If x and a stand for distance, then for what value of n is the given equation dimensionally correct? The equation is
$\int \frac{d x}{\sqrt{a^{2} - x^{n}}} = s i n^{- 1} \frac{x}{a}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $2$
$\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = {sin}^{- 1} \frac{x}{a}$
terms in a equation or terms having arithematic operations of addition and subtraction should have same dimensions.
$\Rightarrow$ In $a^{2} - x^{n}$ $d i m (a^{2}) = d i m (x^{n})$
$L^{2} = L^{n} \Rightarrow n = 2$

Suggest Corrections

Similar questions

x

a

n

\int \frac{d x}{\sqrt{a^{2} - x^{n}}} = {sin}^{- 1} \frac{x}{a}

x

a

n

\int \frac{d x}{\sqrt{a^{2} - x^{n}}} = {sin}^{- 1} \frac{x}{a}

Let x and a stand for distance. Is $\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \frac{1}{a} s i n^{- 1} \frac{a}{x}$ dimensionally correct ?

x

a

\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \frac{1}{a} {sin}^{- 1} \frac{a}{x}

\frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + \frac{a_{2}}{n - 1} + \dots + \frac{a_{n - 1}}{2} + a_{n} = 0

a_{0} x^{n} + a_{1} x^{n - 1} + \dots + a_{n - 1} x + a_{n} = 0

(0, 1)

Join BYJU'S Learning Program