# water drops fall at regular intervals from a roof. at an instant when a drop is about to leave the roof the separations between three successive drops below the roof are in what ratio? please answer fast....

Swapnil Saxena
102 Points
11 years ago

Let the interval be t. Then the first drop of the three has left t sec before, second has 2t sec and the third one at 3t sec. Then the distances covered by them respectively are

0(Drop about to fall), 1/2(t2) (I drop) , 2t2 (II drop), 9/2 t2 (III drop)

The distances b/w them r in ratio (5/2)t2 : (3/2)t2 : (1/2)t2= 5:3:1

goutham 1995 gumm
18 Points
11 years ago

the answer is 1:2:3...can someone tell how?

Vinay
12 Points
6 years ago
let the time interval be t. so for first drop the time elapsed is 3t. for second it’s 2t and for third it’s t.
so distance covered by them be s1,s2,and s3 respectively.

s1 = 1/2(g)((3t)^2) -------(1)
s2 = 1/2(g)((2t)^2) -------(2)
s3 = 1/2(g)(t^2) ---------(3)

so ration of drop1,-2and -3 is
from drop3 to drop1

1:4:9

And this is the right answer indeed!!!!
vaibhav
11 Points
5 years ago
Shreya Nehra
11 Points
5 years ago
the correct answer is 1:3:5 because after we have taken out the three distances from the top , we need to take out the ratio of distances BETWEEN THE DROPS  as this is what has been asked in the question and NOT the ratio of distances of the drops from the top.
Kohav Dey
13 Points
5 years ago
Let the acceleration be a which is same for all drops.
Let time for 1st drop be t, 2nd drop be 2t, 3rd drop be 3t.
therefore
distance covered y 1st drop: u=0, 1/2at^2, 2nd drop : 2at^2, 3rd drop : 9at^2/2.
therefore distance between 1st and 2nd drop :3/2at^2, 2nd and 3rd drop:5/2at^2,1st and 3rd drop: 4at^2.
therefore the ratio is 3:5:4.
Dhruv garg
15 Points
4 years ago
The answer for the question is 1:3:5because it is asking for the particle which has not started to move yet