Question

The weight of the body is
(A) $V \times \rho g$
(B) $\dfrac{V}{{\rho g}}$
(C) $\dfrac{{Vg}}{\rho }$
(D) none of these

Answer 1

Hint : In this problem, we have to equate the SI unit of the parameter which are given in the options. The SI unit of weight is Newton $\left( N \right)$ , and this unit can be written in another form $\left( {kgm{s^{ - 2}}} \right)$ . Keeping this as LHS and we have to equate the SI unit of the given options one by one and the solution can be determined.

Complete Step By Step Answer:
The SI unit of weight is $N$ , in other terms $kgm{s^{ - 2}}$ ,
1. $\left( {V \times \rho g} \right)\,...................\left( 1 \right)$
Where,
$V$ is the volume of the body.
$\rho$ is the density of the body.
$g$ is the acceleration due to gravity.
The SI unit of $V$ is ${m^3}$
The SI unit of $\rho$ is $kg{m^{ - 3}}$
The SI unit of $g$ is $m{s^{ - 2}}$
To calculate the SI unit of $\left( {V \times \rho g} \right)$ ,
So, substitute the SI unit of volume, density and acceleration due to gravity in the equation (1), then,
$\left( {V \times \rho g} \right) = {m^3} \times kg{m^{ - 3}} \times m{s^{ - 2}}$
On multiplying, then the above is written as,
$\left( {V \times \rho g} \right) = kgm{s^{ - 2}}$
Then, comparing this SI unit with the SI unit of weight, both are not equal, it does not satisfy.
2. $\dfrac{V}{{\rho g}}\,.........................\left( 2 \right)$
Substitute the SI unit of volume, density and acceleration due to gravity in the equation (2), then,
$\dfrac{V}{{\rho g}} = \dfrac{{{m^3}}}{{kg{m^{ - 3}} \times m{s^{ - 2}}}}$
Multiplying the terms in denominator, then the above equation be,
$\dfrac{V}{{\rho g}} = \dfrac{{{m^3}}}{{kg{m^{ - 2}}{s^{ - 2}}}}$
Taking the terms in denominator to the numerator, then the above equation is written as,
$\dfrac{V}{{\rho g}} = {m^3} \times k{g^{ - 1}}{m^2}{s^2}$
On multiplying the above equation,
$\dfrac{V}{{\rho g}} = k{g^{ - 1}}{m^5}{s^2}$
Now, comparing the above SI unit with the SI unit of weight, both are not equal, so, it does not satisfy.
3. $\dfrac{{Vg}}{\rho }\,.................\left( 3 \right)$
Substitute the SI unit of volume, density and acceleration due to gravity in the equation (3), then,
$\dfrac{{Vg}}{\rho } = \dfrac{{{m^3} \times m{s^{ - 2}}}}{{kg{m^{ - 3}}}}$
On multiplying the terms in numerator, then,
$\dfrac{{Vg}}{\rho } = \dfrac{{{m^4}{s^{ - 2}}}}{{kg{m^{ - 3}}}}$
Taking the terms in denominator to the numerator, then the above equation is written as,
$\dfrac{{Vg}}{\rho } = {m^4}{s^{ - 2}} \times k{g^{ - 1}}{m^3}$
On multiplying the above equation,
$\dfrac{{Vg}}{\rho } = k{g^{ - 1}}{m^7}{s^{ - 2}}$
Then, comparing these SI units of the three options with the SI unit of weight, the first option $V \times \rho g$ is matched.
Hence, the option (A) is correct.

Note :
The SI unit of the volume, density and acceleration due to gravity is very important to solve this problem. While calculating the SI units we have to give more concentration in the power of the units. And when we take the denominator terms to the numerator the sign of the power will be reversed.