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The dimension formula for the product of two physical quantities \[P\] and \[Q\] \[M{L^2}{T^{ - 2}}\].The dimension formula of \[\dfrac{P}{Q}\]is \[M{T^{ - 2}}\]. Then \[P\] and \[Q\] respectively are:
A. Force and velocity
B. Momentum and displacement
C. Force and displacement
D. Work and velocity

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Last updated date: 13th Jun 2024
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Answer
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Hint:-
- Dimension of a quantity is to find the equation\[{M^a}{L^b}{T^c}\],\[a\], \[b\], and \[c\] are just numbers.
- From the question one of the quantity should depends up on mass \[M\]
- From product of these two ie, \[PQ \times (\dfrac{P}{Q})\] will get the \[P\] and \[PQ \times (\dfrac{Q}{P})\] will get \[Q\]

Complete step by step solution:-
We could do this question by two ways
One is checking each option.
Force and velocity
Force \[F = ma\]
Dimension of acceleration \[a = \dfrac{{{d^2}x}}{{d{t^2}}} \equiv L{T^{ - 2}}\]
\[x\] is the displacement \[x \equiv L\]and time \[t \equiv {T^1}\]
Here P, Dimension of force \[F = ma \equiv {M^1}{L^1}{T^{ - 2}}\]
Q , Dimension of velocity \[V = \dfrac{{dx}}{{dt}} \equiv {L^1}{T^{ - 1}}\]
\[PQ \equiv {M^1}{L^1}{T^{ - 2}} \times {L^1}{T^{ - 1}} \equiv M{L^2}{T^{ - 3}}\] Not correct.
(B) Momentum and displacement
Here, \[P\] Dimension of momentum \[p = mv \equiv {M^1}{L^1}{T^{ - 1}}\]
\[Q\] , Dimension of displacement \[S \equiv {L^1}\]
\[PQ = {M^1}{L^1}{T^{ - 1}} \times {L^1} = {M^1}{L^2}{T^{ - 1}}\]Not correct.
(C) .Force and displacement
Here P, Dimension of force \[F = ma \equiv {M^1}{L^1}{T^{ - 2}}\]
\[Q\] , Dimension of displacement \[S \equiv {L^1}\]
\[PQ = {M^1}{L^1}{T^{ - 2}} \times {L^1} = {M^1}{L^2}{T^{ - 2}}\]
\[\dfrac{P}{Q} = {M^1}{L^1}{T^{ - 2}}/{L^1} = {M^1}{L^0}{T^{ - 2}}\]
Option is correct.
(D).Work and velocity
Work \[W = F \bullet s\]
Dimension of displacement \[s \equiv {L^1}\]
Here, P Dimension of \[W \equiv {M^1}{L^1}{T^{ - 2}} \times {L^1}{T^0} \equiv {M^1}{L^2}{T^{ - 2}}\]
Q, Dimension of velocity \[V = \dfrac{{dx}}{{dt}} \equiv {L^1}{T^{ - 1}}\]
\[PQ = {M^1}{L^2}{T^{ - 2}} \times {L^1}{T^{ - 1}} = {M^1}{L^3}{T^{ - 3}}\] Not correct.

Other way is
\[PQ = {M^1}{L^2}{T^{ - 2}}\]
\[\dfrac{P}{Q} = {M^1}{L^0}{T^{ - 2}}\]
\[PQ \times \dfrac{P}{Q} = {P^2} = {M^1}{L^2}{T^{ - 2}} \times {M^1}{L^0}{T^{ - 2}} = {M^2}{L^2}{T^{ - 4}}\]
Take root, \[P = {M^1}{L^1}{T^{ - 2}}\]
\[PQ \times \dfrac{Q}{P} = {Q^2} = {M^1}{L^2}{T^{ - 2}}/{M^1}{L^0}{T^{ - 2}} = {M^0}{L^2}{T^0}\]
Take root, \[Q = {M^0}{L^1}{T^0}\]
From the option \[P\]is the Dimension of Force.
\[Q\] is the dimension of length or displacement.

So the answer is (C) .Force and displacement

Note:-
- Unit of work is Joule \[\left( J \right).\]
- Unit of force is Newton \[\left( N \right)\].
- In dimensional analysis we couldn’t consider the dimensionless constants like \[\pi ,\theta \],etc.
- From dimensional analysis we can’t distinguish scalar or vector, for example distance and displacement have the same dimension of length.
- If the Dimensions of both sides of the equation are not matching the equation is not correct.