Question

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 $g = 9.8 \text{ m/s}^2$)

• A. 1 drop/second
• B. 2 drops/second
• C. 3 drops/2 seconds
• D. 1 drop/7 seconds

Hint:

In droplet problems, if $T$ is the interval, the $n$-th drop from the bottom has fallen for $(n-1)T$ seconds if the first drop is just reaching the ground.
Always verify if the calculated interval allows the required number of drops to have fallen.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let the interval between two successive drops be \(T\).
If a drop is observed at time \(t\), the next drop has been falling for a time \((t - T)\).
The distance traveled by a freely falling body from rest in time \(t\) is \(s = \frac{1}{2}gt^2\).
Step 2: Key Formula or Approach:
Distance of the \(n\)-th drop: \(s_n = \frac{1}{2}gt_n^2\).
Distance of the \((n+1)\)-th drop: \(s_{n+1} = \frac{1}{2}g(t_n - T)^2\).
Spacing: \(\Delta s = s_n - s_{n+1}\).
Step 3: Detailed Explanation:
According to the question, we observe the 4th drop 4 seconds after it started falling.
Time of flight for the first drop in consideration: \(t_1 = 4 \text{ s}\).
Time of flight for the next drop: \(t_2 = (4 - T) \text{ s}\).
The spacing is given as \(34.3 \text{ m}\).
\[ \frac{1}{2}g(4)^2 - \frac{1}{2}g(4 - T)^2 = 34.3 \]
\[ \frac{1}{2} \times 9.8 \times [16 - (16 + T^2 - 8T)] = 34.3 \]
\[ 4.9 \times [8T - T^2] = 34.3 \]
Divide both sides by 4.9:
\[ 8T - T^2 = \frac{34.3}{4.9} = 7 \]
\[ T^2 - 8T + 7 = 0 \]
Factoring the quadratic equation:
\[ (T - 7)(T - 1) = 0 \]
So, \(T = 7 \text{ s}\) or \(T = 1 \text{ s}\).
Since the droplet is being observed 4 seconds after its fall, the interval \(T\) must be less than 4 seconds for a "next droplet" to have already started its fall.
Therefore, \(T = 1 \text{ s}\).
The rate of droplets is \(1/T = 1 \text{ drop/second}\).
Step 4: Final Answer:
The droplets are coming from the tap at a rate of 1 drop/second.