Question

The mass of a body whose volume is $2{m^3}$and relative density is 0.52 is $x$ kg. find the value of $x$.
A. 0
B. \[1040\]
C. \[2080\]
D. \[3000\]

Answer 1

Hint:
-To solve the problem we have to use the concept of relative density and density.
- Relative density of any object is defined as the ratio between the density of that object and the density of water.
- Again density of any object is defined as the ratio between mass and volume of that object.
- we have to use the data : density of water is $1000kg/{m^3}$.

Complete step by step answer:
Density of any object is defined as the ratio between the mass and volume of the object.
$d = \dfrac{m}{V}$
Where $d$ is the density, $m$ is the mass and $V$is the volume.

Relative density of any object is defined as the ratio between the density of that object and the density of that water.
$r.d. = \dfrac{d}{{{d_w}}}$
Where $r.d.$ is relative density, $d$ is the density of that object and ${d_w}$ is the density of water.

Now, in this question, we have -
Mass of the object is $x$ kg.
Volume of the object is $2{m^3}$.
So, density of the object is, $d = \dfrac{x}{2}$
Now, we have to use the data: density of water(${d_w}$) is $1000kg/{m^3}$.
From the question we got, relative density of the object($r.d.$) is \[.\]
So, using the equation of relative density we can write-
$ 0.52 = \dfrac{x/2}{{1000}}$
$ \Rightarrow $ $0.52 = \dfrac{x}{{2000}}$
$ \Rightarrow $ $x = 2000 \times 0.52$
$ \Rightarrow $ $x = 1040$

So, the mass of the object is $1040$ kg.

So, the correct answer is “Option B”.

Note:
- Students should take care about the unit.
- We should convert every quantity in the SI unit system before putting those values in equations.
- Density of any object depends on temperature. Here, we have considered the situation at room temperature.