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

Motion in A Straight Line PYQ — JEE Main 2021 · 27 Aug · Shift 2 · Q44 with Solution

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Correct answer: D

Time taken by the first drop to reach the floor

The drop falls from height $h=9.8\,\text{m}$ under gravity.

Using

h=\frac{1}{2}gt^2

with $g=9.8\,\text{m/s}^2$ ,

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

9.8=4.9t^2

t^2=2 \Rightarrow t=\sqrt{2}\,\text{s}

So, the first drop takes $\sqrt{2}$ s to reach the floor.

Find the interval between successive drops

The drops fall at regular intervals. Let the interval be $\Delta t$ .

Given: when the first drop strikes the floor, the third drop just begins to fall.

That means from the release of the first drop to the release of the third drop, elapsed time is exactly the falling time of the first drop.

Now,

first drop released at $t=0$
second drop released at $t=\Delta t$
third drop released at $t=2\Delta t$

Since the third drop begins to fall when the first hits the floor,

2\Delta t = \sqrt{2}

\Delta t = \frac{\sqrt{2}}{2}\,\text{s}

Time for which the second drop has been falling

At the instant the first drop hits the floor, the second drop has been falling for

\sqrt{2}-\Delta t

Substitute $\Delta t=\frac{\sqrt{2}}{2}$ :

\sqrt{2}-\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\,\text{s}

Distance fallen by the second drop

Using

s=\frac{1}{2}gt^2

for $t=\frac{\sqrt{2}}{2}$ ,

s=\frac{1}{2}(9.8)\left(\frac{\sqrt{2}}{2}\right)^2

s=4.9\cdot \frac{2}{4}=4.9\cdot \frac{1}{2}=2.45\,\text{m}

So the second drop has fallen $2.45$ m from the nozzle.

Position of the second drop from the floor

Total height = $9.8$ m, so height above floor is

9.8-2.45=7.35\,\text{m}

Check options

A: $4.18$ m ❌
B: $2.94$ m ❌
C: $2.45$ m ❌ (this is distance fallen from nozzle, not distance from floor)
D: $7.35$ m ✅

Therefore, the correct option is D.

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View written solutionFree

Correct answer: D

Time taken by the first drop to reach the floor

The drop falls from height $h=9.8\,\text{m}$ under gravity.

Using

h=\frac{1}{2}gt^2

with $g=9.8\,\text{m/s}^2$ ,

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

9.8=4.9t^2

t^2=2 \Rightarrow t=\sqrt{2}\,\text{s}

So, the first drop takes $\sqrt{2}$ s to reach the floor.

Find the interval between successive drops

The drops fall at regular intervals. Let the interval be $\Delta t$ .

Given: when the first drop strikes the floor, the third drop just begins to fall.

That means from the release of the first drop to the release of the third drop, elapsed time is exactly the falling time of the first drop.

Now,

first drop released at $t=0$
second drop released at $t=\Delta t$
third drop released at $t=2\Delta t$

Since the third drop begins to fall when the first hits the floor,

2\Delta t = \sqrt{2}

\Delta t = \frac{\sqrt{2}}{2}\,\text{s}

Time for which the second drop has been falling

At the instant the first drop hits the floor, the second drop has been falling for

\sqrt{2}-\Delta t

Substitute $\Delta t=\frac{\sqrt{2}}{2}$ :

\sqrt{2}-\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\,\text{s}

Distance fallen by the second drop

Using

s=\frac{1}{2}gt^2

for $t=\frac{\sqrt{2}}{2}$ ,

s=\frac{1}{2}(9.8)\left(\frac{\sqrt{2}}{2}\right)^2

s=4.9\cdot \frac{2}{4}=4.9\cdot \frac{1}{2}=2.45\,\text{m}

So the second drop has fallen $2.45$ m from the nozzle.

Position of the second drop from the floor

Total height = $9.8$ m, so height above floor is

9.8-2.45=7.35\,\text{m}

Check options

A: $4.18$ m ❌
B: $2.94$ m ❌
C: $2.45$ m ❌ (this is distance fallen from nozzle, not distance from floor)
D: $7.35$ m ✅

Therefore, the correct option is D.

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