Question

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

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

Answer 1

Answer: (B)

$F=a+bt+\frac{1}{c+dx}+A\sin \left(\omega t+\varphi \right).$
As $\sin \left(\omega t+\varphi \right)$ is dimensionless, therefore A has dimensions of force.
$\therefore \left[A\right]=\left[F\right]=\left[ML{T}^{-2}\right]$
As each term on RHS represents force
$\therefore \left[\frac{1}{c+dx}\right]=\left[F\right]$
$\left[\frac{1}{c}\right]=\left[F\right]$
$\therefore \left[c\right]=\frac{1}{\left[F\right]}=\frac{1}{\left[ML{T}^{-2}\right]}=\left[M^{-1}{L}^{-1}{T}^2\right]$
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 $A\cdot d$ is
$\left[A\cdot d\right]=\left[ML{T}^2\right]\left[M^{-1}{L}^{-2}{T}^{-2}\right]$ $=\left[L^{-1}\right]$