Search any question & find its solution
Question:
Answered & Verified by Expert
If the rate of change in volume of spherical soap bubble is uniform, then the rate of change of surface area varies as
Options:
Solution:
2291 Upvotes
Verified Answer
The correct answer is:
inversely proportional to radius
Let volume $=\mathrm{V}=\frac{4}{3} \pi r^{3}$ and surface area $=\mathrm{S}=4 \pi r^{2}$
Now, $(1) \Rightarrow \frac{d v}{d t}=\frac{4}{3} \times 3 \pi r^{2} \times \frac{d r}{d t}$
$\quad=4 \pi r^{2} \frac{d r}{d t}$
(2) $\Rightarrow \frac{d s}{d t}=4 \pi \times 2 \times r \frac{d r}{d t}=\frac{8 \pi r^{2}}{r} \frac{d r}{d t}$
$=\frac{2}{r}\left[4 \pi r^{2} \frac{d r}{d t}\right]=\frac{2}{r} \frac{d v}{d t} \quad($ from 3$)$
Now, $(1) \Rightarrow \frac{d v}{d t}=\frac{4}{3} \times 3 \pi r^{2} \times \frac{d r}{d t}$
$\quad=4 \pi r^{2} \frac{d r}{d t}$
(2) $\Rightarrow \frac{d s}{d t}=4 \pi \times 2 \times r \frac{d r}{d t}=\frac{8 \pi r^{2}}{r} \frac{d r}{d t}$
$=\frac{2}{r}\left[4 \pi r^{2} \frac{d r}{d t}\right]=\frac{2}{r} \frac{d v}{d t} \quad($ from 3$)$
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.