Leaky bucket

A bucket is lying on the ground with water up to the brim. It is lifted and a hole is made in it, right at the bottom, through which the water can escape. Find the time taken to empty the bucket (in seconds).

Height of the bucket = 30 cm =30 \ \text{cm}

Area of the bucket = 60 cm 2 =60 \ \text{cm}^2

Area of the hole = 10 cm 2 = 10 \ \text{cm}^2

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


The answer is 1.463850109.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ronak Agarwal
Aug 8, 2014

At any time let the water level in the bucket be h

Then according to torricelli's law we have :

v e f f l u x = 2 g h { v }_{ efflux }=\sqrt { 2gh }

Also by conservation of mass of water we have

A h o l e v e f f l u x = A b u c k e t ( d h d t ) { A }_{ hole }{ v }_{ efflux }={ A }_{ bucket }(\frac { -dh }{ \quad dt } )

Finally we have :

0 t A h o l e A b u c k e t 2 g d t = h 0 d h h \int _{ 0 }^{ t }{ \frac { -{ A }_{ hole } }{ { A }_{ bucket } } \sqrt { 2g } dt } =\int _{ h }^{ 0 }{ \frac { dh }{ \sqrt { h } } }

Solving we have t = A b u c k e t A h o l e 2 h g t=\frac { { A }_{ bucket } }{ { A }_{ hole } } \sqrt { \frac { 2h }{ g } }

Put the values to get t = 1.484 s e c \boxed { t=1.484\quad sec }

Shouldn't we consider the area of holes and use the equation v e f f l u x = 2 g h 1 A 2 a 2 { v }_{ efflux }=\sqrt { \frac { 2gh }{ 1-\frac { { A }^{ 2 } }{ { a }^{ 2 } } } } where A A =Area of bucket and a a =area of hole.

Anandhu Raj - 6 years ago

Log in to reply

ans would come the same i guess !! i tried that too

Log in to reply

Yes..Just asked.

Anandhu Raj - 6 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...