Question 172 of 308

Water drops are falling from a nozzle of a shower onto the floor from a height of $9.8m$. 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.

A$2.94m$B$4.18m$C$2.45m$D$7.35m$

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Answer: (D) $7.35m$

For first drop.
$h=\dfrac{1}{2}g(2n{)}^{2}$
For ${2}^{\text{nd}}$ drop
${h}^{'}=\dfrac{1}{2}g(n{)}^{2}$
$\dfrac{h}{{h}^{'}}=\dfrac{4}{1}$
${h}^{'}=\dfrac{h}{4}=\dfrac{9.8}{4}$
so height of ${2}^{nd}$ drop
$H=h-{h}^{'}=9.8-\dfrac{9.8}{4}=\dfrac{3}{4}\times 9.8=7.35m$