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
In 4 sec. 1<sup>st</sup> drop will travel<br><br>$\Rightarrow {1 \over 2}$ $\times$ (9.8) $\times$ (4)<sup>2</sup> = 78.4 m<br><br>$\therefore$ 2<sup>nd</sup> drop would have travelled<br><br>$\Rightarrow$ 78.4 $-$ 34.3 = 44.1 m.<br><br>Time for 2<sup>nd</sup> drop<br><br>$\Rightarrow {1 \over 2}$(9.8)t<sup>2</sup> = 44.1<br><br>t = 3 sec<br><br>$\therefore$ each drop have time gap of 1 sec<br><br>$\therefore$ 1 drop per sec
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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