Question

water drops are falling from regular intervals of time from a tap which is at a height of 5m above the ground. the first drop reaches when the fifth drop is about to fall. find the seperation between 2nd and 4th drops at this instant.

Answer 1

Explanation:

17. Match the description (a) to (d) below with appropriate

term from the list A to H.

A Acidic oxide

B. Alkali

C. Amphoteric oxide

D. Basic oxide

E. Deliquescence

F. Efflorescence

G. Electrolysis

H. Electrolyte

(a) The property of spontaneously giving up water of

crystalisation to the atmosphere.

(b) A liquid or a solution which conducts electricity

with accompanying chemical change.

(c) A compound which is soluble in water and the only

negative ions in the solution are hydroxide ions.

(d) An oxide which forms salts when it reacts with both

acids and alkalis.

Answer 2

let the time after which 2nd drop is released be   t sec

then position of first drop can be calculated

From

$S=ut+\frac{1}{2}at^{2} \\here\: \\u =0 \\S=0+\frac{1}{2}\times 10\times t^{2}$

$S=5t^{2]$

when 3rd drop is released time will be =2t

then position of 1st will be  

$S=0+\frac{1}{2} \times 10\times (2t)^{2}\\S=20t^{2}$

when 4th drop is released the position of 1st drop will be

$S=45t^{2}$

ratio obtained is

1:4:9 :16:25:36:49

this states position of the drops after specific times

and

$\red{all\:the\:drops\:follow\:same\:trajectory\:}$

and are found at the same positions after specific time i.e.,

when t=2t

1st drop will be at 4 position

and 2nd will be at position

REFER THE ATTACHMENT

HENCE when 5 drop is touching the ground at 25x

3rd drop is at 9x

so

position of 4th drop is 16s

and position of 2nd drop will be 4s

distance b|w two can be calculated as$16-4=12s$

but s =????

we know that at 25s drop reaches ground

hence 25s=5m

s=1/5m

hence distance3 b|w 4th and 2nd drop is

$\frac{12}{5} m$