See the Figure 1 below and compare with the given problem.
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)]#
