Consider the configuration of a stationary water tank of cross-section area A 0 and a small bucket as shown in figure below; What should be the speed v of the bucket, so that the water leaking out of a hole of cross-section area A (as shown) from the water tank does not fall outside the bucket? (Take, h = 5 m , H = 5 m , g = 10 m / s 2 , A = 5 cm 2 and A 0 = 500 cm 2 ).

Step-by-step solution

Velocity of outflow (efflux) from hole = 2 g h . Range of water jet, x = 2 g h × Time of fall = 2 g h × 2 H g = 2 h H Velocity v of bucket is v = d x d t = d d t 2 h H = 2 H . d h d t ⇒ v = 2 H . 1 2 h . d h d t = H h . d h d t . . . ( i ) Now, using equation of continuity at area A 0 and area A , we have A v 1 = A 0 V where, v 1 = 2 g h and V = d h d t . ∴ A 2 g h = A 0 d h d t Substituting for d h d t from Eq ( i ) , we get v = H h . A A 0 . 2 g h = 5 5 . 5 500 . 2 × 10 × 5 = 1 10 ms - 1 = 0 . 1 ms - 1