Question

Which formula is used to find the relative density of solid in Liquids other than water?
A) $\text{R}\text{.D}\text{.}={\displaystyle \frac{w_1}{w_1-w_2}}$
B) $\text{R}\text{.D}\text{.}={\displaystyle \frac{w_1}{w_2-w_1}}$
C) $$\text{R}\text{.D}\text{. = }{\displaystyle \frac{w_1}{w_2-w_1}}\times \text{R}\text{.D}\text{. of liquid}$$
D) $\text{R}\text{.D}\text{.}={\displaystyle \frac{w_2}{w_1-w_2}}\times \text{R}\text{.D}\text{. of liquid}$

Answer 1

Hint: Relative density of a solid is defined as the ratio of the weight of the object in the air to its apparent weight loss when put inside water. If we are to replace water with some other liquid, then there must be a term that relates the density of the liquid to the density of water.

Formula used:
$\text{Relative density = }{\displaystyle \frac{\text{Density of the object}}{\text{Density of water}}}$

Complete step by step solution:
The concept of relative density arises from Archimedes’ principle, and it is defined in terms of water itself (at $4^{\circ }C$).
R.D. is defined as-
$\text{Relative density = }{\displaystyle \frac{\text{Density of the object}}{\text{Density of water}}}$

If we assume that we have a unit volume of water as well as the solid substance, the formula can be rewritten as-
$R.D.={\displaystyle \frac{{\rho }_{\text{substance}}}{{\rho }_{\text{water}}}}={}_{V=\text{Const}}^{}{\displaystyle \frac{M_{\text{Substance}}}{M_{\text{Water}}}}$
Where, $M_{\text{Substance}}$ and $M_{\text{Water}}$are the mass of the substance and the water respectively.

In practical terms, however, we use a different approach, Instead of measuring masses of equal volumes of water and sample substance, we measure the apparent weight loss that the sample feels when submerged in water. So here,
$R.D.={\displaystyle \frac{\text{weight of the sample in air}}{\text{weight loss caused by submerging in water}}}$
Let the weight of the sample in the air be ($w_1$) and its weight in water be ($w_2$).
Then, the apparent weight loss is given $w_1-w_2$.
Thus the formula becomes-
$R.D.={\displaystyle \frac{w_1}{w_1-w_2}}$
But if here, instead of water we use a different liquid, the weight loss would be different. This weight loss depends on the density of the liquid which is used. So we use the pre measured values of R.D. of that liquid relative to water.

So, multiplying the observed R.D. with the R.D. of the liquid used will give us the R.D. of the object relative to water, without even using water in the process.
So, the formula becomes-
$\text{R}\text{.D}\text{. = }{\displaystyle \frac{w_1}{w_1-w_2}}\times \text{R}\text{.D}\text{. of liquid}$

Hence option (D) is correct.

Note: The formula $R.D.={\displaystyle \frac{w_1}{w_1-w_2}}$ is used in experimentally determining the relative density, and we know that if an object has $R.D.<1$, it floats. So in this case the experiment fails as calculation of weight loss is not possible.