Question

Given that $\tan \theta =\frac{v^2}{rg}$, gives the angle of banking of the cyclist going round the curve. Here $v$ is the speed of cyclist, $r$ is the radius of the curve and $g$ is acceleration due to gravity. Which of the following statements about this relation is true?

• A. It is both dimensionally as well as numerically correct.
• B. It is neither dimensionally correct nor numerically correct.
• C. It is dimensionally correct but not numerically.
• D. It is numerically correct but not dimensionally.

Answer 1

The correct option is A It is both dimensionally as well as numerically correct.

$\frac{V^2}{rg}=tan\theta$
Dimensionally, $tan\theta$ is dimensionless.
$\Rightarrow \frac{V^2}{rg}=\frac{{\left({LT}^{-1}\right)}^2}{L.{LT}^{-2}}=M^0{L}^0{T}^0$
Hence dimensionally it is correct.
Numerically also it is correct.