In a book, the answer for a particular question is expressed as `b=(ma)/(k)[sqrt(1+(2kl)/(ma))]` here m represents mass, a represent acceleration, l represent length. The unit of b should be :-

Options

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

Answer

To find the unit of \( b \) given the expression:

\[ b = \frac{ma}{k} \sqrt{1 + \frac{2kl}{ma}} \]

where \( m \) represents mass, \( a \) represents acceleration, and \( l \) represents length, we will follow these steps:

### Step 1: Analyze the expression inside the square root

The term inside the square root is \( 1 + \frac{2kl}{ma} \). For the addition to be valid, both terms must have the same dimensions. Since \( 1 \) is dimensionless, \( \frac{2kl}{ma} \) must also be dimensionless.

### Step 2: Determine the dimensions of \( k \)

Let's express the dimensions of each variable:
- Mass \( m \): \( [M] \)
- Acceleration \( a \): \( [L T^{-2}] \)
- Length \( l \): \( [L] \)

Now, we can express the dimensions of \( \frac{2kl}{ma} \):

\[ \frac{2kl}{ma} = \frac{K \cdot [L]}{[M] \cdot [L T^{-2}]} \]

This simplifies to:

\[ \frac{K \cdot [L]}{[M] \cdot [L T^{-2}]} = \frac{K}{[M] \cdot [T^{-2}]} \]

For this to be dimensionless, the dimensions of \( K \) must be:

\[ K = [M] \cdot [T^{-2}] \]

### Step 3: Substitute the dimensions back into the expression for \( b \)

Now we can find the dimensions of \( b \):

\[ b = \frac{ma}{k} \]

Substituting the dimensions:

\[ b = \frac{[M] \cdot [L T^{-2}]}{[M] \cdot [T^{-2}]} \]

### Step 4: Simplify the dimensions

The \( [M] \) cancels out:

\[ b = [L] \]

This means that the dimension of \( b \) is simply length.

### Step 5: Determine the unit of \( b \)

Since the dimension of \( b \) is length, the unit of \( b \) is:

\[ \text{Unit of } b = \text{meter (m)} \]

### Conclusion

Thus, the unit of \( b \) is meter (m).

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