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 $[M^0L^{-1/2}T^2]$.