The surface of water in a water tank of cross section area 750 cm² on the top of a house is h m above the tap level. The speed of water coming out through the tap of cross section area 500 mm² is 30 cm / s. At that instant, (d h)/(d t) is x × 10⁻³ m / s. The value of x will be ____________.

Solution

Physics sir · 4 steps

Water leaving the tap must equal water lost from the tank, so match the volume flow rates.

1. Same volume flow. Each second, the volume of water leaving the tap equals the volume lost from the tank. The water surface falls at speed |dh/dt|, so the equation of continuity applies to the surface and the tap.

   Aₜₐₙₖ|dh/dt| = Aₜₐₚ vₜₐₚ

2. Match the units. Convert everything to centimetres. 500 mm² = 5 cm², because 1 cm² = 100 mm². The tank area is 750 cm² and vₜₐₚ = 30 cm s⁻¹.
3. Put in numbers. |dh/dt| = (5 × 30)/750 = 0.2 cm s⁻¹ = 2 × 10⁻³ m s⁻¹. The level is falling, so h is decreasing at this rate.
4. So. The surface falls at 2 × 10⁻³ m s⁻¹, so x = 2.

The common slip: 0.2. Stopping at 0.2 cm s⁻¹ and reporting 0.2 leaves the speed in cm/s. The question asks for it in m/s, which is 2 × 10⁻³ m s⁻¹, so x = 2.

The idea to keep. A draining tank is a continuity problem: area times speed of the falling surface equals area times speed at the outlet. Put all areas in one unit before you divide.

Answer from the NTA final answer key. Solution by PrepWiser.