Question

A block of wood floats in brine solution of density 1.2g.cm^-3 such that 3/8 of it's volume is above brine . Calculate the density of the wood​

Answer 1

Answer:

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Answer 2

$\boxed{\textbf{Answer}} \\ \mathrm{The \: density \: of \: the \: brine \: solution, } \\ \rho_{b} = 1.2g/ {cm}^{3} = \frac{1.2 \times {10}^{ - 3} }{( {10}^{ - 2})^{3} } = 1.2 \times {10}^{3} kg/ {m}^{3} \\ \mathrm{Let \: the \: volume \: of \: the \: wood\: be \: V } \\ \mathrm{The \: volume \: of \: the \: Wood \: inside \: the \: brine \: solution,} \\ V' = (1 - \frac{3}{8} ) V = \frac{5}{8} V$

$\mathrm{The \: weight \: of \: the \: wood \: is \: balanced \: by \: the \: weight \: of \: the \: displace \: volume \: brine \: solution. } \\ \therefore \: weight \: of \: the \: wood = weight \: of \: the \: displace \: volume \: brine \: solution. \\ \implies \: \rho_{w}Vg = \rho_{b}V'g \\ \implies \: \rho_{w}Vg = \rho_{b} \times \frac{5}{8} Vg \\ \implies \: \rho_{w} = 1.2 \times 10^{3} \times \frac{5}{8} \\ = 0.75 \times {10}^{3} kg/ {m}^{3}$