If W be the weight of a body of density ρ in vacuum then its apparent weight in air of density ρ is

The correct option is W(ρ-σ/ρ).

Here is the detailed step-by-step derivation to find the apparent weight of the body in air:

Step 1: Understand the true weight of the body in vacuum
The true weight of a body, $W$, in a vacuum is equal to the gravitational force acting on its mass. If $V$ represents the volume of the body and $\rho$ represents its density, the mass of the body is $V\rho$. Thus, the weight in vacuum is given by:

$W=V\rho g$

where $g$ is the acceleration due to gravity.

From this equation, we can express the volume $V$ of the body as:

$V=\frac{W}{\rho g}$

Step 2: Determine the buoyant force (upthrust) exerted by the air
When the body is placed in air of density $\sigma$, it experiences an upward buoyant force (upthrust), $U$, according to Archimedes' principle. The buoyant force is equal to the weight of the air displaced by the body:

$U=V\sigma g$

Substituting the expression for volume $V$ from Step 1 into this formula, we get:

$U=\left(\frac{W}{\rho g}\right)\sigma g=\frac{W\sigma }{\rho }$

Step 3: Calculate the apparent weight in air
The apparent weight of the body in air, $W_{\text{apparent}}$, is the true weight in vacuum minus the upward buoyant force exerted by the air:

$W_{\text{apparent}}=W-U$

Substituting the expression for $U$ into the equation:

$W_{\text{apparent}}=W-\frac{W\sigma }{\rho }$

Factoring out $W$ from the right-hand side:

$W_{\text{apparent}}=W\left(1-\frac{\sigma }{\rho }\right)$

Simplifying the expression inside the parentheses by taking a common denominator:

$W_{\text{apparent}}=W\left(\frac{\rho -\sigma }{\rho }\right)$

Thus, the apparent weight of the body in air is indeed $W(\rho -\sigma /\rho )$.