A tube with cross sections at A and B , having areas of 4 mm ^2 and 2 mm ^2 respectively, is kept vertical with B positioned upward. The distance separating A and B is 15 16 cm . Water enters through B at a rate of 1 cm ^3 s ⁻¹ . Determine the speed of the water at B . Note that the speed decreases as the water falls down. Take g = 10 m s ⁻² .

Step-by-step solution

Given the volume flow rate of water entering through cross-section B is Q = 1 cm ^3 s ⁻¹ . The area of cross-section at B is given as A_B = 2 mm ^2 . Converting the area into cm ^2 : A_B = 2 10⁻² cm ^2 = 0.02 cm ^2 The volume flow rate is the product of the cross-sectional area and the speed of the fluid. Therefore, the speed of water at B , v_B , can be calculated using the relation: Q = A_B v_B Substituting the given values: 1 cm ^3 s ⁻¹ = 0.02 cm ^2 v_B v_B = 1 0.02 cm s ⁻¹ = 50 cm s ⁻¹ The other given parameter