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# A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v : tan θ = v and checks that the relation has a correct limit: as v →0, θ →0.

Q 2.25: 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 $v: tan\Theta = v$ 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.

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The derived formula $tan\Theta = v$ is dimensionally incorrect.

We know, Trigonometric functions are dimensionless.

Hence , [$tan\Theta$ ] = $M^0L^0T^0$

and [v] = $M^0L^1T^{-1} = LT^{-1}$.

$\therefore$ To make it dimensionally correct, we can divide v by $v_{r}$ (where $v_{r}$ is the speed of rain)

Thus, L.H.S and R.H.S are both dimensionless and hence dimensionally satisfied.

The new formula is : $tan\Theta = v / v_{r}$

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