Question

Find the dimension formula of \(\frac{a}{b}\) in the equation \( F = a\sqrt{x} + bt^2 \), where \(F\) is a force, \(x\) is distance and \(t\) is time.

• A. \([M^0L^{-1/2}T^2]\)
• B. \([M^0L^0T^{3/2}]\)
• C. \([M^0L^1T^{-4}]\)
• D. \([M^{0}L^{3/2}T^{4}]\)

Hint:

When calculating dimensional formulas, balance the units on both sides of the equation to isolate the unit of the unknown variable.

Answer 1

The Correct Option is A

We start by analyzing the given equation: \[ F = a\sqrt{x} + bt^2 \] The dimensions of force \( F \) are \([MLT^{-2}]\). 
Let us denote the dimensions of \( a \) as \([A]\), \( x \) as \([L]\) (since \( x \) is a distance), and \( t \) as \([T]\) (since \( t \) is time). For the term \( a\sqrt{x} \), we have: \[ [a][L]^{1/2} = [MLT^{-2}] \] Solving for \([a]\), we get: \[ [a] = [MLT^{-2}][L]^{-1/2} \] \[ [a] = [M^0L^{-1/2}T^2] \] This shows that the dimensions of \( a \) are \( [M^0L^{-1/2}T^2] \). 
Thus, the dimension formula for \( \frac{a}{b} \) will be: \[ \frac{[a]}{[b]} = \frac{[M^0L^{-1/2}T^2]}{[M^0L^0T^0]} = [M^0L^{-1/2}T^2] \] Hence, the correct answer is \([M^0L^{-1/2}T^2]\).