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A rectangular bar of soap has density 80...

A rectangular bar of soap has density `800kg//m^(3)` floats in water density `1000kg//m^(3)`. Oil of density `300kg//m^(3)` is slowly added, forming a layer that does not mix with water. When the top surface of thhe oil is the some level as the top surface of the soap. What is the ratio of the oil layer thickness to the soap's thickness. `x//L`?

A

`3/7`

B

`2/7`

C

`3/10`

D

`3/8`

Text Solution

Verified by Experts

The correct Answer is:
B
Let A be the area of the soap bar.

`PA-P_0A=mgimplies(P-P_0)A=mg`
`implies [300gx+1000g(L-x)A]=AL800g`
`implies x/L=2/7`
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