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Water drops are falling from a nozzle of a shower onto the floor, from a height of $9.8 \mathrm{~m}.$ The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.
Option: 1 $2.45\: m$
Option: 2 $4.18\: m$
Option: 3 $2.94\: m$
Option: 4 $7.35\: m$

Answers (1)

best_answer

Let $t$ be the time interval between two succesive drop.

The first drop has been falling for $2t\: s$ whereas the second drop for $t\: s$ .

By Galileo 's ratio, the position of $2nd$ and $3rd$ will be $3s$ and $4s$ from the surface.

$s= \frac{9.8}{4}\mathrm{m}\\$

$4s= 9.8\mathrm{m}\\$ .......(1)

For $2nd \\$ drop

$s= ut+\frac{1}{2}at^{2}\\$

$s= 0+\frac{1}{2}\left ( 9.8 \right )t^{2}\\$

$\frac{9.8}{4}= \frac{9.8}{2}\times t^{2}\\$

$t= \frac{1}{\sqrt{2}}s\\$

Position of second drop from floor is $3s= \frac{3\times 9.8}{4}=7 .35\mathrm{m}$ .

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vishal kumar

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