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)]#