Question

A wooden cube of side 10 cm and specific gravity 0.8 floats in water with its upper surface horizontal. What depth of the cube remains immersed? what mass of aluminum of specific gravity 2.7 must be attached to

(i)The upper surface

(ii) The lower surface so that the cube will be just immersed

Answer 1

$V_s{d}_{liq}g=V_s{d}_{s}g$

$d_s=0.8$

$d_l=1$

$V_s\times 1={10}^3\times 0.8$

$V_s=800c{m}^3$

$V_s=800=Ah=10\times 10\times h$

$h=8cm$

$\left(ii\right)incaseofbothsurfaceneedtobeimmerse$

$V_{s}d=(m_a+m_s)g$

$V_s=V_S+V_a$

${10}^3+V_a\times 1=V_a\times 2.7+0.8\times {10}^3$

$m_a=\frac{200}{1.7}\times 2.7=317$