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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Physics
Chapter
Mechanical Properties of Fluids

JIPMER 2019 · Physics · Mechanical Properties of Fluids

If ratio of terminal velocity of two drops falling in air is 3: 4 ,...

If ratio of terminal velocity of two drops falling in air is 3:43: 4, then what is the ratio of their surface area?

Options

  1. A23\frac{2}{3}
  2. B34\frac{3}{4}
  3. C43\frac{4}{3}
  4. D32\frac{3}{2}

Correct answer

B. 34\frac{3}{4}

Step-by-step solution

Terminal velocity is given by v=2r2(ρ−σ)9ηgv=\frac{2 r^2(\rho-\sigma)}{9 \eta} g where, r=r= radius of drop, ρ=\rho= density of medium of drop, σ=\sigma= density of surrounding medium η=\eta= coefficient of viscocity of drop medium. ⇒v∝r2\Rightarrow \mathrm{v} \propto \mathrm{r}^2, we know area of drop ∝r2\propto \mathrm{r}^2 ⇒v∝A\Rightarrow \quad \mathrm{v} \propto \mathrm{A} (area) ⇒v1v2=A1 A2=34\Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{3}{4} ⇒A1 A2=34\Rightarrow \quad \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{3}{4}
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