Question

An object of weight $W$ and density $\rho$ is submerged in the fluid of density $\rho 1$. Its apparent weight will be
$\left( A \right)W\left( {\rho - \rho 1} \right)$
$\left( B \right)\dfrac{{\left( {\rho - \rho 1} \right)}}{W}$
$\left( C \right)W\left( {1 - \dfrac{{\rho 1}}{\rho }} \right)$
$\left( D \right)W\left( {\rho 1 - \rho } \right)$

Answer 1

First we need to calculate the volume of the object then equate volume of the fluid displaced with the volume of the object.Now using the upthrust formula on the solid we can find the upthrust on the liquid. We know that apparent weight is the difference of rue weight of the solid and upthrust.Now in this way we can find the apparent weight of the solid.

Complete step by step solution:
From the problem we know that,
Weight of the object is equal to $W$.
The density of the object is equal to $\rho$.
The density of the fluid is equal to $\rho 1$.
We know ,
Weight is equal to mass multiplied with acceleration due to gravity.
Mathematically,
$W = mg$
Hence we can write mass of the object as,
$m = \dfrac{W}{g} \ldots \ldots \left( 1 \right)$
Where,
Mass of the object = m
Weight of the object = W
Acceleration due to gravity = g
We know,
Volume is equal to mass divided by density.
Matematiccly,
$V = \dfrac{m}{\rho }$
Putting equation $\left( 1 \right)$ in the above equation we will get,
$V = \dfrac{{\dfrac{W}{g}}}{\rho } \ldots \ldots \left( 2 \right)$
As per the archimedes principle,
Volume of fluid displaced = Volume of object
Hence, Using equation $\left( 2 \right)$ we will get,
Volume of fluid displaced = $\dfrac{W}{{g\rho }}$
$\Rightarrow VL = \dfrac{W}{{g\rho }} \ldots \ldots \left( 3 \right)$
Now,
We know the formula for upthrust,
Upthrust on the object is the multiplication of volume of fluid displaced, Density of liquid and acceleration due to gravity.
Mathematically,
$Upthrust = VL \times \rho 1 \times g$
Putting equation $\left( 3 \right)$ in the above equation we will get,
$Upthrust = \dfrac{W}{{g\rho }} \times \rho 1 \times g$
Cancelling the common terms we will get,
$Upthrust = \dfrac{W}{\rho } \times \rho 1$
Now we need to calculate apparent weight
Where,
Apparent weight = True weight of the object – Upthrust on the object
Now putting the correcting given values of the true weight and equation $\left( 4 \right)$ respectively we will get,
Apparent weight = $W - \dfrac{W}{\rho } \times \rho 1$
Taking out W as the common term we will get,
Apparent weight = $W\left( {1 - \dfrac{{\rho 1}}{\rho }} \right)$
Therefore the correct option is $\left( C \right)$ .

Note:
Remember weight and mass of a body are not the same things. Weight changes its value as it depends on the value of acceleration due to gravity whereas mass never changes its value it remains the same whatever tha souring it is. Another important thing in this type of problem is that the apparent loss in weight would be equal to the weight of the fluid displaced.