Question

Force on a particle in one-dimensional motion is given by $F=Av+Bt+\frac{Cx}{At+D^{\prime }}$, where $F$ is force, $v$ is speed, $t$ is time, $x$ is position and $A,B,C$ and $D$ are constants. Dimensions of $C$ are $\left[M^p{L}^q{T}^r\right]$. Find $(p-q-r)$

Answer 1

Given, force on a particle in one-dimensional motion, $F=Av+Bt+\frac{Cx}{At+D}$
According to the principle of dimensional homogeneity,
$\left[F\right]=\left[Av\right]=\left[Bt\right]=\left[\frac{cx}{At+D}\right]$
So, dimension of $A$,
$\left[A\right]=\left[\frac{F}{v}\right]$
Dimension of $C$,
$\left[C\right]=\frac{\left[F\right]\left[At\right]}{\left[x\right]}$
$\left[C\right]=\frac{\left[F\right]\left[\frac{Ft}{v}\right]}{\left[x\right]}$
$\left[C\right]=\left[\frac{F^2}{v^2}\right]$
$\left[C\right]={\frac{\left[ML{T}^{-2}\right]}{{\left[L{T}^{-1}\right]}^2}}^2=\left[M^2{L}^0{T}^{-2}\right]$
Compare above equation with $M^p{L}^q{T}^r$ $p=2,q=0$ and $r=-2$
So, $(p-q-r)=2-0+2=4$
Final answer: 4