Question

You may not know integration. But using dimensional analysis you can check or prove results. In the integral $\int_{}^{}\frac{dx}{(2ax - x^{2})^{1/2}} = \text{ }\text{a}^{n}\text{ si}\text{n}^{\text{-1}}$ $\left( \frac{x}{a} - 1 \right)$the value of n should be–

• A. 1
• B. -1
• C. 0
• D. ½

Answer 1

Answer: (C)

0

Answer 2

$\int_{}^{}\frac{dx}{(2ax - x^{2})}$ = aⁿsin^–1 $\left\lbrack \frac{x}{a} - 1 \right\rbrack$.

From R.H.S. dimension of [a] = [L],

Since $\left\lbrack \frac{x}{a} \right\rbrack$ should be dimensionless.

Dimension L.H.S. : $\frac{\lbrack L\rbrack}{\lbrack L\rbrack} = \text{ }\text{L}^{0}$

Since dimension of $\lbrack\text{2ax - }\text{x}^{2}\rbrack^{\text{1/2}}\ = \ \lbrack L\rbrack$

& dimension of [dx] = L

Equating dimensions of L.H.S. & R.H.S. $L^{0}\ = \text{ Ln }$

n = 0

$L^{0}\ = \text{ Ln }$

n = 0