The potential energy of a particle moving along x-direction varies as V = A x 2 x + B . The dimensions of A 2 B are [2024]

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Solution

Q.
The potential energy of a particle moving along x-direction varies as $V = \frac{A x^{2}}{\sqrt{x} + B} .$ The dimensions of $\frac{A^{2}}{B}$ are [2024]

1 $[M^{3 / 2} L^{1 / 2} T^{- 3}]$

2 $[M^{1 / 2} L T^{- 3}]$

3 $[M^{2} L^{1 / 2} T^{- 4}]$

4 $[M L^{2} T^{- 4}]$

Ans.
(3)

$V = \frac{A x^{2}}{\sqrt{x} + B}$

Dimensions of $V = [M^{1} L^{2} T^{- 2}]$

Dimensions of $x^{2} = [L^{2}]$

Dimension of $\sqrt{x} = [L^{1 / 2}]$

From the principle of homogeneity $B = [L^{1 / 2}]$

$[M^{1} L^{2} T^{- 2}] = \frac{A [L^{2}]}{[L^{1 / 2}]}$

$\Rightarrow A = [M^{1} L^{1 / 2} T^{- 2}] \Rightarrow A^{2} = [M^{2} L^{1} T^{- 4}]$

$∴$ $\frac{A^{2}}{B} = \frac{[M^{2} L^{1} T^{- 4}]}{[L^{1 / 2}]} = [M^{2} L^{1 / 2} T^{- 4}]$

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