Question

A layer of oil with density $724\, kg/m^3$ floats on water of density $1000\, kg/m^3$. A block floats at the oil-water interface with $1/6$ of its volume in oil and $5/6$ of its volume in water, as shown in the figure. What is the density of the block?

• A. $776\, kg/m^3$
• B. $954\, kg/m^3$
• C. $1024\, kg/m^3$
• D. $1276\, kg/m^3$

Answer 1

Answer: (B)

Given,
Density of oil $\left(\rho_{1}\right)=724\, kg / m ^{3}$
Density of water $\left(\rho_{2}\right)=1000\, kg / m ^{3}$
According to Archemedes' principle,
Upthrust = Weight of the liquid displaced
$V \rho_{b} g=V_{0} \rho_{1} g +V_{w} \rho_{2} g$
$\left(V_{0}\right.$ and $V_{w}$ are volume of oil and water displaced respectively)
$V \rho_{b} g=\frac{V}{6}(724) g+\frac{5 V}{6}(1000) g$
where, $\rho_{b}=$ density of block
and $V_{0} =\frac{V}{6}$ and $V_{w}=\frac{5 V}{6}$
$\Rightarrow V \rho_{b} g =\left(\frac{724 V}{6}+\frac{5000 V}{6}\right) g$
$\Rightarrow V_{b} g =\left(\frac{724 V+5000 V}{6}\right) g$
$\Rightarrow V \rho_{b} g =\frac{5724}{6} V g$
$\rho_{b} =954\, kg / m ^{3}$