The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec. Find the rate of increase of its surface area, when the radius is 7 cm.

Asked by Aaryan | 1 year ago |  42

##### Solution :-

We know that a soap bubble will be in sphere shape

Consider the radius of soap bubble as $$r \dfrac{dr}{dt}$$ = 0.2 cm/s

The surface area of soap bubble = 4πr2

So the rate of change of surface area = 8πr

By substituting the values = 8 × 3.14 × 7 × 0.2

We get,

$$\dfrac{dS}{dt}$$= 35.2 cm2/s

Answered by Sakshi | 1 year ago

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