If a body weighs 60g in air and 40g in water then the specific gravity of the body is
\[\begin{align}
& A)3 \\
& B)6 \\
& C)1.5 \\
& D)4.5 \\
\end{align}\]
Answer
606.3k+ views
Hint: Weight of a body in two different mediums is given. We need to find the specific gravity of the body. Specific gravity or relative density of an object is the ratio of weight of the body in air to apparent loss of weight in water. By using this equation, we can find the specific gravity of the body.
Formula used:
\[\text{Specific gravity = }\dfrac{\text{Weight of object in air}}{\text{Apparent loss of weight in water}}\]
Complete step by step solution:
Given that,
Weight of the body in air, \[\text{= 60g}\]
Weight of the body in water, \[\text{= 40g}\]
We know,
\[\text{Specific gravity = }\dfrac{\text{Weight of object in air}}{\text{Apparent loss of weight in water}}\]
\[\text{Apparent loss of weight in water = }\!\!~\!\!\text{ Weight in air - Weight in water = 60-40 =20g}\]
Therefore, \[\text{Specific gravity of the body =}\dfrac{60}{20}=3\]
Answer is option A.
Additional information:
Density of a substance is defined as the ratio of mass to volume. A change in volume of the substance will result in change in its density. It is the measure of matter within a given area. Density of a substance is a unique number. Wherein, relative density or specific gravity of an object is defined as the ratio of density of the object in air to the density of water at \[\text{4}{}^\circ \text{C}\]. It can be measured in many ways.
\[\text{Specific gravity = }\dfrac{\text{Density of object in air}}{\text{Density of water at 4}{}^\circ \text{C}}\]
Or,
\[\text{Specific gravity = }\dfrac{\text{Weight of object in air}}{\text{Apparent loss of weight in water}}\]
Or
\[\text{Specific gravity = }\dfrac{\text{Mass of substance}}{\text{Mass of equal volume of water at 4}{}^\circ \text{C}}\]
Note:
Value of relative density is the same in all systems of measurement. Since, it is the ratio of two similar quantities, the relative density doesn’t have a unit. It is a pure number. It tells us how many times an object is heavier than water. Relative density of a substance can be found or compared with numerous other reference bodies. If the relative density of a substance is less than one it will float on that reference object.
Formula used:
\[\text{Specific gravity = }\dfrac{\text{Weight of object in air}}{\text{Apparent loss of weight in water}}\]
Complete step by step solution:
Given that,
Weight of the body in air, \[\text{= 60g}\]
Weight of the body in water, \[\text{= 40g}\]
We know,
\[\text{Specific gravity = }\dfrac{\text{Weight of object in air}}{\text{Apparent loss of weight in water}}\]
\[\text{Apparent loss of weight in water = }\!\!~\!\!\text{ Weight in air - Weight in water = 60-40 =20g}\]
Therefore, \[\text{Specific gravity of the body =}\dfrac{60}{20}=3\]
Answer is option A.
Additional information:
Density of a substance is defined as the ratio of mass to volume. A change in volume of the substance will result in change in its density. It is the measure of matter within a given area. Density of a substance is a unique number. Wherein, relative density or specific gravity of an object is defined as the ratio of density of the object in air to the density of water at \[\text{4}{}^\circ \text{C}\]. It can be measured in many ways.
\[\text{Specific gravity = }\dfrac{\text{Density of object in air}}{\text{Density of water at 4}{}^\circ \text{C}}\]
Or,
\[\text{Specific gravity = }\dfrac{\text{Weight of object in air}}{\text{Apparent loss of weight in water}}\]
Or
\[\text{Specific gravity = }\dfrac{\text{Mass of substance}}{\text{Mass of equal volume of water at 4}{}^\circ \text{C}}\]
Note:
Value of relative density is the same in all systems of measurement. Since, it is the ratio of two similar quantities, the relative density doesn’t have a unit. It is a pure number. It tells us how many times an object is heavier than water. Relative density of a substance can be found or compared with numerous other reference bodies. If the relative density of a substance is less than one it will float on that reference object.
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