Solveeit Logo

Question

Physics Question on Hydrostatics

A layer of oil with density 724kg/m3724\, kg/m^3 floats on water of density 1000kg/m31000\, kg/m^3. A block floats at the oil-water interface with 1/61/6 of its volume in oil and 5/65/6 of its volume in water, as shown in the figure. What is the density of the block?

A

776kg/m3776\, kg/m^3

B

954kg/m3954\, kg/m^3

C

1024kg/m31024\, kg/m^3

D

1276kg/m31276\, kg/m^3

Answer

954kg/m3954\, kg/m^3

Explanation

Solution

Given,
Density of oil (ρ1)=724kg/m3\left(\rho_{1}\right)=724\, kg / m ^{3}
Density of water (ρ2)=1000kg/m3\left(\rho_{2}\right)=1000\, kg / m ^{3}
According to Archemedes' principle,
Upthrust = Weight of the liquid displaced
Vρbg=V0ρ1g+Vwρ2gV \rho_{b} g=V_{0} \rho_{1} g +V_{w} \rho_{2} g
(V0\left(V_{0}\right. and VwV_{w} are volume of oil and water displaced respectively)
Vρbg=V6(724)g+5V6(1000)gV \rho_{b} g=\frac{V}{6}(724) g+\frac{5 V}{6}(1000) g
where, ρb=\rho_{b}= density of block
and V0=V6V_{0} =\frac{V}{6} and Vw=5V6V_{w}=\frac{5 V}{6}
Vρbg=(724V6+5000V6)g\Rightarrow V \rho_{b} g =\left(\frac{724 V}{6}+\frac{5000 V}{6}\right) g
Vbg=(724V+5000V6)g\Rightarrow V_{b} g =\left(\frac{724 V+5000 V}{6}\right) g
Vρbg=57246Vg\Rightarrow V \rho_{b} g =\frac{5724}{6} V g
ρb=954kg/m3\rho_{b} =954\, kg / m ^{3}