\ & {\left[\mathrm{M}^0 \mathrm{ L}^0 \mathrm{ T}^1\right]=\left[\mathrm{MT}^{-2}\right]^\alpha\left[\mathrm{L}^2\right]^\beta\left[\mathrm{ML}^{-3} \mathrm{ V}^\gamma[\mathrm{L}]^\delta\right.} \\ & {\left[\mathrm{M}^0 \mathrm{ L}^0 \mathrm{ T}^1\right]=\left[\mathrm{M}^{\alpha+\gamma}, \mathrm{L}^{2 \beta+\delta-3 \gamma}, \mathrm{ T}^{-2 \alpha}\right]} \\ & \Rightarrow-2 \alpha=1 \\ & \Rightarrow \alpha=-\frac{1}{2} \\ & \Rightarrow \alpha+\gamma=0 \\ & 2 \beta+\gamma-3 \gamma=0 \\ & 2 \beta+\delta-3\left(\frac{1}{2}\right)=0 \\ & 2 \beta+\delta=\frac{3}{2}\end{aligned}
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