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 \[Q\]

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.

- 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.

- 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.

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