Question

A sample of metal weighs 210 gm in air, 180 gm in water and 120 gm in a liquid. Then the:
(A) relative density of metal is 3
(B) relative density of metal is 7
(C) relative density of liquid is 1010
(D) relative density of liquid is $\dfrac{1}{3}$

Answer 1

Hint
The relative density of any metal is given by the ratio of the density of the metal in the air to the density of water and the relative density of the unknown liquid is given by the ratio of the density of that fluid to the density of water, where we can find the density of any substance by
$\Rightarrow \rho = \dfrac{{mass}}{{volume}}$
To solve this problem we use the following formula,
The relative density of metal, $R.D{._{metal}} = \dfrac{{{\text{density of metal}}}}{{{\text{density of water}}}}$
and the relative density of the unknown liquid, $R.D{._{Liq}} = \dfrac{{{\text{Density of liquid}}}}{{{\text{Density of water}}}}$
Where the density of any substance is given by, $\rho = \dfrac{{mass}}{{volume}}$.

Complete step by step answer
The relative density of any substance is given by the ratio of the density of that substance to the density of another standard substance like water.
So in the question, we have the mass of the metal in the air ${m_a} = 210gm$.
The mass of the metal in water is given by ${m_w} = 180gm$. So the loss of the mass of the metal in water is,
$\Rightarrow {m_a} - {m_w} = 210 - 180 = 30gm$.
This is also the upthrust of water. From here we can find the volume of water that is displaced, as $volume = \dfrac{{upthrust}}{{density}}$
Now we all know that the density of water is $1gm/c{m^3}$.
So by substituting this value, we get the volume of the water displaced as,
$\Rightarrow volume = \dfrac{{30}}{1}c{m^3} = 30c{m^3}$
The volume of the water displaced is the same as the volume of the metal. Hence, from here we get the volume of the metal as $30c{m^3}$.
Now we have both the mass and the volume of the metal. So its density is given by,
$\Rightarrow {\rho _{metal}} = \dfrac{{mass}}{{volume}} = \dfrac{{210}}{{30}}gm/c{m^3}$
By calculating this value we get the density of metal as,
$\Rightarrow {\rho _{metal}} = 7gm/c{m^3}$
Now the relative density of metal will be,
$\Rightarrow R.D{._{metal}} = \dfrac{{{\text{density of metal}}}}{{{\text{density of water}}}}$
$\Rightarrow R.D{._{metal}} = \dfrac{{7gm/c{m^3}}}{{1gm/c{m^3}}} = 7$
So the relative density of the metal is 7.
To find the density of the unknown liquid, we need to find the upthrust of the liquid as, $\Rightarrow density = \dfrac{{upthrust}}{{volume}}$ where the volume of the liquid displaced will be the same as the volume of the metal, that is $30c{m^3}$. The mass of the metal in the liquid is ${m_l} = 120gm$ and that in the air is ${m_a} = 210gm$. The upthrust of the liquid is the loss of the mass of the metal in the liquid, that is, ${m_a} - {m_l} = 210 - 120 = 90gm$
So the density of the liquid, ${\rho _{liquid}} = \dfrac{{90gm}}{{30c{m^3}}} = 3gm/c{m^3}$
Therefore, the relative density of the liquid is,
$\Rightarrow R.D{._{Liq}} = \dfrac{{{\text{Density of liquid}}}}{{{\text{Density of water}}}}$
$\Rightarrow R.D{._{Liq}} = \dfrac{{3gm/c{m^3}}}{{{\text{1gm/c}}{{\text{m}}^3}}} = 3$
So the relative density of the unknown liquid is 3 and the relative density of the metal is 7.
Hence, the correct option will be (B).

Note
We can also solve this problem alternatively by,
The relative density of metal is also given by the formula, $R.D{._{metal}} = \dfrac{{{\text{weight of metal in air}}}}{{{\text{loss of weight in water}}}}$
So, by substituting the values, we get
$\Rightarrow R.D{._{metal}} = \dfrac{{210gm}}{{\left( {210 - 180} \right)gm}}$
$\Rightarrow R.D{._{metal}} = \dfrac{{210}}{{30}} = 7$
And for the relative density of the liquid, $R.D{._{Liq}} = \dfrac{{{\text{Loss of mass in liquid}}}}{{{\text{Loss of mass in water}}}}$
By substituting the values,
$\Rightarrow R.D{._{Liq}} = \dfrac{{\left( {210 - 120} \right)gm}}{{\left( {210 - 180} \right)gm}}$
$\Rightarrow R.D{._{Liq}} = \dfrac{{90}}{{30}} = 3$