InterviewSolution
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InterviewSolution
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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. |
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Answer» 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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