In the equation ∫ a 2 + x 2 d x = a n 1 tan − 1 a x ( x and a are distances), find the value of n .

LHS has dimensions [ L 2 ] [ L ] = [ L − 1 ] ; RHS has dimensions [ L − n ] . Hence n = 1 .

Physics - Unit & Dimension Question with Solution | TestHub

PhysicsUnit & DimensionMiscellaneousEasy2 minQB

PhysicsEasyinteger

In the equation $\int \frac{d x}{a ^{2} + x ^{2}} = \frac{1}{a ^{n}} tan^{- 1} \frac{x}{a}$ ( $x$ and $a$ are distances), find the value of $n$ .

Answer:

Solution:

LHS has dimensions $\frac{[ L ]}{[ L ^{2} ]}$
$= [L^{- 1}]$ ;
RHS has dimensions $[L^{- n}]$ .
Hence $n = 1$ .

Stream:JEESubject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

Loading discussions...