Question

A 2.0-cm-thick bar of soap is floating on a water surface

Answer 1

Answer:

Explanation:

Given

thickness of soap bar t=2 cm

thickness underwater t'=1.5 cm

oil is poured in the water with specific gravity=0.6

Soap top surface is just level  with the upper surface of oil i.e.

weight of soap is balance by buoyancy effect of oil and water

initially soap is balanced by water

$\rho _s\times A\times 2\times g=\rho _w\times A\times 1.5\times g$

$2\times \rho _s=1.5\rho _w$

if $\rho _s=density\ of\ soap$

$\rho _w=density of water$

$A=area\of\ cross-section$

$d=depth\ of\ soap\ in\ oil$

$\rho _s\times A\times 2\times g=0.6\rho _w\times A\times d\times g+\rho _w\times A\times (2-d)\times g$

$1.5\rho _w=0.6d\rho _w+\rho _w(2-d)$

$1.5=0.6d+2-d$

$0.4d=0.5$

$d=\frac{5}{4}$

$d=1.25\ cm$