Question

A force is represented by \( F = ax^2 + b t^{1/2} \) Where \( x = \) distance and \( t = \) time. The dimensions of \( \frac{b^2}{a} \) are:

• A. \( [ML^3T^{-3}] \)
• B. \( [MLT^{-2}] \)
• C. \( [ML^{-1}T^{-1}] \)
• D. \( [ML^2T^3] \)

Answer 1

The Correct Option is A

To determine the dimensions of \( \frac{b^2}{a} \), we first analyze the dimensions of the given force equation: \(F = ax^2 + b t^{1/2}\).

The dimensions of force \( [F] \) are \( [MLT^{-2}] \), where M represents mass, L represents length, and T represents time. The dimension of distance \( x \) is \( [L] \), and the dimension of time \( t \) is \( [T] \).

For the equation to be dimensionally consistent, both terms on the right side must have the same dimensions as \( F \).

1. For the term \( ax^2 \):

\([a][L^2] = [MLT^{-2}]\)

Therefore, the dimensions of \( a \) are:

\([a] = [MLT^{-2}][L^{-2}] = [ML^{-1}T^{-2}]\)

2. For the term \( b t^{1/2} \):

\([b][T^{1/2}] = [MLT^{-2}]\)

Therefore, the dimensions of \( b \) are:

\([b] = [MLT^{-2}][T^{-1/2}] = [MLT^{-3/2}]\)

Now, we calculate the dimensions of \( \frac{b^2}{a} \):

\(\frac{b^2}{a} = \frac{[b]^2}{[a]}\)

Substituting the derived dimensions:

\(\frac{b^2}{a} = \frac{[MLT^{-3/2}]^2}{[ML^{-1}T^{-2}]}\)

The dimensions of the numerator are:

\([b]^2 = [M^2L^2T^{-3}]\)

Plugging this into the fraction:

\(\frac{[M^2L^2T^{-3}]}{[ML^{-1}T^{-2}]} = [M^{2-1}L^{2-(-1)}T^{-3-(-2)}] = [ML^3T^{-1}]\)

The dimensions of \( \frac{b^2}{a} \) are \( [ML^3T^{-1}] \).

The final answer is \( [ML^3T^{-1}] \).