Question

Match the column according to dimensions. \[ \begin{array}{c|c|c} \text{Column A} & & \text{Column B} \\ \hline (A) & \sqrt{\dfrac{GM}{C}} & (1)\; M^{-1}L^{2}T^{-3/2} \\ (B) & \sqrt{\dfrac{h^{2}}{GC}} & (2)\; M^{0}L^{2}T^{-3/2} \\ (C) & \sqrt{GMC} & (3)\; M^{3/2}L^{0}T^{0} \\ (D) & \sqrt{\dfrac{GC}{M}} & (4)\; LT^{-1/2} \\ \end{array} \]

• A. A-1, B-2, C-3, D-4
• B. A-4, B-3, C-1, D-2
• C. A-4, B-3, C-2, D-1
• D. A-1, B-3, C-4, D-2

Hint:

When solving dimensional matching problems, first write dimensions of constants such as: \[ [G],\; [h],\; [c] \] Then simplify algebraically before taking square roots.

Answer 1

The Correct Option is C

Concept: Dimensions of important quantities: \[ [G] = M^{-1}L^{3}T^{-2} \] \[ [h] = ML^{2}T^{-1} \] \[ [C] = LT^{-1} \]
Step 1: Dimension of (A). \[ \sqrt{\frac{GM}{C}} \] \[ = \left(\frac{M^{-1}L^{3}T^{-2}\cdot M}{LT^{-1}}\right)^{1/2} \] \[ = (L^{2}T^{-1})^{1/2} \] \[ = LT^{-1/2} \] Thus, \[ A \rightarrow (4) \]
Step 2: Dimension of (B). \[ \sqrt{\frac{h^2}{GC}} \] \[ = \left(\frac{M^2L^4T^{-2}}{(M^{-1}L^3T^{-2})(LT^{-1})}\right)^{1/2} \] \[ = (M^{3})^{1/2} \] \[ = M^{3/2}L^{0}T^{0} \] Thus, \[ B \rightarrow (3) \]
Step 3: Dimension of (C). \[ \sqrt{GMC} \] \[ = (M^{-1}L^{3}T^{-2}\cdot M \cdot LT^{-1})^{1/2} \] \[ = (L^{4}T^{-3})^{1/2} \] \[ = L^{2}T^{-3/2} \] Thus, \[ C \rightarrow (2) \]
Step 4: Dimension of (D). \[ \sqrt{\frac{GC}{M}} \] \[ = \left(\frac{M^{-1}L^{3}T^{-2}\cdot LT^{-1}}{M}\right)^{1/2} \] \[ = (M^{-2}L^{4}T^{-3})^{1/2} \] \[ = M^{-1}L^{2}T^{-3/2} \] Thus, \[ D \rightarrow (1) \] Final matching: \[ A-4,\; B-3,\; C-2,\; D-1 \] \[ \boxed{(3)} \]