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Explanation for concept of relative density

Physics Asked by mnulb on June 20, 2021

I am in class $9^{th}$, I’m reading my physics book in which it is written that: $$mathrm{Relative Density} = frac{mathrm{Weight of solid in air}}{mathrm{Loss of weight of solid in water}}$$
As I mention my level that I’m in class $9^{th}$, Plaese help me to understand this in easy method.enter image description here

2 Answers

The 'catch point' is that Relative density of a substance is relative to another material-say water; strictly speaking its the ratio of two densities ; if RD=1, the densities are equal; RD is a number , if RD<1 the body floats in water.

In Laboratories if somebody weighs a particular Volume of a substance say V (V.density.g)and again weighs the sample immersed in water -the loss in weight is equal to weight of the same volume of water(V.density of water.g) as the sample is - thereby one gets the RD as given in the book.

Actually the reference provided is a book on conducting experiment to determine the relative density of a material/substance with respect to a reference substance and in the above case its taken as 'water'.

The relative density of a substance is a ratio of density of particular substance with that of a reference . Sometimes we also use the term "specific gravity " when the reference material is taken as water.

So the term RD is ratio of two densities but many a time it can be easily calculated /measured by taking a chosen quantity (sample) of the material and using good physical balance

The sample is weighed first in air and then in water and use the following relationship-

Relative Density = Weight in air /( Weight in air - weight in water)

(the referred formula talks of denominator as "loss of weight in water"

This technique can also be used in the labs to find RD of substances which float in water , meaning thereby whose RD is less than 1 using Nicholson Hydrometer (e.g. RD of cork)

Just for info:

Another practical method uses three measurements. The sample is weighed dry. Then a container filled to the brim with water is weighed, and weighed again with the sample immersed, after the displaced water has overflowed and been removed. Subtracting the last reading from the sum of the first two readings gives the weight of the displaced water. The relative density result is the dry sample weight divided by that of the displaced water. This method works with scales that can't easily accommodate a suspended sample, and also allows for measurement of samples that are less dense than water. for details visit :https://en.wikipedia.org/wiki/Relative_density

Answered by drvrm on June 20, 2021

Relative density of a solid is the ratio of density of solid to that of water:

$$rho_r=frac{rho_{solid}}{rho_{water}}$$

Now, according to Archimedes' principle, the weight of an object in a liquid seems to be reduced due to the upthrust, and this reduced weight is termed as apparent weight. There is no actual loss in mass of the solid. The buoyant force exerted on the object will be equal to the weight of the water displaced.

At this point, you should carefully note that the weight of the water displaced never equals the weight of the body, but equal to the buoyant force acting upwards.

$$W_A=W-W'$$

where $W_A$ is the apparent weight of the body, $W$ is the weight of the body in air, and $W'$ is the weight of the fluid displaced by the object. Now, the buoyant force is given by (assuming that the object is fully submerged in water)

$$F_B=W'=rho_{water}V'g$$

where $V'$ is the volume of water submerged by the solid and $g$ is the acceleration due to gravity.

Weight of the solid in air

$$W=rho_{solid}Vg$$

Since the object is completely immersed in water, $V'=V$.

Now, $W/W'$ gives

$$frac{W}{W'}=frac{rho_{solid}V}{rho_{water}V'}=rho_r$$

Now, to determine relative density experimentally, you need a spring balance. The weight you observe for the solid in air is $W$ and that you measure when the solid is completely immersed in water will be $W_A$. Rewriting $W'$ as $W-W_A$,

$$rho_r=frac{W}{W-W_A}$$

Hence, the formula in the textbook is right.

Answered by UKH on June 20, 2021

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