Water is being filled at the rate of 1 cm 3 sec - 1 in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm . When the height of the water level is 10 cm , the rate (in cm 2 sec - 1 ) at which the wet conical surface area of the vessel increases is

Step-by-step solution

Let the volume of the cone be V   c . c . Given d v d t = 1 cc / sec ,   h = 35 cm ,   r = 7 cm i.e. h r = 5 We know for a cone l 2 = r 2 + h 2 Lateral surface area, S = π r r 2 + h 2 S = π h 5 h 2 25 + h 2 = π 26 25 h 2 Also V = 1 3 π r 2 h = 1 3 π h 5 2 h = π 75 h 3 ⇒ d V d t = π 25 h 2 d h d t ⇒ π 25 h 2 d h d t = 1 ⇒ d h d t = 25 π h 2           … i Now d S d t = π 26 25 × 2 h d h d t = 2 26 h