\[(a)\;3 km/hr\quad (b)\;\frac{3}{\sqrt 2} km/hr \quad (c)\;5km/hr \quad (d)\;3 \sqrt 2 km/hr\]

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Let $\hat i$ and $ \hat j$ be unit vector in horizontal and vertical respectively.

Let velocity of rain $v_r=a \hat i+b \hat j$

speed of rain$=\sqrt {a^2+b^2}$

Case I:

when relative velocity of rain w.r.t man is vertical

$v_{rm}=\bar v_r-\bar v_m$

$v_m=3 \hat i$

$v_{rm}=(a-3) \hat i+b \hat j$

Since $v_{rm}$ is vertical

$a-3=0=>a=3$

Case II:

when relative velocity is at $45 ^{\circ}$

$\bar v_m=6 \hat i$

$v_{rm}=(a-6) \hat i+b \hat j$

$\qquad=- 3 \hat i+b \hat j$

Since $\tan \theta=\large\frac{b}{-3}$

tan 45=1

=> $|b|=3$

Therefore $Speed=\sqrt {3^2+3^2}=3 \sqrt 2$

Hence d is the correct answer.

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