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

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

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

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\frac{d_{0}}{d}

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

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Solution

The correct option is D $\frac{d - d_{0}}{d}$
Let x be the fraction of volume of object floating above the surface of the liquid.
As weight of liquid displaced = weight of object
$∴ (V_{0} - x V_{0}) d g = V_{0} d_{0} g$
$(1 - x) d = d_{0}$ or $x = 1 - \frac{d_{0}}{d} = \frac{d - d_{0}}{d}$

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Similar questions

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

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

a < 0

D < 0

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

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

The density of liquid at any temperature t is given $d_{t} = d_{0} [1 - γ t]$ . This equation applies to :

d_{0} = 2 d_{i}

d_{0}

d_{i}

d_{0}

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