Question

Let the dimensional formula for acceleration, velocity and length are $\alpha\beta^{-2}$, $\alpha \beta^{-1}$ and $\alpha\gamma$ respectively. What should be the dimensional formula for the coefficient of friction.
$\text {A. }\alpha\beta\gamma$
$\text {B. }\alpha^{-1}\beta^{0}\gamma^{0}$
$\text {C. }\alpha^{0}\beta^{0}\gamma^{-1}$
$\text {D. }\alpha^{0}\beta^{-1}\gamma^{0}$

Answer 1

Hint: Friction is a force and coefficient of friction is a constant denoted as $\mu$. Coefficient of friction is a constant of proportionality relating force of friction (f) to the normal reaction (N) of the surface. We should remember that an equation should be dimensionally the same on both sides. The dimension of friction is the same as that of normal reaction. Write dimensional formulas and try to get the dimension of $\mu$.

Formula used: $f=\mu N$

Complete step by step answer:
Friction is a force that exists between two surfaces in contact. This force always tries to oppose relative motion between the surfaces. It has been observed that friction depends on the Normal reaction of the surface. Means the harder an object is pressed against a surface, greater will be friction. In other words, we can say:
$f \propto N$
To convert this relation to equation, we can multiply a constant of proportionality $\mu$ on the right-hand side to get:
$f=\mu N$
$\Rightarrow \dfrac{f}{N}=\mu$
As f and N are of the same dimension, dividing them would cancel out the dimensions. And we can conclude that $\mu$ must be dimensionless. That is the power of all dimensions becomes zero.
We know that the dimension of acceleration is $LT^{-2}$. Thus, we can write:
$LT^{-2}=\alpha\beta^{-2}$
Comparing both sides, we can say:
$\alpha =L,\beta =T$
Also, the dimension of length is $L^{1}$. Thus, we can write:
$L^{1}=\alpha\gamma$
But we already found that $\alpha=L$. So, we can write:
$\begin{aligned} & {{L}^{1}}=L\gamma \\\ & \Rightarrow \dfrac{L}{L}=\gamma \\\ & \Rightarrow {{L}^{0}}=\gamma \\\ \end{aligned}$
That means $\gamma$ is also dimensionless.
Let’s say the dimensional formula of $\mu$ can be obtained by multiplying $\alpha, \beta$ and $\gamma$ raised to powers p, q and r respectively. So, we can write:
$\left[ \mu \right]={{\alpha }^{p}}{{\beta }^{q}}{{\gamma }^{r}}\cdots \cdots \cdots (i)$
now, substitute the dimensions of $\mu$ as $M^{0}L^{0}T^{0}$ and $\alpha=L, \beta=T$ and $\gamma=L^{0}$.
${{M}^{0}}{{L}^{0}}{{T}^{0}}={{L}^{p}}\cdot {{T}^{q}}\cdot {{({{L}^{0}})}^{r}}={{L}^{p}}{{T}^{q}}$
By comparing both sides, we get:
$p=0,q=0$
Also, whatever is the value of ‘r’, it will not affect the dimension of $\mu$ as $\gamma$ itself is dimensionless. Thus, ‘r’ can take any value. Substitute the final values of ‘p’ and ‘q’ back in equation (i) to obtain dimensional formula for $\mu$.
$\left[ \mu \right]={{\alpha }^{0}}\cdot {{\beta }^{0}}\cdot {{\gamma }^{r}}={{\alpha }^{0}}{{\beta }^{0}}{{\gamma }^{-1}}$
As ‘r’ can take any value without affecting dimension of the coefficient of friction, option C. Is correct.

Note:
1. Students may get confused regarding the power of $\gamma$, which is given as ‘-1’ in the correct option. Keep in mind that the power of $\gamma$ does not affect the dimension of coefficient of friction in any way. Thus, it can be any value and not necessarily need to be 0.
2. Alternatively, after finding the relation of $\alpha, \beta$ and $\gamma$ with fundamental dimensions, M, L and T, students may directly check the given options. Eliminate the options, which does not result in $M^{0}L^{0}T^{0}$.