The figure shows a cylindrical tank of base radius r = 7 10 m . The height h (in m ) of water in the tank is maintained by controlling the inlet volume flow rate V i (in m 3 m i n ) and outlet volume flow rate V o (in m 3 m i n ) of water. V i and V o are recorded during time t ∈ ( 0 , 6 ) (in m i n ) as V i = 1 50 ( t 2 + 14 t ) and V 0 = 1 50 ( − t 2 − 6 t ) respectively. Given that for a small time interval V ′ = rate of change of volume of water in the tank h ′ = rate of change of height of water in the tank V ′ = π × r 2 h ′ = inlet rate of water − outlet rate of water If h ′ = k ( a 2 t 2 + a 1 t + a 0 ) , then h at t would be h = k ( a 2 t 3 3 + a 1 t 2 2 + a 0 t ) , where k is a constant, a 0 , a 1 , a 2 are real numbers. Find h (till two decimal places) at t = 2 . (Take π = 22/ 7 )
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The answer is 0.59
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