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Question

Physics Question on physical world

In the following equation, x, t and F represent respectively, displacement, time and force : F=a+bt+1c+dx+Asin(ωt+ϕ).F = a+bt + \frac{1}{c+dx} + A \,sin \left(\omega t + \phi\right). The dimensional formula for A?dA?d is

A

T1T^{-1}

B

L1L^{-1}

C

M1M^{-1}

D

TL1TL^{-1}

Answer

L1L^{-1}

Explanation

Solution

F=a+bt+1c+dx+Asin(ωt+ϕ).F = a+bt + \frac{1}{c+dx} + A \,sin \left(\omega t + \phi\right).
As sin(?t+ϕ)sin\left(?t + \phi\right) is dimensionless, therefore A has dimensions of force.
[A]=[F]=[MLT2]\therefore\quad\left[A\right] = \left[F\right] = \left[MLT^{-2}\right]
As each term on RHS represents force
[1c+dx]=[F]\therefore \quad \left[\frac{1}{c+dx}\right] = \left[F\right]
[1c]=[F]\left[\frac{1}{c}\right] = \left[F\right]
[c]=1[F]=1[MLT2]=[M1L1T2]\therefore \quad \left[c\right] = \frac{1}{\left[F\right]} = \frac{1}{\left[MLT^{-2}\right]} = \left[M^{-1}L^{-1}T^{2}\right]
As c is added to dx, therefore dimensions of c is same that of dx.
[dx]=[c]\therefore \quad \left[dx\right] = \left[c\right]
or [d]=[c][x]=[M1L1T2][L]=[M1L2T2]\quad\left[d\right] = \frac{\left[c\right]}{\left[x\right]} = \frac{\left[M^{-1}L^{-1}T^{2}\right]}{\left[L\right]} = \left[M^{-1}L^{-2}T^{-2}\right]
The dimensional formula for A?dA?d is
[A?d]=[MLT2][M1L2T2]\left[A?d\right] = \left[MLT^{2}\right]\left[M^{-1}L^{-2}T^{-2}\right] =[L1]= \left[L^{-1}\right]