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 aluminimum 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

$V_{solidInvested}\times d_{liquid}\times g=V_{solid}\times d_{solid}\times g$

Given: $V_{solid}={10}^3$

$d_{solid}=0.8$

$d_{lig}=1$

$V_{si}\times 1={10}^3\times 0.8$

$V_{si}=800c{m}^3$

$V_{si}=800=A\times h=10\times 10\times h$

$h=\frac{800}{100}=8cm$

As in first case, only cube needs to be immersed so,

$V_{si}deg=(m_a+m_s)g$

Where $m_a=$ mass of aluminium

$m_s=$ mass of solid

$V_{si}={10}^3$ as completely immersed

${10}^3\times 1=m_a+{10}^3\times 0.8$ $(\because m_s=V_s{d}_s)$

Hence both needs to be immersed

So $V_{si}=(V_s+V_a)$

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

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