A piece of wood of uniform cross section and height 15 cm floats vertically with its height 10 cm in water and 12 cm in spirit. Find the density of (i) wood and (ii) spirit.

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Solution

Let densities of water, wood and spirit are ρ, ρ_w and ρ_s respectively.

Let A be the area of the wooden block.

Total volume of the wooden block is, V = 15A

1. Mass of wood = Mass of water displaced by the wood

Vρ_wg = A×10×ρ×g

$rho subscript w equals fraction numerator A cross times 10 cross times rho cross times g over denominator 15 cross times A cross times g end fraction equals 0.67 space g c m to the power of negative 3 end exponent$

2. Mass of the wood = Mass of then spirit displaced

Vρ_wg = A×12×ρ_s×g
15× Aρ_wg = A×12×ρ_s×g
ρ_s= $\frac{15 \times 0.67}{12}$ = 0.8375 g/ $c m^{3}$

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