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Chapter 1: Problem 73

\(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\frac{1}{a} \sin ^{-1} \frac{a}{x}\) (A) is dimensionally correct. (B) dimensionally incorrect. (C) such mathematical relations cannot be tested. (D) cannot say.

Short Answer

Expert verified
(B) dimensionally incorrect.

Step by step solution

01

Analyze the integral equation

We are given the integral equation: \( \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\frac{1}{a} \sin ^{-1} \frac{a}{x} \) We can break down this equation into two parts: the left-hand side (LHS) and the right-hand side (RHS). We will analyze the dimensions of both parts separately.
02

Analyze the left-hand side (LHS)

The LHS of the equation is: \( \int \frac{d x}{\sqrt{a^{2}-x^{2}}} \) Here, dx is the infinitesimal change in x, which has a dimension of length (L). The expression under the square root sign is also of dimension length squared (L虏), as both \(a^2\) and \(x^2\) are length squared. Now, the entire denominator \(\sqrt{a^{2}-x^{2}}\) has a dimension of length (L). As a result, the integrand has the dimensions: \(\frac{L}{L} = 1\). Since the integral of a dimensionless quantity is still dimensionless, we conclude that the LHS has no dimensions.
03

Analyze the right-hand side (RHS)

The RHS of the equation is: \( \frac{1}{a} \sin^{-1} \frac{a}{x} \) The sin鈦宦� function (inverse sine) does not have any dimensions, as it simply converts a ratio of lengths to an angle in radians. The ratio \(\frac{a}{x}\) is also dimensionless, since both a and x have dimensions of length (L). Finally, the factor \(\frac{1}{a}\) has the dimensions \(\frac{1}{L}\). So, the RHS has the dimensions of \(\frac{1}{L}\).
04

Compare the dimensions of LHS and RHS

We found that the dimensions of the LHS are dimensionless while the dimensions of the RHS are \(\frac{1}{L}\). Since these dimensions are not the same, the integral equation is not dimensionally correct. Therefore, the correct answer is: (B) dimensionally incorrect.

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Most popular questions from this chapter

The dimensions of \(\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\) are (A) \(\left[I^{0} M^{0} L^{1} T^{-1}\right]\) (B) \(\left[I^{2} M^{4} L^{-3} T^{1}\right]\) (C) \(\left[I^{-2} T^{-2} L^{1} M^{1}\right]\) (D) \(\left[I^{0} M^{0} L^{-1} T^{1}\right]\)

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