Answer
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Hint: To solve this question, you need to know the concept of buoyancy and buoyant force acting on a body which is submerged in a liquid. Buoyant force is basically the force that acts upwards on a body submerged in a liquid and is equal to the weight of the volume of the liquid which is displaced by the submerged body. Hence, apparent weight in liquid can be given as:
${W_{app}} = {W_{air}} - B$ , where $B$ is the buoyant force acting on the body.
Complete step by step answer:
As explained in the hint section, to solve the question, we need to have a basic understanding of the concept Buoyant force and buoyancy.
Buoyant force can be explained as the force that acts on a body which is submerged in a liquid. This force is equal to the weight of the volume of the liquid which is displaced by the submerged body in magnitude. Hence, the buoyant force is actually independent of the density and properties of the body except for the dimensional property, volume. Two bodies having the same volume will experience exactly the same amount of buoyant force by the liquid if they are submerged in the same liquid.
Buoyant force can be mathematically given as:
$B = {M_l}g$
Where, $g$ is the acceleration due to gravity and,
${M_l}$ is the mass of the displaced liquid, which can be given as the product of the density of the liquid and volume of the liquid displaced, mathematically, it can be represented as:
${M_l} = {d_l} \times {V_b}$
Where, ${d_l}$ is the density of the liquid while ${V_b}$ is the volume of the body or the object submerged into the liquid.
In our case, let us assume that the body has a fixed volume of magnitude $V$
The density of the body, which we need to find out, can be assumed to be $d$
Hence, the weight of the body in air can be given as: ${W_{air}} = Vdg$
Now, the question has told us that this body is submerged in water, which has a density of ${d_l} = 1\,g/cc$
We also know that the volume of water displaced would be exactly the same as the volume of the body submerged, hence, the buoyant force acting upwards on the body can be given as:
$B = V{d_l}g$
Or, $B = Vg$ since ${d_l} = 1\,g/cc$
The apparent weight of the body can be given as:
${W_{app}} = {W_{air}} - B$
The question has told us that:
${W_{app}} = \dfrac{{{W_{air}}}}{3}$
Substituting the values of ${W_{air}}$ and $B$ , we get:
$\Rightarrow \dfrac{{Vdg}}{3} = Vdg - Vg$
Cancelling out $\left( {Vg} \right)$ from both sides of the equation, we get:
$\Rightarrow \dfrac{d}{3} = d - 1$
After transposing, we get:
$\Rightarrow d = 1.5\,g/cc$
Hence, the answer is: $d = 1.5\,g/cc$.
Note: Main error that numerous students make is that when reporting the answer, they report the density dimensionless, if you are reporting the final answer unitless, remember to mention that you have found out the relative density and not the density of the material of the body.
${W_{app}} = {W_{air}} - B$ , where $B$ is the buoyant force acting on the body.
Complete step by step answer:
As explained in the hint section, to solve the question, we need to have a basic understanding of the concept Buoyant force and buoyancy.
Buoyant force can be explained as the force that acts on a body which is submerged in a liquid. This force is equal to the weight of the volume of the liquid which is displaced by the submerged body in magnitude. Hence, the buoyant force is actually independent of the density and properties of the body except for the dimensional property, volume. Two bodies having the same volume will experience exactly the same amount of buoyant force by the liquid if they are submerged in the same liquid.
Buoyant force can be mathematically given as:
$B = {M_l}g$
Where, $g$ is the acceleration due to gravity and,
${M_l}$ is the mass of the displaced liquid, which can be given as the product of the density of the liquid and volume of the liquid displaced, mathematically, it can be represented as:
${M_l} = {d_l} \times {V_b}$
Where, ${d_l}$ is the density of the liquid while ${V_b}$ is the volume of the body or the object submerged into the liquid.
In our case, let us assume that the body has a fixed volume of magnitude $V$
The density of the body, which we need to find out, can be assumed to be $d$
Hence, the weight of the body in air can be given as: ${W_{air}} = Vdg$
Now, the question has told us that this body is submerged in water, which has a density of ${d_l} = 1\,g/cc$
We also know that the volume of water displaced would be exactly the same as the volume of the body submerged, hence, the buoyant force acting upwards on the body can be given as:
$B = V{d_l}g$
Or, $B = Vg$ since ${d_l} = 1\,g/cc$
The apparent weight of the body can be given as:
${W_{app}} = {W_{air}} - B$
The question has told us that:
${W_{app}} = \dfrac{{{W_{air}}}}{3}$
Substituting the values of ${W_{air}}$ and $B$ , we get:
$\Rightarrow \dfrac{{Vdg}}{3} = Vdg - Vg$
Cancelling out $\left( {Vg} \right)$ from both sides of the equation, we get:
$\Rightarrow \dfrac{d}{3} = d - 1$
After transposing, we get:
$\Rightarrow d = 1.5\,g/cc$
Hence, the answer is: $d = 1.5\,g/cc$.
Note: Main error that numerous students make is that when reporting the answer, they report the density dimensionless, if you are reporting the final answer unitless, remember to mention that you have found out the relative density and not the density of the material of the body.
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