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Question: A block of wood is floating on oil with half of its volume submerged. If the density of oil \[840\]\...

A block of wood is floating on oil with half of its volume submerged. If the density of oil 840$$$$kg{m^{ - 3}}, the relative density of wood (relative to water) is :
A. 0.840.84
B. 0.420.42
C. 0.210.21
D. 1.001.00

Explanation

Solution

Buoyant force is generally a pressure experienced by object when it is immersed into the fluid.
In general, buoyancy is positive, because we assume that an object has less density than fluid or liquid displaced. But in Physics, other types of buoyancy are negative and neutral.
When the density of an immersed object is greater than fluid displaced, then resultant buoyancy is called negative buoyancy.
Similarly, when the density or weight of an immersed object is equal to the density of fluid displaced, neutral buoyancy occurs in this case. Any object floats or sinks in the fluid on the basis of buoyant force experienced by the object immersed.
The direction of buoyant force is always upward in direction. So we can use equilibrium conditions to find the density of wooden blocks with respect to water.

Complete step by step answer:
At equilibrium, the buoyant force acting upwards is equal to the weight of the body (which acts downward direction)
Numerically, we can write that
Vimmersed×ρoil×g=mg{V_{immersed}} \times {\rho _{oil}} \times g = mg
where, Vimmersed{V_{immersed}} = immersed volume of wood block in oil, ρoil{\rho _{oil}} = density of oil, VV = volume of wooden block, mm = mass of wooden block
We know that m=V×dwoodm = V \times {d_{wood}}
⇒Vimmersed×ρoil×g=V×dwood×g\Rightarrow {V_{immersed}} \times {\rho _{oil}} \times g = V \times {d_{wood}} \times g
∴V2×ρoil=V×dwood\therefore \dfrac{V}{2} \times {\rho _{oil}} = V \times {d_{wood}} [Vimmersed=V2{V_{immersed}} = \dfrac{V}{2}According to question]
⇒dwood=ρoil2\Rightarrow {d_{wood}} = \dfrac{{{\rho _{oil}}}}{2}
⇒dwood=8402\Rightarrow {d_{wood}} = \dfrac{{840}}{2}
∴dwood=420\therefore {d_{wood}} = 420
⇒mg=V×dwood×g\Rightarrow mg = V \times {d_{wood}} \times g
⇒Vimmersed×ρoil×g=V×dwood×g\Rightarrow {V_{immersed}} \times {\rho _{oil}} \times g = V \times {d_{wood}} \times g
∴Vimmersed×ρoil=V×dwood\therefore {V_{immersed}} \times {\rho _{oil}} = V \times {d_{wood}}
V2×ρoil=V×dwood\dfrac{V}{2} \times {\rho _{oil}} = V \times {d_{wood}}
⇒dwood=ρoil2\Rightarrow {d_{wood}} = \dfrac{{{\rho _{oil}}}}{2}
⇒dwood=8402\Rightarrow {d_{wood}} = \dfrac{{840}}{2}
\therefore {d_{wood}} = 420$$$$kg{m^{ - 3}}
We know that, density{\text{ }}of{\text{ }}water = 1000$$$$kg{m^{ - 3}}
Hence, the density of wood relative to water=4201000=0.42\dfrac{{420}}{{1000}} = 0.42.

So, the correct answer is “Option B”.

Additional Information:
The concept of buoyancy or buoyant force is defined by Archimedes’ principle. According to Archimedes’ principle, when an object is immersed in a fluid or liquid, then the object experiences a force in upward direction. This force is applied by fluid. The magnitude of this force is equal to the volume of fluid or liquid displaced by the object immersed. The force applied by fluid on an object is known as buoyant force.
The buoyant force always opposes gravity.

Note:
In question, the direction of different forces acts on an object plays an important role for calculation. For getting an accurate answer, students must take carefully the direction of forces acting on objects. Important point to remember that direction of weight of object is always downwards and direction of buoyant force is always upward.