
A wooden block floating in a bucket of water has $\dfrac{4}{5}$ of its volume submerged. If a specific amount of an oil is poured into the bucket, it is visible that the block is just under the oil surface with half of its volume under water and half in oil. What will be the density of oil relative to that of water?
$\begin{align}
& A.0.5 \\
& B.0.7 \\
& C.0.6 \\
& D.0.8 \\
\end{align}$
Answer
557.4k+ views
Hint: The weight will be a constant. The weight in the liquid is given by the product of the volume of the body, density of the body and acceleration due to gravity of the body. Find the ratio of the submerged volume of the block to the volume of the block. Find the equation for the weight of the body which is equal in equilibrium condition. This all will help you in answering the question.
Complete step by step answer:
For the first situation, we can write that,
${{V}_{b}}{{\rho }_{b}}g={{V}_{s}}{{\rho }_{w}}g$
Where ${{V}_{b}}$ be the volume of the block, ${{\rho }_{b}}$ be the density of the wooden block, $g$ be the acceleration due to gravity, ${{V}_{s}}$ be the volume of the submerged block and ${{\rho }_{w}}$ be the density of the water. Rearranging the equation will give,
$\dfrac{{{V}_{s}}}{{{V}_{b}}}=\dfrac{{{\rho }_{b}}}{{{\rho }_{w}}}=\dfrac{4}{5}$
In the equilibrium condition, the condition can be written as,
${{V}_{b}}{{\rho }_{b}}g=\dfrac{{{V}_{b}}}{2}{{\rho }_{o}}g+\dfrac{{{V}_{b}}}{2}{{\rho }_{w}}g$
Where ${{\rho }_{o}}$be given as the density of the oil.
Cancelling the common terms in the equation will give,
$2{{\rho }_{b}}={{\rho }_{o}}+{{\rho }_{w}}$
From this we will get that,
$\dfrac{{{\rho }_{o}}}{{{\rho }_{w}}}=\dfrac{3}{5}=0.6$
Therefore the density of the oil which is relative to the water is given as $0.6$.
The correct answer is mentioned in the option as option C.
Note:
Fluid pressure is defined as a measurement of the force per unit area on a body kept in the fluid or on the surface of a closed container. This pressure can be because of the gravity, acceleration, or by any other forces outside a closed container.
Complete step by step answer:
For the first situation, we can write that,
${{V}_{b}}{{\rho }_{b}}g={{V}_{s}}{{\rho }_{w}}g$
Where ${{V}_{b}}$ be the volume of the block, ${{\rho }_{b}}$ be the density of the wooden block, $g$ be the acceleration due to gravity, ${{V}_{s}}$ be the volume of the submerged block and ${{\rho }_{w}}$ be the density of the water. Rearranging the equation will give,
$\dfrac{{{V}_{s}}}{{{V}_{b}}}=\dfrac{{{\rho }_{b}}}{{{\rho }_{w}}}=\dfrac{4}{5}$
In the equilibrium condition, the condition can be written as,
${{V}_{b}}{{\rho }_{b}}g=\dfrac{{{V}_{b}}}{2}{{\rho }_{o}}g+\dfrac{{{V}_{b}}}{2}{{\rho }_{w}}g$
Where ${{\rho }_{o}}$be given as the density of the oil.
Cancelling the common terms in the equation will give,
$2{{\rho }_{b}}={{\rho }_{o}}+{{\rho }_{w}}$
From this we will get that,
$\dfrac{{{\rho }_{o}}}{{{\rho }_{w}}}=\dfrac{3}{5}=0.6$
Therefore the density of the oil which is relative to the water is given as $0.6$.
The correct answer is mentioned in the option as option C.
Note:
Fluid pressure is defined as a measurement of the force per unit area on a body kept in the fluid or on the surface of a closed container. This pressure can be because of the gravity, acceleration, or by any other forces outside a closed container.
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