4y  /v  '     3  sin2<9

.'.  the  above  inequality  holds  if  2  cos2  6  —  cos2  a  -  cos  a  cos  6  is
positive.
But  cos  6  >  cos  a.
.'.  this  is  true.
.-.  the  greatest  depression  exceeds  the  greatest  elevation.
18.  Let  h  be  the  height  above  the  base  of  the  cylinder  of  the  top
of  the  solid  at  first,  and  k  when  there  is  equilibrium.
Let  z  be  the  greatest  height  above  the  top  of  the  solid  of  the  surface
of  the  liquid  above  it,  and  x  the  depth  below  the  top  of  the  solid  of  the
paraboloidal  surface  continued.
Then  wa*h  —  \ndl .  -= —  =  volume  of  liquid

—na2k  -  hnci1 .  — \-  volume  of  liquid  above,

.  \  na2  (A  -k)  =  irah  -  hta2  (z  +  x)  +  \tc  -f  x .  x,
but  a2=-f  («+#),


. ,  wa2    a2©2         fra2  a2co2      jr    2ff  .        )

Also         rx--y'-iTV.i'3!#"*K.
...  *t-£^?g  and  A-*=ay/l-  V"-^}"-
p     4g2  4g   \        V      p   J


116  MISCELLANEOUS   PROBLEMS.      II.
19.  Let  v  be  the  volume  of  each  sphere,  p,  p  their  densities,  o-  that
of  the  fluid.
The  straight  line  joining  the  spheres  must  intersect  the  axis  of
rotation  at  right  angles.
If  a  and  b  are  the  distances  from  the  axis,
ava>2a  +  T=pva>2a,
(rVco2b  —  T=p'vo>2b ;
so  that,  if  I  be  the  length  of  the  string,

o-  (a  +  b)=pa  +  p'b,         a-b  =  l.
If  we  take  account  of  gravity  the  line  of  the  string  must  intersect
the  axis  and  may  be  inclined  to  it  at  an  angle  6.
We  then  have  the  equations

aVco2a  +  T  sin  6=pvco2a}

ava)2b  -TsinB= p'va)2bf

g<rv  +  T  cos  6 = gpv,

guv  —  Tcob  6=gp'v.
