Search any question & find its solution
Question:
Answered & Verified by Expert
If ratio of terminal velocity of two drops falling in air is $3: 4$, then what is the ratio of their surface area?
Options:
Solution:
2519 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{4}$
Terminal velocity is given by $v=\frac{2 r^2(\rho-\sigma)}{9 \eta} g$
where, $r=$ radius of drop,
$\rho=$ density of medium of drop,
$\sigma=$ density of surrounding medium
$\eta=$ coefficient of viscocity of drop medium.
$\Rightarrow \mathrm{v} \propto \mathrm{r}^2$, we know area of drop $\propto \mathrm{r}^2$
$\Rightarrow \quad \mathrm{v} \propto \mathrm{A}$ (area) $\Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{3}{4}$
$\Rightarrow \quad \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{3}{4}$
where, $r=$ radius of drop,
$\rho=$ density of medium of drop,
$\sigma=$ density of surrounding medium
$\eta=$ coefficient of viscocity of drop medium.
$\Rightarrow \mathrm{v} \propto \mathrm{r}^2$, we know area of drop $\propto \mathrm{r}^2$
$\Rightarrow \quad \mathrm{v} \propto \mathrm{A}$ (area) $\Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{3}{4}$
$\Rightarrow \quad \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{3}{4}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.