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An object floats in three immiscible liq...

An object floats in three immiscible liquids A, B and C of densities `3"g cm"^(-3),2"g 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 `cm^(2)and5cm^(2)`, respectively. Calculate the density of the object.

Text Solution

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Weight of the floating body = Weight of the liquids A,B and C displaced.
Volume of the body immersed in a liquid = (Rise in level) `xx` (area of cross-section of the container)
`V_(A)=(3cm)xx(10cm^(2))`
`V_(A)+V_(B)=(5cm)xx(10cm^(2))`
`V_(A)+V_(B)+V_(C)=(8cm)xx(10cm)`= Total volume of the body
Weight of A displaced `=V_(A)xxd_(A)`
Similarly, determine the weights of liquids B and C displaced.
Determine the density from the definition,
`"Density"=("mass")/("volume")`
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