Water droplets are coming from an open tap at particular rate. spacing between a droplet observed at

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2.Motion in Straight Line

hard

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

A

$1 \,drop/ second$

B

$2 \,drops / second$

C

$1\, drops / 7 seconds$

D

$3\, drops / 2 seconds$

(JEE MAIN-2021)

Solution

In 4 sec. $1^{\text {st }}$ drop will travel $\Rightarrow \frac{1}{2} \times(9.8) \times(4)^{2}=78.4 m$

$\therefore 2^{\text {nd }}$ drop would have travelled $\Rightarrow 78.4-34.3=44.1 m$.

Time for $2^{\text {nd }}$ drop

$\frac{1}{2}(9.8) t ^{2}=44.1$

$t=3 sec$

$\therefore$ each drop have time gap of $1 sec$

$\therefore 1$ drop per sec

Standard 11

Physics

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