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Question: A body of density \( {{d}_{1}} \) is counterpoised by \( Mg \) of weights of density \( {{d}_{2}} \)...

A body of density d1{{d}_{1}} is counterpoised by MgMg of weights of density d2{{d}_{2}} in air of density dd . Then, true mass of the body is
A. M\text{A}\text{. }M
B. M(1−dd2)(1−dd1)\text{B}\text{. }\dfrac{M\left( 1-\dfrac{d}{{{d}_{2}}} \right)}{\left( 1-\dfrac{d}{{{d}_{1}}} \right)}
C. M(1−dd2)\text{C}\text{. }M\left( 1-\dfrac{d}{{{d}_{2}}} \right)
D. M(1−dd1)\text{D}\text{. }M\left( 1-\dfrac{d}{{{d}_{1}}} \right)

Explanation

Solution

Hint: The weight of the body is counterpoised by the weight of another mass. Densities of both objects are different. We will equate the apparent mass of the body and apparent weight of mass. The equation obtained will give the true mass of the body.
Formula used:
Apparent weight of body =mg−Vdg=mg-Vdg

Complete step-by-step answer:
Apparent is a property of objects that corresponds to how heavy an object is. The apparent weight of an object differs from the weight of an object whenever the gravitational force acting on that object is not balanced by the normal force acting in the opposite direction on the body.
We have to calculate the true mass of the body whose weight is balanced by the weight of another mass having different density.
Apparent weight of the body is equal to the apparent weight of mass.
Apparent weight of body =mg−Vdg=mg-Vdg
Where,
mgmg is the actual weight of body
VV is the volume of body
dd is the density of body
We have, V=md1V=\dfrac{m}{{{d}_{1}}}
Therefore, apparent weight of body =mg−(md1)dg=mg-\left( \dfrac{m}{{{d}_{1}}} \right)dg
=m(1−dd1)g=m\left( 1-\dfrac{d}{{{d}_{1}}} \right)g
Apparent weight of body is equal to apparent weight of mass,
M(1−dd2)g==m(1−dd1)g m=M(1−dd2)(1−dd1) \begin{aligned} & M\left( 1-\dfrac{d}{{{d}_{2}}} \right)g==m\left( 1-\dfrac{d}{{{d}_{1}}} \right)g \\\ & m=M\dfrac{\left( 1-\dfrac{d}{{{d}_{2}}} \right)}{\left( 1-\dfrac{d}{{{d}_{1}}} \right)} \\\ \end{aligned}
Thus, the true mass of body is M(1−dd2)(1−dd1)M\dfrac{\left( 1-\dfrac{d}{{{d}_{2}}} \right)}{\left( 1-\dfrac{d}{{{d}_{1}}} \right)}
Hence, the correct option is B.

Additional information:
Real life examples where apparent weight finds its applications:
The apparent weight differs from the actual weight of an object when an object is partially or fully immersed in a fluid, where there is an upthrust from the fluid that is working against the force of gravity.
When a person is riding in an elevator and the elevator begins rising, the person exerts a force in the downward direction. Weight of the person becomes heavier because of the downward force, changing the apparent weight of the person.

Note:
Apparent weight of a body can be positive or negative depending upon the different situations. The apparent weight of an object can be negative in some situations such as, when the force of buoyancy on an object becomes more than the weight of the object when it is placed in a fluid. In this case, external force is required to balance the weight of the object. In the above question, apparent weight of the body was considered positive.