Question

A spherical balloon is pumped at the constant rate of

3 m³/min. The rate of increase of it's surface area at certain instant is found to be 5 m²/ min. At this instant it's radius is equal to

• A. $\frac{1}{5}$m
• B. $\frac{3}{5}$m
• C. $\frac{6}{5}$m
• D. $\frac{2}{5}$m

Answer 1

Answer: (C)

$\frac{6}{5}$m

Answer 2

V = $\frac { 4 } { 3 }$πr³ ⇒ $\frac { d v } { d t } = 4 \pi r ^ { 2 } \cdot \frac { d r } { d t }$

Also S = 4πr² ⇒ $\frac { d s } { d t } = 8 \pi r \cdot \frac { d r } { d t }$

where $\frac { d v } { d t }$ = 3, = 5,

We get $\frac { r } { 2 } = \frac { 3 } { 5 }$ ⇒ r = $\frac { 6 } { 5 }$ m.