A piece of wax floats on brine. What fraction of its volume is immersed?
Density of water $= 0.95$ g $c m^{- 3}$ , Density of brine $= 1.1$ g $c m^{- 3}$ .

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Solution

Given:
Density of max $= 0.95 g c m^{- 3}$
Density of brine $= 1.1 g c m^{3}$
Consider $^{'} V^{'}$ to be net volume of the wax and $^{'} v^{'}$ be the volume of the submerged part
As per the law of floatation, $\frac{ν}{V} = \frac{D e n s i t y o f w a x}{D e n s i t y o f b r i n e}$
$\Rightarrow \frac{ν}{V} = \frac{0.95}{1.1} = 0.86$
$\Rightarrow ν = 0.86 V$
Hence, the piece of wax floats with ${0.86}^{t h}$ part of its volume over the brine surface.

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A piece of wax floats on brine. What fraction of its volume is immersed?Density of water =0.95 g cm−3, Density of brine =1.1g cm−3.

A piece of wax floats on brine. What fraction of its volume is immersed?
Density of water $= 0.95$ g $c m^{- 3}$ , Density of brine $= 1.1$ g $c m^{- 3}$ .