A spherical balloon is being inflated at the rate of 35 cc/min. The rate of increase of the surface area of the balloon, when its diameter is 14 cm, is
1) 7 sq cm/min
2) 10 sq cm/min
3) 17.5 sq cm/min
4) 28 sq cm/min

Hint: In this question, we have to find the rate of increase of the surface area of the balloon. For this first, we differentiate the formula of volume of the sphere as the balloon is sphere-shaped. Then by finding the rate of change of velocity, we put the value in the formula of the surface area of the sphere and find out the correct answer.

Formula used:
A sphere's volume is $V=\dfrac{4}{3} \pi \cdot r^{3}$
A sphere's surface area is $S=4 \pi \cdot r^{2}$

Complete step by step Solution:
A sphere has a volume of. $V=\dfrac{4}{3} \pi \cdot r^{3}$.
Differentiating both sides;
$\dfrac{d V}{d t}=\dfrac{4}{3} \pi \cdot 3 r^{2} \dfrac{d r}{d t}$
The balloon is being inflated at a 35 percent rate.
 So $\dfrac{d V}{d t}=35^{\text {. }}$
Therefore, the diameter is given as 14.
Thus, the radius is determined by. $r=7$.
replacing the values in; $\dfrac{d V}{d t}=\dfrac{4}{3} \pi \cdot 3 r^{2} \dfrac{d r}{d t}$;
$35=\dfrac{4}{3} \pi \cdot 3\left(7^{2}\right) \dfrac{d r}{d t}$
$35=4 \pi(49) \dfrac{d r}{d t}$
$\dfrac{d r}{d t}=\dfrac{5}{28 \pi}$
The surface area is given by $S=4 \pi \cdot r^{2}$.
Differentiating on both sides;
$\dfrac{d S}{d t}=4 \pi(2 r) \dfrac{d r}{d t}$
Substituting the values;
$\dfrac{d S}{d t}=4 \pi(2 \cdot 7)\left(\dfrac{5}{28 \pi}\right)$
$\dfrac{d S}{d t}=10$
The rate at which the balloon's surface area is growing $10 \mathrm{sq} \mathrm{cm} / \mathrm{min}$.
Hence, the correct option is 2.

Additional information:
The quantity of space occupied within a sphere is referred to as its volume. Every point on the surface of the sphere is equally spaced from its centre, making it a three-dimensional round solid object. The fixed point and fixed distance are referred to as the sphere's centre and radius, respectively

Note:
In these types of questions, Students must know the formula of volume and surface area, then they are able to solve the questions. Students must know how to differentiate the equation so that they are able to find the answer in less time.