In the equation -dt-2at-t-2-a-x sin-1 - ta-1 - where -t- is time- The value of x is-

The quantity ta 1 is dimensionless i.e., [a]=[t] ∴ [ √(2at t^2) ]=[t] or [ dt√(2at t^2) ]= [ tt ]=[M^0L^0T^0] i.e., ax should also be dimensionless. or x=0 ...

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Byju's Answer

Standard XII

Physics

Dimensional Analysis

In the equati...

Question

In the equation $\int \frac{d t}{\sqrt{2 a t - t^{2}}} = a^{x} {sin}^{- 1} [\frac{t}{a} - 1]$ , where ' $t$ ' is time. The value of $x$ is

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Solution

The quantity $\frac{t}{a} - 1$ is dimensionless i.e.,
$[a] = [t]$
$∴ [\sqrt{2 a t - t^{2}}] = [t]$
or $[\frac{d t}{\sqrt{2 a t - t^{2}}}] = [\frac{t}{t}] = [M^{0} L^{0} T^{0}]$
i.e., $a x$ should also be dimensionless.
or $x = 0$

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In the equation ∫dt√2at−t2=axsin−1[ta−1], where 't' is time. The value of x is

The quantity ta−1 is dimensionless i.e., [a]=[t] ∴[√2at−t2]=[t] or [dt√2at−t2]=[tt]=[M0L0T0] i.e., ax should also be dimensionless. or x=0

In the equation $\int \frac{d t}{\sqrt{2 a t - t^{2}}} = a^{x} {sin}^{- 1} [\frac{t}{a} - 1]$ , where ' $t$ ' is time. The value of $x$ is

The quantity $\frac{t}{a} - 1$ is dimensionless i.e.,
$[a] = [t]$
$∴ [\sqrt{2 a t - t^{2}}] = [t]$
or $[\frac{d t}{\sqrt{2 a t - t^{2}}}] = [\frac{t}{t}] = [M^{0} L^{0} T^{0}]$
i.e., $a x$ should also be dimensionless.
or $x = 0$