Question

Two communicating vessels contain mercury. The

diameter of one vessel is four times larger than the

diameter of the other. A column of water of height h0 =

70 cm is poured into the left hand vessel (the narrower

one). How much will be mercury level rise in the right

hand vessel ? (Specific density of mercury = 13.6)

Answer 1

Let us assume that , if the level of narrow tube goes down by $h_1$ then, wider tube goes up by $h_2$
then, volume decreases in case of narrow tube = volume increases in case of wider tube

Let radius of narrow tube is r
then, radius of wider tube is 4r

so, $\pi r^2h_1=\pi(4r)^2h_2$

$h_1=16h_2$.......(1)

now, pressure at point A = pressure at point B
$h\rho g=(h_1+h_2)\rho' g$

because specific density , s = $\rho'/\rho$
so, $h=(h_1+h_2)s$

$h=16h_2+h_2$

$h_2=\frac{h}{17s}=\frac{70}{17\times13.6}=0.3$

Answer 2

Answer:

Let us assume that , if the level of narrow tube goes down by h_1h

1

then, wider tube goes up by h_2h

2

then, volume decreases in case of narrow tube = volume increases in case of wider tube

Let radius of narrow tube is r

then, radius of wider tube is 4r

so, \pi r^2h_1=\pi(4r)^2h_2πr

2

h

1

=π(4r)

2

h

2

h_1=16h_2h

1

=16h

2

.......(1)

now, pressure at point A = pressure at point B

h\rho g=(h_1+h_2)\rho' ghρg=(h

1

+h

2

)ρ

′

g

because specific density , s = \rho'/\rhoρ

′

/ρ

so, h=(h_1+h_2)sh=(h

1

+h

2

)s

h=16h_2+h_2h=16h

2

+h

2

h_2=\frac{h}{17s}=\frac{70}{17\times13.6}=0.3h

2

=

17s

h

=

17×13.6

70

=0.3