1. Let the time interval between successive droplets be $T$ seconds.

2. Consider two successive droplets:

   • The lower droplet has been falling for $4$ s when observed.
   • The next droplet was released $T$ seconds later, so at that same instant it has been falling for $(4-T)$ s.

3. Distance fallen by a droplet in free fall from rest after time $t$ is $s=\frac{1}{2}gt^2$

   Hence,

   • Distance fallen by the first droplet: $s_1=\frac{1}{2}g(4)^2=\frac{1}{2}\cdot 9.8 \cdot 16$
   • Distance fallen by the next droplet: $s_2=\frac{1}{2}g(4-T)^2$

4. The spacing between them is given as $34.3$ m: $s_1-s_2=34.3$ $\frac{1}{2}g\left[16-(4-T)^2\right]=34.3$

5. Substitute $g=9.8$: $4.9\left[16-(4-T)^2\right]=34.3$ $16-(4-T)^2=\frac{34.3}{4.9}=7$ $(4-T)^2=9$

6. So, $4-T=\pm 3$ giving $T=1 \text{ s} \quad \text{or} \quad T=7 \text{ s}$

7. Now check physically: Since the second droplet must already have fallen at the observation instant, we need $4-T\ge 0 \Rightarrow T\le 4$ Therefore, $T=1 \text{ s}$ is valid, while $T=7$ s is not possible.

8. Thus droplets are coming at the rate of $1 \text{ drop per second}$

9. Checking options:

   • A: $3$ drops / $2$ s $=1.5$ drops/s ❌
   • B: $2$ drops / s ❌
   • C: $1$ drop / s ✅
   • D: $1$ drop / $7$ s ❌

Therefore, the correct option is C.