A water tank has a shape of inverted right circular cone whose semi-vertical angle is ⁻¹ ( 1 2 ) . Water is poured into it at constant rate of 5 cubic meter/minute. The rate in meter/minute at which level of water is rising. at the instant when depth of water in the tank is. 10 ~m is

Step-by-step solution

Semi-vertical angle = ⁻¹ ( 1 2 ) Let = ⁻¹ ( 1 2 ) = 1 2 r h = 1 2 r = h 2 Given, d ~V dt =5 ~m ^3 / min . V = Volume of cone Volume of cone = 1 3 r^2 ~h V = 1 3 ( h 2 )^2 h V = 1 12 h ^3 Differentiating w, r.t. t, we get dV dt = 1 12 3 ~h ^2 dh dt 5= 1 4 h^2 dh dt dh dt = 20 h ^2 Now, h =10 ... [Given] dh dt = 20 (10)^2 dh dt = 1 5 Rate of change of water level is 1 5 m / min .