An unknown quantity `''alpha''` is expressed as `alpha = (2ma)/(beta) log(1+(2betal)/(ma))` where m = mass, a = acceleration `l` = length, The unit of `alpha` should be

Options

A. meter
B. m/s
C. `m//s^(2)`
D. `s^(-1)`

Explanation

To determine the unit of the unknown quantity \(\alpha\) given by the expression:

\[ \alpha = \frac{2ma}{\beta} \log\left(1 + \frac{2\beta l}{ma}\right) \]

we need to analyze the components involved in this equation.

### Step 1: Analyze the logarithmic term
The term inside the logarithm, \(\frac{2\beta l}{ma}\), must be dimensionless. This means that the dimensions of the numerator must equal the dimensions of the denominator.

### Step 2: Determine the dimensions of \(\beta\)
Let's break down the dimensions of each component:

- \(m\) (mass) has the dimension \([M]\)
- \(a\) (acceleration) has the dimension \([L T^{-2}]\)
- \(l\) (length) has the dimension \([L]\)

Now, the denominator \(ma\) has dimensions:
\[ [m][a] = [M][L T^{-2}] = [M L T^{-2}] \]

The numerator \(2\beta l\) has dimensions:
\[ [\beta][l] = [\beta][L] \]

For the term \(\frac{2\beta l}{ma}\) to be dimensionless, we set the dimensions equal:
\[ [\beta][L] = [M L T^{-2}] \]

From this, we can solve for the dimensions of \(\beta\):
\[ [\beta] = \frac{[M L T^{-2}]}{[L]} = [M T^{-2}] \]

### Step 3: Determine the dimensions of \(\alpha\)
Now we can find the dimensions of \(\alpha\) using the expression \(\frac{2ma}{\beta}\):

Substituting the dimensions we have:
\[ \alpha = \frac{2ma}{\beta} = \frac{[M][L T^{-2}]}{[M T^{-2}]} = \frac{[M L T^{-2}]}{[M T^{-2}]} \]

Here, the \(M\) cancels out:
\[ \alpha = [L] \]

Thus, the dimension of \(\alpha\) is simply:
\[ \alpha = [L] \]

### Conclusion
The unit of \(\alpha\) is that of length, which is meters (m).