Answer
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Hint To solve this question, we need to consider the free body diagram of the plastic box, which is released under water. We have to use the Archimedes’ principle for determining the upward buoyant force on the box.
Formula Used: The formula used to solve this question is given by
${F_B} = {V_i}\sigma g$, here ${F_B}$ is the buoyant force acting on a body whose ${V_i}$ portion of the volume is immersed inside a fluid of density $\sigma $.
Complete step by step answer
Consider a plastic box, of density $\rho $ released under water, of density $\sigma $ as shown below.
As we can see from the free body diagram of the plastic box, the forces acting on it are:
The weight of the box in downward direction$\left( W \right)$- The weight of a body is given by
$W = mg$
The mass of the box can be written in terms of its density as
$m = V\rho $ (1)
$\therefore W = V\rho g$ (2)
The buoyant force in the upward direction $\left( {{F_B}} \right)$- We know from the Archimedes’ principle that the buoyant force on a body immersed in a fluid is given by
${F_B} = {V_i}\sigma g$
As the box is completely immersed inside the water, so ${V_i} = V$. Therefore the buoyant force on the box is
${F_B} = V\sigma g$ (3)
So, the net force in the upward direction on the box is
$F = {F_B} - W$
From (1) and (3) we have
$F = V\sigma g - V\rho g$
$ \Rightarrow F = V\left( {\sigma - \rho } \right)g$ (4)
We know that the plastic is lighter than water. So, its density is less than that of water. Therefore we have
$\rho < \sigma $
Or $\sigma > \rho $
So from (4) we see that the net force in upward direction is positive. Thus the plastic box, which was released under the water, experiences a net force in the upward direction, which accelerates it upwards.
Hence, it comes up to the surface of the water until the upward buoyant force reduces to the weight of the box.
Note
The buoyant force acts in the upward direction. This does not mean that any substance which is released inside the water will come up to the surface. We must note that the magnitude of the upward buoyant force must be greater than its weight. For this, the density of the substance should be less than that of water. If the density of a substance is more than that of water, then that substance would rather sink. In our case, the substance was plastic, whose density is less than that of water.
Formula Used: The formula used to solve this question is given by
${F_B} = {V_i}\sigma g$, here ${F_B}$ is the buoyant force acting on a body whose ${V_i}$ portion of the volume is immersed inside a fluid of density $\sigma $.
Complete step by step answer
Consider a plastic box, of density $\rho $ released under water, of density $\sigma $ as shown below.
As we can see from the free body diagram of the plastic box, the forces acting on it are:
The weight of the box in downward direction$\left( W \right)$- The weight of a body is given by
$W = mg$
The mass of the box can be written in terms of its density as
$m = V\rho $ (1)
$\therefore W = V\rho g$ (2)
The buoyant force in the upward direction $\left( {{F_B}} \right)$- We know from the Archimedes’ principle that the buoyant force on a body immersed in a fluid is given by
${F_B} = {V_i}\sigma g$
As the box is completely immersed inside the water, so ${V_i} = V$. Therefore the buoyant force on the box is
${F_B} = V\sigma g$ (3)
So, the net force in the upward direction on the box is
$F = {F_B} - W$
From (1) and (3) we have
$F = V\sigma g - V\rho g$
$ \Rightarrow F = V\left( {\sigma - \rho } \right)g$ (4)
We know that the plastic is lighter than water. So, its density is less than that of water. Therefore we have
$\rho < \sigma $
Or $\sigma > \rho $
So from (4) we see that the net force in upward direction is positive. Thus the plastic box, which was released under the water, experiences a net force in the upward direction, which accelerates it upwards.
Hence, it comes up to the surface of the water until the upward buoyant force reduces to the weight of the box.
Note
The buoyant force acts in the upward direction. This does not mean that any substance which is released inside the water will come up to the surface. We must note that the magnitude of the upward buoyant force must be greater than its weight. For this, the density of the substance should be less than that of water. If the density of a substance is more than that of water, then that substance would rather sink. In our case, the substance was plastic, whose density is less than that of water.
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