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When a car is at rest, its driver sees raindrops falling on it vertically. When driving the car with speed V, he sees that raindrops are coming at an angle $60^{0}$ from the horizontal on further increasing the speed of the car to $(0.732 + β) V$ this angle changes to $45^{0}$ . The value of $β$ is close to

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Detailed Solution

The given situation is shown in diagram ,

Here, $V_{2}$ be the velocity of rain drop

When car is moving with speed $V_{1}$

$\tan 60^{0} = \frac{V_{r}}{V} - - - (1)$

When car is moving with speed $(0.732 + β) V$

$\tan 45^{0} = \frac{V_{r}}{{(0.732 + β)}^{2}} - - - - (2)$

Eq (1)/ eq (2) $\sqrt{3} = 0.732 + β$

$1.732 = 0.732 + β$

$β = 1$

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