Question

Kindly answer this question
an object float in three immiscible liquids a,b and c of densities 3 g cm-3 , 2 g cm-3 and 1g cm-3 respectively as shown in the figure.when the object is placed in the liquids the levels of liquid a,b and c rise by 3 cm,5 cm and 8 cm respectively. the areas of cross-sections of the container and the object are 10 cm2 and 5 cm2 respectively. calculate the density of the object.

Answer 1

Answer:

Weight of the floating body = weight of the liquids A , B ,C displaced

Volume of the body immersed in a liquid = (Rise in level)×(area of cross section of the container)

VA=3 cm×10 cm

2VB=5 cm×10 cm

2VC=8 cm×10 cm

2VA+VB+VC=Total volume of the body

Weight of A displaced =VA×dA

Similarly,

Weight of B displaced =VB×dBand Weight of C displaced =VC×dC

Now density of the body =

mass/ volume=

(VA×dA)+(VB×dB)+(VC×dC)[(3+5+8) ×5]cm3=(30×3)+(50×2)+(80×1)80

=27080

=3⋅375g