Question.
A man walking briskly in rain with speed $v$ must slant his umbrella forward making an angle $\theta$ with the vertical. A student derives the following relation between $\theta$ and $v \cdot \tan \theta=v$ and checks that the relation has a correct limit: as $v \rightarrow 0, \theta \rightarrow 0$, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.
A man walking briskly in rain with speed $v$ must slant his umbrella forward making an angle $\theta$ with the vertical. A student derives the following relation between $\theta$ and $v \cdot \tan \theta=v$ and checks that the relation has a correct limit: as $v \rightarrow 0, \theta \rightarrow 0$, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.
solution:
Incorrect; on dimensional ground
The relation is $\tan \theta=v$.
Dimension of R.H.S $=M^{0} L^{1} T^{-1}$
Dimension of $\mathrm{L} . \mathrm{H} . \mathrm{S}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
( $\because$ The trigonometric function is considered to be a dimensionless quantity)
Dimension of R.H.S is not equal to the dimension of L.H.S. Hence, the given relation is not correct dimensionally.
To make the given relation correct, the R.H.S should also be dimensionless. One way to achieve this is by dividing the R.H.S by the speed of rainfall.
Therefore, the relation reduces to
$\tan \theta=\frac{v}{v^{\prime}}$. This relation is dimensionally correct.
Incorrect; on dimensional ground
The relation is $\tan \theta=v$.
Dimension of R.H.S $=M^{0} L^{1} T^{-1}$
Dimension of $\mathrm{L} . \mathrm{H} . \mathrm{S}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
( $\because$ The trigonometric function is considered to be a dimensionless quantity)
Dimension of R.H.S is not equal to the dimension of L.H.S. Hence, the given relation is not correct dimensionally.
To make the given relation correct, the R.H.S should also be dimensionless. One way to achieve this is by dividing the R.H.S by the speed of rainfall.
Therefore, the relation reduces to
$\tan \theta=\frac{v}{v^{\prime}}$. This relation is dimensionally correct.