Answer

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**Hint:**This is an example of the motion that occurs in two dimensions. Hence, when we consider each element, we have to look into its horizontal and vertical components as well. Further we have to frame the connection between both the velocities by involving the concept of relative velocity. Using these methods we will be able to solve the problem.

**Complete Step by Step Solution**

From the question we are supposed to consider three main entities; velocity of the man, velocity of the rain and the velocity of the rain with respect to the man.

Velocity of rain: $ {\vec v_r} = a\hat i + b\hat j $ . Let the velocity of man be $ {\vec v_m} $ .

Given that, a man is running on a horizontal road at $ 8m{s^{ - 1}} $ , thus $ {\vec v_m} = 8\hat i $ .

According to the first condition, when the man runs with a speed of $ 8m{s^{ - 1}} $ , he sees the rain falling vertically. It is represented by, $ {\vec v_{rm}} = 0\hat i + b\hat j $ where $ {\vec v_{rm}} $ is the relative velocity of the rain and the man, and it’s value is given by, $ {\vec v_{rm}} = {\vec v_r} - {\vec v_m} $ .

Thus, $ {\vec v_{rm}} = {\vec v_r} - {\vec v_m} = b\hat j $ .

We can write, $ a\hat i + b\hat j - 8\hat i = b\hat j $ . Simplifying, we get,

$ \left( {a - 8} \right)\hat i + b\hat j = b\hat j $ .

Thus, from this, it can be written,

$ \left( {a - 8} \right) = 0 $

$ \Rightarrow a = 8 $

According to the second condition, when the speed of the man is increased to $ 12m{s^{ - 1}} $ , he finds that drops make $ {30^ \circ } $ angle with the vertical.

$ {\vec v_m} = 12\hat i $

$ {\vec v_{rm}} = {\vec v_r} - {\vec v_m} = \left( {a\hat i + b\hat j} \right) - 12\hat i $

Simplifying, we get,

$ \left( {a - 12} \right)\hat i + b\hat j = b\hat j $ .

Thus, from this, it can be written,

$ \left( {a - 12} \right) = 0 $

$ \Rightarrow a = 12 $

Since the angle with the vertical is $ {30^ \circ } $ , we can write,

$ \dfrac{b}{{a - 12}} = \tan {30^ \circ } $

Since $ a = 4 $ , $ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{b}{{4 - 12}} $ .

Simplifying, $ b = - \dfrac{4}{{\sqrt 3 }} $ .

As we’ve got the values of $ a $ and $ b $ , the velocity of rain with respect to road is,

$ {\vec v_r} = 8\hat i - \dfrac{4}{{\sqrt 3 }}\hat j $

$ \Rightarrow \left| {{v_r}} \right| = \sqrt {{8^2} + {{\left( { - \dfrac{4}{{\sqrt 3 }}} \right)}^2}} = 8.32m{s^{ - 1}}. $

**Hence, option D is the correct answer among the given options.**

**Note**

The sign of the relative velocity shows variations in accordance to the direction of the motion. If both the bodies are in the same directions, we get a positive value. Else it is a negative value. The students should read the question well and get an idea about the sign of the relative velocity.

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