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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 velocityB. Momentum and displacement C. Force and displacementD. Work and velocity

Last updated date: 13th Jun 2024
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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.