A solid of density 'D' has weight 'W'. It is completely immersed in a liquid of density 'd', then apparent weight of the solid is :

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Question

$W (1 + \frac{d}{D})$

$W (1 + \frac{D}{d})$

$W (1 - \frac{D}{d})$

$W (1 - \frac{d}{D})$

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The weight of body W = VDg
Where V = volume, D =density
The buoyant force exerted up by liquid W¢ = Vdg
∴ Apparent weight $= W - W^{'} = W (1 - \frac{W^{'}}{W})$
$= W (1 - \frac{Vdg}{VDg}) = W (1 - \frac{d}{D})$