The fraction of a floating object of volume $V_{0}$ and density $d_{0}$ above the surface of a liquid of density d will be

\frac{d_{0}}{d}

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\frac{d d_{0}}{d + d_{0}}

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\frac{d - d_{0}}{d}

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\frac{d d_{0}}{d - d_{0}}

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Solution

The correct option is C $\frac{d - d_{0}}{d}$
For the fraction of flotation of an object in the liquid is given by
$V_{0} d_{0} g = V_{i n} d g \Rightarrow V_{i n} = V_{0} \frac{d_{0}}{d}$
$∴ V_{o u t} = V_{0} - V_{i n} = V_{0} - V_{0} \frac{d_{0}}{d} = V_{0} [\frac{d - d_{0}}{d}]$
$\Rightarrow \frac{V_{o u t}}{V_{0}} = \frac{d - d_{0}}{d}$

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V_{0} a n d d e n s i t y d_{0}

V_{0}

d_{0}

d

D = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

D_{0} = ∣ ∣ ∣ ∣ \begin{matrix} k a_{1} & k b_{1} & k c_{1} k a_{2} & k b_{2} & k c_{2} k a_{3} & k b_{3} & k c_{3} \end{matrix} ∣ ∣ ∣ ∣

D_{0} = k^{3} D

What will be the fraction of volume of a floating object of volume V and density d above the surface of a liquid of density D?

a < 0

D < 0

y = a x^{2} + b x + c

(where D = b^{2} - 4 a c)

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