A unifrom rod has a mass attached to one end to make it float upright in liquild. If 3.0 cm of the rod is immersed when floats in water and 3.5 cm when it floats in a liquid of sp. Gravity 0.9 , what length of it will be immersed, when it floats in a liquid of sp. gravity 1.2?

Let a be the fo crosssection of the rod and V be the volume of mass attached. When rod is floating in water, then

weight of float = weight of waterd is placed \(={\left({a}\times{3}+{V}\right)}\times{1}\times{g}\) …(i)

When rod is floating in a liquid,

weight of liquid displaced = \({\left({a}\times{3.5}+{V}\right)}{0.9}\times{g}\) ..(ii)

From (i) and (ii), we have \({\left({a}\times{3}+{V}\right)}\times{1}\times{g}={\left({a}\times{3.5}+{V}\right)}\times{0.9}\times{g}\)

or, \({3}{a}+{V}={3.15}{a}+{0.9}{V}{\quad\text{or}\quad}{V}={1.5}{a}\) ...(iii)

Let x be the depth of the rod immersed in a liquid of sp. gravity 1.2.

then weight of liquid displaced = \({\left({x}{a}+{V}\right)}\times{1.2}\times{g}\)

\(\therefore{\left({x}{a}+{V}\right)}\times{1.2}\times{g}={\left({a}\times{3}+{V}\right)}{g}{\quad\text{or}\quad}{\left({x}{a}+{1.5}{a}\right)}\times{1.2}={3}{a}+{1.5}{a}={4.5}{a}\)

or \({1.2}{x}+{1.8}={4.5}{\quad\text{or}\quad}{x}={2.25}{c}{m}\).

weight of float = weight of waterd is placed \(={\left({a}\times{3}+{V}\right)}\times{1}\times{g}\) …(i)

When rod is floating in a liquid,

weight of liquid displaced = \({\left({a}\times{3.5}+{V}\right)}{0.9}\times{g}\) ..(ii)

From (i) and (ii), we have \({\left({a}\times{3}+{V}\right)}\times{1}\times{g}={\left({a}\times{3.5}+{V}\right)}\times{0.9}\times{g}\)

or, \({3}{a}+{V}={3.15}{a}+{0.9}{V}{\quad\text{or}\quad}{V}={1.5}{a}\) ...(iii)

Let x be the depth of the rod immersed in a liquid of sp. gravity 1.2.

then weight of liquid displaced = \({\left({x}{a}+{V}\right)}\times{1.2}\times{g}\)

\(\therefore{\left({x}{a}+{V}\right)}\times{1.2}\times{g}={\left({a}\times{3}+{V}\right)}{g}{\quad\text{or}\quad}{\left({x}{a}+{1.5}{a}\right)}\times{1.2}={3}{a}+{1.5}{a}={4.5}{a}\)

or \({1.2}{x}+{1.8}={4.5}{\quad\text{or}\quad}{x}={2.25}{c}{m}\).

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