# Question 20dd3

Jan 22, 2017

$W = 3.449 \left[{10}^{4} {\cos}^{- 1} \left(\frac{100 - h}{100}\right) - \left(100 - h\right) \sqrt{200 h - {h}^{2}}\right] N$, where $h$ is in cm.

#### Explanation:

See the Figure 1 below and compare with the given problem.

Using CGS system of units
Weight of Benzene $W = \text{Mass} \times g$
$= \text{Volume"xx"Density, (Mass per unit volume)} \times g$ ...........(1)
where $g = 981 c m \cdot {s}^{-} 2$ is acceleration due to gravity.

To calculate Volume of Benzene shown in Blue.
Volume $V = \text{Area of circular segment (Blue)"xx"length } L$ .....(2)

Let $R$ be the radius of cylinder, $h$ is the height of Benzene as shown in the figure given in the problem, Let $\theta$ be $\angle A O B$ in the figure below.

From the above figure we see that
$\text{Segment area (the blue area)"= "Sector area (Red and blue areas)"- "Area } \Delta A O B$

Now $\text{Sector area"="Area of circle} \times$
$\text{Angle subtended by the sector at the center"/"Total angle at centre} = \frac{{R}^{2} \theta}{2}$ ......(3)
From $\Delta A O E$ we have
$\theta = 2 {\cos}^{-} 1 \left(\frac{O E}{A O}\right)$
$\implies \theta = 2 {\cos}^{-} 1 \left(\frac{R - h}{R}\right)$
Inserting above in equation (3) we get
$\text{Sector area} = \frac{{R}^{2} \times 2 {\cos}^{-} 1 \left(\frac{R - h}{R}\right)}{2}$
$\text{Sector area} = {R}^{2} {\cos}^{-} 1 \left(\frac{R - h}{R}\right)$ .....(4)

And $\text{Area } \Delta A O B = 2 \times \Delta O E B$
$\implies \text{Area } \Delta A O B = 2 \left(\frac{1}{2} O E \times E B\right)$
=>"Area "Delta AOB=(R-h)xxsqrt(OB^2-OE^2))
=>"Area "Delta AOB=(R-h)xxsqrt(R^2-(R-h)^2))# .......(5)

Inserting equations (4) and (5) in equation (2) for volume we have

$V = L \left[{R}^{2} {\cos}^{- 1} \left(\frac{R - h}{R}\right) - \left(R - h\right) \sqrt{2 R h - {h}^{2}}\right]$ .......(6)
Inserting this value of $V$ in equation (1) and given values of $L \mathmr{and} R$ we get

$W = 400 \left[{100}^{2} {\cos}^{- 1} \left(\frac{100 - h}{100}\right) - \left(100 - h\right) \sqrt{2 \times 100 h - {h}^{2}}\right] \times 0.879 \times 981$
$W = 3.449 \times {10}^{5} \left[{10}^{4} {\cos}^{- 1} \left(\frac{100 - h}{100}\right) - \left(100 - h\right) \sqrt{200 h - {h}^{2}}\right] \text{ dyne}$
We know that One dyne is equal to ${10}^{-} 5$ newtons. Hence we get

$W = 3.449 \left[{10}^{4} {\cos}^{- 1} \left(\frac{100 - h}{100}\right) - \left(100 - h\right) \sqrt{200 h - {h}^{2}}\right] N$, where $h$ is in cm.

Note. This derivation is valid for $h \le R$
.-.-.-.-.-.-.-.-.-.-.-.-.--.-.
In case of $h \ge R$, as referred to Figure 1 above, volume now becomes volume of uncolored part. It may be instructive to put both figures upside down to visulize.

In this case
Volume $V ' = \text{Area of circular segment (uncolored)"xx"length } L$

From Figure 2 we have
$\text{Segment area (uncolored area)"= "Area of the circle"-"Sector area (Red and blue areas)"+ "Area } \Delta A O B$
We know that $\text{Area of the circle} = \pi {R}^{2}$
Steps for remaining two terms are same as for obtaining equations (4) and (5).

Therefore, if $h ' = 2 R - h$ is height of empty space above the fluid level
$V ' = L \left[\pi {R}^{2} - {R}^{2} {\cos}^{- 1} \left(\frac{R - h '}{R}\right) + \left(R - h '\right) \sqrt{2 R h ' - {\left(h '\right)}^{2}}\right]$