Let a be the fo crosssection of the rod and V be the volume of mass attached. When rod is floating in water, then
weight of float = weight of waterd is placed `= (axx3 +V) xx 1 xxg ` …(i)
When rod is floating in a liquid,
weight of liquid displaced = `(a xx 3.5 + V) 0.9 xxg` ..(ii)
From (i) and (ii), we have `(a xx 3 + V) xx 1 xx g = (a xx 3.5 +V) xx 0.9 xxg`
or, `3a + V = 3.15 a + 0.9 V or V=1.5 a` ...(iii)
Let x be the depth of the rod immersed in a liquid of sp. gravity 1.2.
then weight of liquid displaced = `(xa +V) xx 1.2 xx g`
`:. (xa + V) xx 1.2 xx g = (a xx 3 +V)g or (xa+1.5 a) xx 1.2 = 3a+1.5 a = 4.5 a`
or `1.2 x + 1.8 = 4.5 or x=2.25cm`.