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 find the dimensions of \( \frac{b^2}{a} \), we need to understand the dimensional analysis of the force equation given:

The force equation is: \(F = ax^2 + b t^{1/2}\), where \( F \) is the force, \( x \) is distance, and \( t \) is time.

The dimension of force \([F]\) in terms of mass (M), length (L), and time (T) is known to be \([MLT^{-2}]\). 

The dimensional formula for distance \( x \) is \([L]\), and for time \( t \) is \([T]\).

Since both terms on the right-hand side of the equation must have the same dimensions as \( F \), we can equate the dimensions:

1. For the term \(ax^2\):
\([a][L^2] = [MLT^{-2}]\).
Thus, the dimension of \( a \) is:
\([a] = [MLT^{-2}][L^{-2}] = [ML^{-1}T^{-2}]\).

2. For the term \(b t^{1/2}\):
\([b][T^{1/2}] = [MLT^{-2}]\).
Thus, the dimension of \( b \) is:
\([b] = [MLT^{-2}][T^{-1/2}] = [MLT^{-3/2}]\).

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

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

Substitute the dimensions of \( b \) and \( a \):

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

Calculate dimensions in the numerator:

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

Plug into the fraction:

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

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

The correct answer is, therefore, \([ML^3T^{-3}]\).

Answer 2

Given:

\[ F = ax^2 + bt^{1/2} \]

The dimensions of \( a \) are given by:

\[ [a] = \left[ \frac{F}{x^2} \right] = [MLT^{-2}][L]^{-2} = [ML^{-1}T^{-2}] \]

The dimensions of \( b \) are given by:

\[ [b] = \left[ \frac{F}{t^{1/2}} \right] = [MLT^{-2}][T^{-1/2}] = [MLT^{-5/2}] \]

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

\[ \left[ \frac{b^2}{a} \right] = \frac{[M^2L^2T^{-5}]}{[ML^{-1}T^{-2}]} = [ML^3T^{-3}] \]