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A block of wood floats in a liquid with four-fifths of its volume submerged. If the relative density of wood is 0.8 , what is the density of the liquid in units of $\mathrm{\mathrm{kg} \, \mathrm{m}^{-3}}$ ?
Option: 1
750

Option: 2
1000

Option: 3
1250

Option: 4
1500

Answers (1)

best_answer

$\mathrm{\text { Let the volume of the block be } V \mathrm{~m}^3 \text {. }}$

$\mathrm{\text { Volume of block under liquid }=\frac{4 \mathrm{~V}}{5} \mathrm{~m}^3}$

$\mathrm{\therefore \quad \text { Volume of liquid displaced }=\frac{4 V}{5} \mathrm{~m}^3}$

$\mathrm{\text { Now let the density of the liquid be } \rho \ \ \mathrm{kg} \mathrm{m}^{-3} \text {. }}$

$\mathrm{\text { Mass of liquid displaced }}$

$\mathrm{=(\text { volume of liquid displaced }) \times(\text { density of liquid })}$

$\mathrm{=\frac{4 V}{5} \rho \mathrm{kg}}$

$\mathrm{\text { Weight of liquid displaced }=\frac{4 V}{5} \times \rho \times g \text { newton }}$

$\mathrm{\text { Relative density of wood }=0.8}$

$\mathrm{\therefore \quad \text { Density of wood }=0.8 \times 1000}$

$\mathrm{=800 \mathrm{~kg} \mathrm{~m}^{-3}}$

$\mathrm{\therefore \text { Mass of the block }=800 \times V \mathrm{~kg}}$

$\mathrm{\text { Weight of the block }=800 \times V \times g \text { newton }}$

From the law of flotation,

$\mathrm{\text { Weight of block }=\text { weight of liquid displaced }}$

$\mathrm{\text { or } \quad 800 \times V \times g=\frac{4 V}{5} \times \rho \times g}$

$\mathrm{\text { or } \quad \rho=800 \times \frac{5}{4}}$

$\mathrm{=1000 \mathrm{~kg} \mathrm{~m}^{-3}}$

Posted by

manish painkra

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