The dimensional formula for acceleration, velocity and length are \[\alpha {{\beta }^{-2}},\alpha {{\beta }^{-1}}\text{ and }\alpha y\] what is the dimensional formula for the coefficient of friction.
\[\begin{align}
  & \text{A}\text{. }\alpha \beta y \\
 & \text{B}\text{. }{{\alpha }^{-1}}{{\beta }^{10}}{{y}^{0}} \\
 & \text{C}\text{. }{{\alpha }^{0}}{{\beta }^{-1}}{{y}^{0}} \\
 & \text{D}\text{. }{{\alpha }^{0}}{{\beta }^{0}}{{y}^{0}} \\
\end{align}\]

Hint: Here we will do dimension analysis of the given quantity to find the dimension of coefficient of friction in terms of α, β and y. First we can compare the dimensions of the acceleration, velocity and length given here with the actual dimensions which are in terms of M, L and T. Then by analysis which dimensions α, β and y are representing we can write the dimension for the coefficient of the friction.

Formula used:
\[f=\mu N\]

Complete answer:
The actual dimensions for the acceleration, velocity and length are given as follows
\[\begin{align}
  & a={{L}^{1}}{{T}^{-2}} \\
 & v={{L}^{1}}{{T}^{-1}} \\
 & l={{L}^{1}} \\
\end{align}\] ………….(i)

Where a is acceleration, v is velocity, l is length, L is the dimension for length and T is dimension for time.
And according to question, the dimensions of acceleration, velocity and length are given as
\[\begin{align}
  & a={{\alpha }^{1}}{{\beta }^{-2}} \\
 & v={{\alpha }^{1}}{{\beta }^{-1}} \\
 & l={{\alpha }^{1}}{{y}^{1}} \\
\end{align}\] ……….. (ii)
Comparing the set of equations (i) and (ii) we get
\[\begin{align}
  & \alpha =L \\
 & \beta =T \\
 & y=1 \\
\end{align}\]
Which shows that α has dimension of length L and β has dimension of time and y is dimensionless quantity.
Now the coefficient of friction is also a dimensionless quantity, as the static friction is given as
\[f=\mu N\]
Where μ is coefficient of friction and N is the normal force. So μ can be given as
\[\mu =\dfrac{f}{N}\]
And static friction and normal force have the same dimension of force. Hence in terms of dimension coefficient of friction is given as
\[\mu =1\]
That is μ is also a dimensionless quantity. Hence the power of α and β will be zero and y is already a dimensionless quantity.
Hence coefficient of friction in term of α, β and y can be given as \[{{\alpha }^{0}}{{\beta }^{0}}{{y}^{0}}\] which will be numerically equal to one, that is dimensionless quantity.

So, the correct answer is “Option D”.

Note:
As y is also a dimensionless quantity, therefore dimension of coefficient of friction can be equal to y. But here as there is no option like that we consider the option D. Remember that any variable having power zero has value one. Therefore option D satisfies the answer to the question.