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

The Correct Option is B

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