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(A) Let the time interval between two consecutive drops be $T$. At $t = 4 \, s$, the first drop has fallen for $4 \, s$. The distance covered by the f

Vedclass · Accepted Answer

$1 \, 	ext{drop/second}$

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(A) Let the time interval between two consecutive drops be $T$. At $t = 4 \, s$, the first drop has fallen for $4 \, s$. The distance covered by the first drop is $h_1 = \frac{1}{2} g (4)^2 = \frac{1}{2} 	imes 9.8 	imes 16 = 78.4 \, m$. The second drop was released $T$ seconds after the first, so its time of fall is $(4 - T) \, s$. The distance covered by the second drop is $h_2 = \frac{1}{2} g (4 - T)^2$. The spacing between the two drops is given as $h_1 - h_2 = 34.3 \, m$. Substituting the values: $78.4 - \frac{1}{2} 	imes 9.8 	imes (4 - T)^2 = 34.3$. $78.4 - 4.9(4 - T)^2 = 34.3$. $4.9(4 - T)^2 = 44.1$. $(4 - T)^2 = 9$. $4 - T = 3 \Rightarrow T = 1 \, s$. Since the time interval between drops is $1 \, s$, the rate is $1 \, 	ext{drop/second}$.

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at $4^{th} \, s$ after its fall and the next droplet is $34.3 \, m$. At what rate are the droplets coming from the tap? (Take $g = 9.8 \, m/s^2$)

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