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Question: Calculate the mass of air enclosed in a room of length, breadth and height equal to \(5m,3m\)and \(4...

Calculate the mass of air enclosed in a room of length, breadth and height equal to 5m,3m5m,3mand 4m4mrespectively. Density of air is1.3kg/m31.3kg/{m^3}.
A. 78kg78kg
B. 38kg38kg
C. 42kg42kg
D. 87kg87kg

Explanation

Solution

As in the question, we are given density and to calculate mass from it, volume is required. So, by finding the volume of the room and then density dependence on mass and volume, the mass of air enclosed in the room can be calculated.

Formula used:
(1) Volume of room == lbh
Where l is length, b is breadth, and h is height
(2) Density =MassVolume = \dfrac{{Mass}}{{Volume}}
⇒mass=density×volume\Rightarrow mass = density \times volume

Complete step by step answer:
We know that the shape of the room is cuboidal with
Length (l) =5m = 5m
Breadth (b) =3m = 3m and height (h) =4m = 4m
The volume of air enclosed in room
== Volume of the room
== Volume of cuboid
=lbh =5×3×4 60m3  = lbh \\\ = 5 \times 3 \times 4 \\\ 60{m^3} \\\
So, volume of air enclosed in room
=60m3= 60{m^3}
So, volume of air enclosed in room =60m3 = 60{m^3}
Now, as density is mass per unit volume
So, mass of air =(volume×density) = \left( {volume \times density} \right)of air
Mass of air=60m3×1.3kg/m3 = 60{m^3} \times 1.3kg/{m^3}
Mass of air =78kg = 78kg
So, mass of air enclosed in room is 78kg78kg

So, the correct answer is “Option A”.

Additional Information:
→\to The term density is generally denoted by the symbol ‘ρ\rho ’
i.e. ρ=MV\rho = \dfrac{M}{V}
Where M is mass and V is volume.
→\to Different substances have different densities,
→\to To Simplify, it is sometimes replaced by a dimensionless quantity i.e. “relative density” or “specific gravity”.
→\to “Relative density” or “specific gravity” is the density of any substance with respect to hot water. It is the ratio of density of substance to the density of water.
→\to The density of a material varies with temperature and pressure.
→\to Density is an intensive property i.e. increasing the amount of any substance does not increase its density. Rather, it increases its mass.
→\to Its units in SI system are kg−m−3kg - {m^{ - 3}} and in CGS are g−cm−3g - c{m^{ - 3}}
→\to Its dimensional formula is [ML−3]\left[ {M{L^{ - 3}}} \right]

Note:
This can also be solved by unitary method which is mass of air corresponding to 1m3=1.3kg1{m^3} = 1.3kg
Mass of air corresponding to 60m360{m^3}
=60×1.3 =78kg.  = 60 \times 1.3 \\\ = 78kg. \\\