Question

From the equation, $\tan \theta = \dfrac{{rg}}{{{v^2}}}$, one can obtain the angle of banking $\theta $ for traversing a curved banked road such that no friction acts on the vehicle (the symbols have their usual meanings). Then the equation is
(A) Dimensionally correct only.
(B) Both dimensionally and numerically correct.
(C) Neither numerically nor dimensionally correct.
(D) Numerically correct only.

Answer 1

Hint
From the equation of the velocity of a vehicle in a curved road we have ${v^2} = rg\tan \theta $ which is numerically correct. So by comparing we can find if this given equation is numerically correct. And then by taking the dimensions of the individual components, we can find if the LHS and the RHS match.
In this solution we will be using the following formula,
$\Rightarrow {v^2} = rg\tan \theta $
where $v$ is the velocity of the vehicle
$r$ is the radius of the curved road.
$g$ is the acceleration due to gravity and
$\theta $ is the angle of banking.

Complete step by step answer
For a vehicle that is traversing on a curved banked road, the velocity of the vehicle when there is no friction is given by the formula,
$\Rightarrow {v^2} = rg\tan \theta $
Now from this equation, we can find the angle of banking $\theta $ for minimum wear and tear as,
$\Rightarrow \tan \theta = \dfrac{{{v^2}}}{{rg}}$
This is the numerically correct equation
But from the question we are given the equation for the angle of banking $\theta $ is given as,
$\Rightarrow \tan \theta = \dfrac{{rg}}{{{v^2}}}$
By comparing these two equations we see that they are not the same. Hence, the equation given in the question is not numerically correct.
Now to check if the equation is dimensionally correct, we take the dimensions of the individual components, that is,
$\Rightarrow r = \left[ L \right]$
$\Rightarrow g = \left[ {L{T^{ - 2}}} \right]$
$\Rightarrow v = \left[ {L{T^{ - 1}}} \right]$
So, ${v^2} = \left[ {{L^2}{T^{ - 2}}} \right]$
And $\tan \theta $ is dimensionless as it is a numeric value.
Therefore, by substituting the values in the RHS of the equation given in the question, we get
$\Rightarrow \tan \theta = \dfrac{{rg}}{{{v^2}}} = \dfrac{{\left[ L \right]\left[ {L{T^{ - 2}}} \right]}}{{\left[ {{L^2}{T^{ - 2}}} \right]}}$
From here we can see that the values on the numerator and the denominator cancel each other out and we are left with no dimensions. Hence the LHS and the RHS of the equation match dimensionally. So the equation is only dimensionally correct.
Therefore the correct answer is option (A).

Note
The system of raising the outer edges of a curved road more than the inner edges is known as banking of the road. The angle through which the outer edge is raised is called the angle of banking. The curved roads are banked in order to reduce the wear and tear of the tyres caused due to friction.