Gravitation

A body floats with ( 1 3 ) \left(\frac{1}{3}\right) of its volume outside water and ( 3 4 ) \left(\frac{3}{4}\right) of its volume outside another liquid.The density of the another liquid is:

( 8 3 ) × 1 0 3 \left(\dfrac{8}{3}\right)×10^3 k g kg m 3 m^{-3} ( 3 9 ) × 1 0 3 \left(\dfrac{3}{9}\right)×10^3 k g kg m 3 m^{-3} ( 9 4 ) × 1 0 3 \left(\dfrac{9}{4}\right)×10^3 k g kg m 3 m^{-3} ( 4 9 ) × 1 0 3 \left(\dfrac{4}{9}\right)×10^3 k g kg m 3 m^{-3}

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1 solution

Peter Macgregor
Apr 14, 2016

We will use Archimedes principle that a floating body displaces its weight.

If the body has volume V V and weight W W and the densities of the liquid and of water are ρ L \rho_L and ρ W \rho_W we can write

2 3 V ρ W = M \frac{2}{3}V\rho_W=M

1 4 V ρ L = M \frac{1}{4}V\rho_L=M

2 3 ρ W = 1 4 ρ L \implies\frac{2}{3}\rho_W=\frac{1}{4}\rho_L

ρ L ρ W = 8 3 \implies\frac{\rho_L}{\rho_W}=\frac{8}{3}

Taking the density of water to be 1000 k g m 3 1000 kg m^{-3} gives the answer

ρ L = 8 3 × 1 0 3 k g m 3 \rho_L=\frac{8}{3}\times 10^{3}kg m^{-3}

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