MHT CET · Maths · Application of Derivatives
The rate of change volume of a sphere with respect to its surface area when the radius is 5 meters is
- A \(\frac{2}{5}\)
- B 5
- C \(\frac{5}{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & V=\frac{4}{3} \pi r^3 \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=4 \pi r^2 \frac{\mathrm{d} r}{\mathrm{~d} t} \\ & S=4 \pi r^2 \\ & \Rightarrow \frac{\mathrm{d} s}{\mathrm{~d} t}=8 \pi r \frac{\mathrm{d} r}{\mathrm{~d} t} \\ & \text { Now } \frac{\mathrm{d} v}{\mathrm{~d} s}=\frac{\frac{\mathrm{d} v}{\mathrm{~d} t}}{\frac{\mathrm{ds}}{\mathrm{d} t}}=\frac{r}{2}=\frac{5}{2}\end{aligned}\)
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