See the Figure 1 below and compare with the given problem.
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Using CGS system of units
Weight of Benzene W="Mass"xxg
="Volume"xx"Density, (Mass per unit volume)"xxg ...........(1)
where g=981cmcdot s^-2 is acceleration due to gravity.
To calculate Volume of Benzene shown in Blue.
Volume V="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 AOB in the figure below.
From the above figure we see that
"Segment area (the blue area)"= "Sector area (Red and blue areas)"- "Area "Delta AOB
Now "Sector area"="Area of circle"xx
"Angle subtended by the sector at the center"/"Total angle at centre"=(R^2theta)/2 ......(3)
From DeltaAOE we have
theta=2cos^-1((OE)/(AO))
=>theta=2cos^-1((R-h)/(R))
Inserting above in equation (3) we get
"Sector area"=(R^2xx2cos^-1((R-h)/(R)))/2
"Sector area"=R^2cos^-1((R-h)/(R)) .....(4)
And "Area "Delta AOB=2xxDeltaOEB
=>"Area "Delta AOB=2(1/2OExxEB)
=>"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[R^2cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2)] .......(6)
Inserting this value of V in equation (1) and given values of Land R we get
W=400[100^2cos^(-1)((100-h)/100)-(100-h)sqrt(2xx100h-h^2)]xx0.879xx981
W=3.449xx10^5[10^4cos^(-1)((100-h)/100)-(100-h)sqrt(200h-h^2)]" dyne"
We know that One dyne is equal to 10^-5 newtons. Hence we get
W=3.449[10^4cos^(-1)((100-h)/100)-(100-h)sqrt(200h-h^2)]N, where h is in cm.
Note. This derivation is valid for h<=R
.-.-.-.-.-.-.-.-.-.-.-.-.--.-.
In case of h>=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'="Area of circular segment (uncolored)"xx"length "L
From Figure 2 we have
"Segment area (uncolored area)"= "Area of the circle"-"Sector area (Red and blue areas)"+ "Area "Delta AOB
We know that "Area of the circle"=piR^2
Steps for remaining two terms are same as for obtaining equations (4) and (5).
Therefore, if h'=2R-h is height of empty space above the fluid level
V'=L[piR^2-R^2cos^(-1)((R-h')/R)+(R-h')sqrt(2Rh'-(h')^2)]