 Welcome to the session, let us understand the following question which says, the radius of an air bubble is increasing at the rate of 1 by 2 cm per second at what rate is the volume of the bubble increasing when the radius is 1 cm. Now let us proceed on to the solution. Let R be the radius of the air bubble, B be the volume, it is given to us that radius of an air bubble is increasing at the rate of 1 by 2 cm per second that is dr by dt is equal to 1 by 2 and we have to find at what rate the volume of bubble is increasing that is dv by dt, we know volume of an air bubble that is a square is given by v is equal to 4 by 3 pi r cube, now differentiating it with respect to t we get dv by dt is equal to 4 by 3 pi multiplied by 3 r square dr by dt which is equal to this 3 and 3 gets cancelled so 4 pi r square dr by dt is given to us as 1 by 2 multiplied by 1 by 2, this 4 gets cancelled by 2 and we get here 2 so it is equal to 2 pi r square therefore dv by dt is equal to 2 pi r square, now we have to find dv by dt when radius is equal to 1 cm therefore dv by dt at r is equal to 1 is equal to 2 pi multiplied by 1 multiplied by 1 which is equal to 2 pi hence volume is increasing at the rate of 2 pi cm cube per second which is the required answer I hope you understood this question buy and have a nice day.