
The dimensional formula of $\Delta Q$heat supplied to the system is given by:
$\eqalign{
& {\text{A}}.\;{\text{ }}[{M^1}{L^2}{T^{ - 2}}] \cr
& {\text{B}}.\;{\text{ }}[{M^1}{L^1}{T^{ - 2}}] \cr
& {\text{C}}.{\text{ }}\;[{M^1}{L^2}{T^{ - 1}}] \cr
& {\text{D}}.\;{\text{ }}[{M^1}{L^1}{T^{ - 1}}] \cr} $
Answer
583.2k+ views
Hint: Principle of homogeneity says that only like quantities can be added or subtracted and in all the equations, the dimensions on the right hand side and on the left hand side must be the same. First law of thermodynamics says that the heat supplied to the system directly goes in either increasing the internal energy of the system or doing the work by the gas.
Formula used:
$\Delta Q = W + \Delta U$, $W=\int { F.ds }$, $F=ma$
Where $\Delta U$ is the change in internal energy.
And $W$ is the work done.
Complete step-by-step answer:
As we know by the principle of homogeneity of dimensions, only like quantities can be added or subtracted. Using this principle, we can say internal energy, work is done and heat supplied all have the same dimensions. Now, it is easy to get the dimensions of work done as W=F.s
(s means displacement). So, dimensions of W :
W=F.s
& F=ma
So, W=ma.s
Now, Putting the dimensions of all the terms.
M=$[M^1]$ [mass]
a = [ $L^1 T^{-2}$] [as a=rate of change of velocity ]
s = [$L^{1}$] [displacement means the length traveled]
Now, W = m.a.s
W=$[M^1][L^1T^{-2}][L]$
W=$[M^1 L^{1+1}T^{-2}]$
W=$[M^1 L^{2}T^{-2}]$
So, the correct answer is “Option A”.
Additional Information: It is to be noted that many physical quantities have the same physical dimensions by they can be completely different in every aspect like work and torque, both have the same dimensions $[M^1 L^{2}T^{-2}]$ which are also the dimensions of heat given.
Note: Students are advised to learn basic quantities’ dimensions only (for example displacement, angle, intensity, mass, etc.) and complex quantities’ dimensions (for example torque, pressure, force, etc.) can be easily derived by knowing the formulae and proceeding as shown in the above example.
Formula used:
$\Delta Q = W + \Delta U$, $W=\int { F.ds }$, $F=ma$
Where $\Delta U$ is the change in internal energy.
And $W$ is the work done.
Complete step-by-step answer:
As we know by the principle of homogeneity of dimensions, only like quantities can be added or subtracted. Using this principle, we can say internal energy, work is done and heat supplied all have the same dimensions. Now, it is easy to get the dimensions of work done as W=F.s
(s means displacement). So, dimensions of W :
W=F.s
& F=ma
So, W=ma.s
Now, Putting the dimensions of all the terms.
M=$[M^1]$ [mass]
a = [ $L^1 T^{-2}$] [as a=rate of change of velocity ]
s = [$L^{1}$] [displacement means the length traveled]
Now, W = m.a.s
W=$[M^1][L^1T^{-2}][L]$
W=$[M^1 L^{1+1}T^{-2}]$
W=$[M^1 L^{2}T^{-2}]$
So, the correct answer is “Option A”.
Additional Information: It is to be noted that many physical quantities have the same physical dimensions by they can be completely different in every aspect like work and torque, both have the same dimensions $[M^1 L^{2}T^{-2}]$ which are also the dimensions of heat given.
Note: Students are advised to learn basic quantities’ dimensions only (for example displacement, angle, intensity, mass, etc.) and complex quantities’ dimensions (for example torque, pressure, force, etc.) can be easily derived by knowing the formulae and proceeding as shown in the above example.
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