Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4ᵗʰ second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap ? (Take g = 9.8 m/s²)

Solution

Physics sir · 3 steps

Every drop falls the same way, just one interval later. Find how long the next drop has fallen.

1. First drop. Each drop starts from rest at the tap. After 4 s the first drop has fallen 1/2 × 9.8 × 4² = 78.4 m.
2. Next drop. The next drop is 34.3 m behind, so it has fallen 78.4 - 34.3 = 44.1 m. It left the tap T seconds later, so it has been falling for (4 - T) seconds.

   1/2 × 9.8 × (4 - T)² = 44.1 ⇒ (4 - T)² = 9 ⇒ T = 1 s

3. So. One drop leaves the tap every second, so the rate is 1 drop / second, option (c).

The idea to keep. Drops from a tap all obey s = 1/2gt², each starting one interval T late. The gap between two neighbours is s(t) - s(t - T), so solve that for T.

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