A body with an air cavity loses one-third of its weight when immersed in water. If the relative density of the material of the body is 6, then the ratio of the volume of the cavity and outer volume of the body is
A. \[\dfrac{1}{2}\]
B. \[\dfrac{2}{3}\]
C. \[\dfrac{3}{4}\]
D. \[\dfrac{4}{5}\]
Answer
623.7k+ views
Hint: Weight is defined as the force acting on the body due to gravity. Volume is defined as the quantity of the space that a substance occupies. Relative density is defined as the ratio of the density of a substance to the density of the given material.
Complete step by step answer:
Given data:
The relative density of body = 6
The ratio of the volume of the cavity and outer volume of the body =?
Let the weight of the body be w
Let the volume of the body is V1 and the volume of water is V2
Then when it is immersed in water, it loses one-third of its weight. Hence the weight in water$ = \dfrac{w}{3}$
We know that weight=mg and $\text{density} = \dfrac{{\text{mass}}}{{\text{volume}}}$
Thus \[\text{density}_{\text{body}} = \dfrac{{\dfrac{w}{g}}}{{{V_1}}} = \dfrac{w}{{{V_1}g}}\] and \[\text{density}_{\text{water}} = \dfrac{{\dfrac{w}{{3g}}}}{{{V_2}}} = \dfrac{w}{{3{V_2}g}}\] $\left( {\because m = \dfrac{w}{g}} \right)$
We know that the relative density of water is 1
\[\therefore \dfrac{\text{density}_{\text{body}}}{\text{density}_{\text{water}}} = \dfrac{{\dfrac{w}{{{V_1}g}}}}{{\dfrac{w}{{3{V_2}g}}}} = \dfrac{{3{V_2}}}{{{V_1}}}\]
Substituting the density of body and density of water values, we get
$ \Rightarrow \dfrac{6}{1} = \dfrac{{3{V_2}}}{{{V_1}}}$
\[ \Rightarrow 6{V_1} = 3{V_2}\]
\[ \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{3}{6} = \dfrac{1}{2}\].
Hence, the correct answer is option (A).
Note:
• By using a spring balance, we can measure the weight of the body. The unit is Newton in the SI system and Dye in the CGS system. Weight is a vector quantity. Weight depends on the acceleration due to gravity. Weight value differs from place to place.
• Mass is defined as the amount of matter that is contained in a body. Mass is a scalar quantity. Mass value remains the same everywhere. Mass can be measured by using a beam balance or even with a physical balance.
• Relative density is also known as specific gravity and it is unitless. Relative density depends on the mass, volume, and shape of the substance.
Complete step by step answer:
Given data:
The relative density of body = 6
The ratio of the volume of the cavity and outer volume of the body =?
Let the weight of the body be w
Let the volume of the body is V1 and the volume of water is V2
Then when it is immersed in water, it loses one-third of its weight. Hence the weight in water$ = \dfrac{w}{3}$
We know that weight=mg and $\text{density} = \dfrac{{\text{mass}}}{{\text{volume}}}$
Thus \[\text{density}_{\text{body}} = \dfrac{{\dfrac{w}{g}}}{{{V_1}}} = \dfrac{w}{{{V_1}g}}\] and \[\text{density}_{\text{water}} = \dfrac{{\dfrac{w}{{3g}}}}{{{V_2}}} = \dfrac{w}{{3{V_2}g}}\] $\left( {\because m = \dfrac{w}{g}} \right)$
We know that the relative density of water is 1
\[\therefore \dfrac{\text{density}_{\text{body}}}{\text{density}_{\text{water}}} = \dfrac{{\dfrac{w}{{{V_1}g}}}}{{\dfrac{w}{{3{V_2}g}}}} = \dfrac{{3{V_2}}}{{{V_1}}}\]
Substituting the density of body and density of water values, we get
$ \Rightarrow \dfrac{6}{1} = \dfrac{{3{V_2}}}{{{V_1}}}$
\[ \Rightarrow 6{V_1} = 3{V_2}\]
\[ \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{3}{6} = \dfrac{1}{2}\].
Hence, the correct answer is option (A).
Note:
• By using a spring balance, we can measure the weight of the body. The unit is Newton in the SI system and Dye in the CGS system. Weight is a vector quantity. Weight depends on the acceleration due to gravity. Weight value differs from place to place.
• Mass is defined as the amount of matter that is contained in a body. Mass is a scalar quantity. Mass value remains the same everywhere. Mass can be measured by using a beam balance or even with a physical balance.
• Relative density is also known as specific gravity and it is unitless. Relative density depends on the mass, volume, and shape of the substance.
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