Fluid Mechanics

The problem involves the principle of conservation of mass and Bernoulli’s equation. As the water falls, the velocity increases, but the mass flow rate (product of the cross-sectional area and velocity) remains constant. Let the area of cross-section at the tap be and the velocity at the tap is . At a point 20 cm below, let the cross-sectional area be and the velocity be . The principle of conservation of mass gives: Using the energy conservation principle (Bernoulli’s equation), we have: where is the height difference. Now solving for : Substitute the given values: Now, applying the continuity equation: Thus, the cross-sectional area at 20 cm below the tap is approximately .

The given equation is: To find the value of , we need to analyze the dimensions of both sides of the equation. The dimension of pressure ( ) is: The dimension of density ( ) is: The dimension of velocity ( ) is: Now, applying the dimensions to the equation , we get: Simplifying: Now, comparing the dimensions of mass ( ), length ( ), and time ( ) on both sides, we find: For mass: For length: Thus: For time: Thus: So, the value of is 1, meaning .

When rain drops coalesce, their volume adds up, meaning the volume of the new drop will be the sum of the volumes of all the smaller drops. The volume of a sphere is given by: Thus, if 64 drops coalesce, the volume of the new drop will be: The terminal velocity is proportional to the square of the radius of the drop, i.e., Since the radius increases by a factor of , the new terminal velocity will be: Converting to m/s: Thus, the terminal velocity of the coalesced drop is 0.08 m/s.