Give reason why, a block of plastic when released under water comes up to the surface of the water.
Answer
631.5k+ views
Hint: Recall Archimedes’ principle which states that a body when partially or fully submerged in a liquid, experiences an upward force that is equal to the weight of the liquid displaced by the object.
Complete step by step answer:
Note: Students must remember that the volume of liquid displaced and the body can only be considered equal when the body is fully submerged. If the body is floating or partially submerged, the volume of liquid displaced should be taken as the volume of the body that is below the surface of the liquid.
Complete step by step answer:
The reason why a block of plastic when released under water comes up to the surface of the water is due to the effect of the Archimedes’ principle. Archimedes' principle states that a body when partially or fully submerged in a liquid, experiences an upward force that is equal to the weight of the liquid displaced by the object. This force is known as buoyant force and this phenomenon is known as buoyancy.
The value of the buoyant force is
${{F}_{B}}={{m}_{l}}g$
Where $F_B$ is the magnitude of the upward buoyant force, $m_l$ is the mass of the liquid displaced and $g$ is the acceleration due to gravity.
${{F}_{B}}={{V}_{l}}{{\rho }_{l}}g$
(Using the relation that mass = density volume).
Where $V_l$ and $ρ_l$ are the volume of liquid displaced and its density respectively.
Thus, when a plastic body is released under water, it experiences an upward buoyant force. Simultaneously it also experiences a downward gravitational force due to its own weight which is equal to
${{F}_{g}}={{V}_{p}}{{\rho }_{p}}g$
(Using the relation that mass = density volume).
Where $V_p$ and $ρ_p$ are the volume and density of the plastic body respectively.
Now, when the plastic body is fully submerged in water, then volume of water displaced is equal to the volume of the plastic body (i.e. $V_p=V_l$).
Then, the net force experienced by the body in the upward direction will be equal to its weight subtracted from the buoyant force. (${F_B}-{F_g}$)
Net force = ${{F}_{B}}-{{F}_{g}}$
$={{V}_{l}}{{\rho }_{l}}g-{{V}_{p}}{{\rho }_{p}}g$
$={{V}_{l}}g({{\rho }_{l}}-{{\rho }_{g}})$
$V_l=V_p$ since the body is fully submerged
Now, since ($ρ_l>ρ_g$, i.e., the density of water is more than that of plastic), the above result is positive and the body experiences a net force in the upward direction.
So, a block of plastic when released under water comes up to the surface of the water.
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