Question

If force \( F = \frac{\alpha}{\beta} \), then the dimensional formulae of \( \alpha \) and \( \beta \) are respectively:

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

Hint:

When dealing with dimensional analysis, ensure that the dimensions of both sides of an equation are consistent.

Answer 1

The Correct Option is C

The given equation is \( F = \frac{\alpha}{\beta} \), where \(F\) represents force. The dimensional formula of force is: \[ [F] = [M L T^{-2}]. \] Now, for the dimensional consistency of the equation, we have: \[ \left[\frac{\alpha}{\beta}\right] = [F] = [M L T^{-2}]. \] By equating the powers of the fundamental quantities (M, L, T), we solve for the dimensional formula of \(\alpha\) and \(\beta\), which gives the answer as: \[ \alpha \sim [M^2 L^{-2} T^{-3}],
\beta \sim [M^{1/3} L^{-1} T^3]. \]