Question

A bottle weighs 30g when empty, 53.4g when completely filled with a liquid and 48g when completely filled with water. Calculate the density (in $kg\ m^{-3}$) of the liquid. Take density of water =$1000 kg\ m^{-3}$.
A. 1300
B. 13
C. 245
D. None of these.

Answer 1

Hint: There is a special property of fluids that it attains the shape of the container in which it is placed. Since the bottle is the same, hence the volume of all the fluids being poured in the bottle will remain the same. Hence we can easily calculate the mass of fluid lying inside the bottle by knowing the density of the fluid.

Formula used:
$\rho = \dfrac mV$

Complete step by step answer:
Since the bottle weighs 30g when empty, that means the structure or material of the bottle has a weight of 30g.
Now, when the bottle is filled with water, total weight will be combined weight of both bottle and water, i.e.
$48 = 30 + W_{water}$
$W_{water} = 18g$
Hence mass = $\dfrac{18g}{g} = 18\ kg$
Now, density of water is $1000kg\ m^{-3}$, hence using$\rho = \dfrac mV$, we get
$V = \dfrac{18}{1000} = 18 \times 10^{-3} m^3$
Hence this is also the volume of the bottle.
Now, we are given the total weight when liquid is filled:
Thus, $53.4 = 30 + W_{liq}$
$W_{liq} = 23.4g$
$mass = 23.3\ kg$
Now, using$\rho = \dfrac mV$
$\rho = \dfrac{23.4}{18 \times 10^{-3}} = 1300 kg\ m^{-3}$
Hence the density of liquid is $1300 kg\ m^{-3}$

Additional information: Density may be defined as the ratio of mass and volume and is defined as the mass present per unit volume. There is one more term called ‘Weight density’ which is defined as the weight present per unit volume.

Note:
One should not confuse with the units of the weight given. In the question it is clearly mentioned that the bottle ‘weighs’ 30g. This clearly suggests that the weight is given and not the mass. Hence we have to use $W = mg$ and then the mass will be obtained (in kg). One must not confuse the units of a given question as ‘gram’. S.I. units of density is $kg\ m^{-3}$ and to get the value in these units, we must use the values also in S.I. units only.