Question

The relation between force \(F\) and density \(d\) is \(F \propto \dfrac{x}{\sqrt{d}}\). The dimensions of \(x\) are

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

Hint:

Always convert proportional relations into dimensional equations before solving.

Answer 1

The Correct Option is A

Step 1: Write dimensional formulae.
Force:
\[ [F] = [M L T^{-2}] \] Density:
\[ [d] = [M L^{-3}] \]
Step 2: Write the given relation in dimensional form.
\[ F \propto \frac{x}{\sqrt{d}} \Rightarrow [F] = \frac{[x]}{[d]^{1/2}} \]
Step 3: Substitute dimensions of force and density.
\[ [M L T^{-2}] = \frac{[x]}{[M^{1/2} L^{-3/2}]} \]
Step 4: Solve for dimensions of \(x\).
\[ [x] = [M L T^{-2}] \times [M^{1/2} L^{-3/2}] \] \[ [x] = [M^{3/2} L^{-1/2} T^{-2}] \]
Step 5: Conclusion.
The dimensions of \(x\) are \([L^{-1/2}\, M^{3/2}\, T^{-2}]\).