Given
#"wt of the object in air"=320N#
#"wt of the object in water"=255N#
#"wt of the object in oil"=295N#
We know
#"the density of water"= d_w=1000kgm^-3#
#"acceleration due to gravity "g=9.8ms^-2#
Wt of displaced water by the object
#="Its wt in air"-"wt in water"#
#=320-255=65N#
So mass of displaced water #m_w=65/9.8kg=6.63kg#
Volume of displaced water or volume of the object #v=m_w/d_w=6.63/1000m^3=6.63xx10^-3m^3#
#"mass of the object"=m ="its wt"/g=320/9.8kg=32.65kg#
a) #"density of the object"=d=m/v=32.65/(6.63xx10^-3)kgm^-3=4924.6kgm^-3#
b) The wt of displaced oil by the object
#="Its wt in air"-"wt in oil"#
#=320-295=25N#
If the density of oil be #d_o#
then
#vxxd_o xxg=25#
#=>d_o=25/(vxxg)=25/(6.63xx10^-3xx9.8)kgm^-3#
#=384.77kgm^-3#
