Question

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

• A. $T^{-1}$
• B. $L^{-1}$
• C. $M^{-1}$
• D. $TL^{-1}$

Answer 1

Answer: (B)

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