A cubic body floats on mercury with $0.25$ fraction of its volume below the surface. $\frac{x}{10}$ fraction of the volume of the body will be immersed in the mercury if a layer of water poured on top of the mercury covers the body completely. Find $x$ .

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Solution

$W = (0.25) V ρ_{H g} g$ (i)
Let $x$ fraction of volume is immersed in mercury in the second case. Then,
$W = x V ρ_{H g} g + (1 - x) V ρ_{w} g$ (ii)
Equatiug Eqs. (i) and (ii), we have
$\frac{ρ_{H g}}{4} = x ρ_{H g} + (1 - x) ρ_{w}$
$∴ 12.6 x = 2.4$
or $x = 0.19$

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A cubic body floats on mercury with 0.25 fraction of its volume below the surface. x10 fraction of the volume of the body will be immersed in the mercury if a layer of water poured on top of the mercury covers the body completely. Find x.

W=(0.25)VρHgg (i)Let x fraction of volume is immersed in mercury in the second case. Then,W=xVρHgg+(1−x)Vρwg (ii)Equatiug Eqs. (i) and (ii), we haveρHg4=xρHg+(1−x)ρw∴12.6x=2.4or x=0.19

A cubic body floats on mercury with $0.25$ fraction of its volume below the surface. $\frac{x}{10}$ fraction of the volume of the body will be immersed in the mercury if a layer of water poured on top of the mercury covers the body completely. Find $x$ .

$W = (0.25) V ρ_{H g} g$ (i)
Let $x$ fraction of volume is immersed in mercury in the second case. Then,
$W = x V ρ_{H g} g + (1 - x) V ρ_{w} g$ (ii)
Equatiug Eqs. (i) and (ii), we have
$\frac{ρ_{H g}}{4} = x ρ_{H g} + (1 - x) ρ_{w}$
$∴ 12.6 x = 2.4$
or $x = 0.19$