A body of mass $m$ is immersed in water with $3 / 4 t h$ of volume outside of water. Buoyant force acting on it is $1000 N$ . Total volume of body is

( Take density of water as $1000 k g / m^{3}$ and accelaration due to gravity as $10 m / s^{2}$ )

$1 m^{3}$

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$0.8 m^{3}$

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$10 m^{3}$

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$0.4 m^{3}$

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Solution

The correct option is D
$0.4 m^{3}$

Given buoyant force = $1000 N$

Density of water, $ρ_{w}$ = $1000 k g / m^{3}$

Accelaration due to gravity, $g$ = $10 m / s^{2}$

Volume of water displaced, $v_{w}$ = $\frac{1}{4} v_{b}$ where $v_{b}$ is the volume of the body.

Buoyant force, $f$ = $\frac{v_{b} ρ_{w} g}{4}$

$v_{b}$ = $\frac{4000}{10000}$ = $0.4 m^{3}$

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A body of mass m is immersed in water with 3/4th of volume outside of water. Buoyant force acting on it is 1000N. Total volume of body is ( Take density of water as 1000kg/m3 and accelaration due to gravity as 10m/s2)

The correct option is D 0.4m3 Given buoyant force = 1000N Density of water, ρw = 1000kg/m3 Accelaration due to gravity, g = 10m/s2 Volume of water displaced, vw = 14vb where vb is the volume of the body. Buoyant force, f = vbρwg4 vb = 400010000 = 0.4m3

A body of mass $m$ is immersed in water with $3 / 4 t h$ of volume outside of water. Buoyant force acting on it is $1000 N$ . Total volume of body is

( Take density of water as $1000 k g / m^{3}$ and accelaration due to gravity as $10 m / s^{2}$ )