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(D) Let the time interval between consecutive drops be $\Delta t$. When the first drop reaches the floor,the total time elapsed is $T = \sqrt{\frac{2H

Vedclass · Accepted Answer

$7.35$

Vedclass · Answer

(D) Let the time interval between consecutive drops be $\Delta t$. When the first drop reaches the floor,the total time elapsed is $T = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 	imes 9.8}{9.8}} = \sqrt{2}\, s$. Since the third drop is just starting to fall at this instant,the time elapsed for the third drop is $0$. The second drop has been falling for a time of $\Delta t$,and the first drop has been falling for a time of $2\Delta t$. Thus,$2\Delta t = T = \sqrt{2}\, s$,which gives $\Delta t = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\, s$. The distance fallen by the second drop from the nozzle is $y_2 = \frac{1}{2} g(\Delta t)^2 = \frac{1}{2} 	imes 9.8 	imes (\frac{1}{\sqrt{2}})^2 = 4.9 	imes 0.5 = 2.45\, m$. The position of the second drop from the floor is $H - y_2 = 9.8 - 2.45 = 7.35\, m$.

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