Question

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 ${60}^{\circ }$ from the horizontal. On further increasing the speed of the car to $(1+\beta )v,$ this angle changes to ${45}^{\circ }$. The value of $\beta$ is close to :

• A. $0.50$
• B. $0.41$
• C. $0.37$
• D. $0.73$

Answer 1

The correct option is D $0.73$

The given situation is shown in the diagram. Here $v_r$ be the velocity of rain drop.


When car is moving with speed $v,$

$\tan {60}^{\circ }=\frac{v_r}{v}...\left(i\right)$

When car is moving with speed $(1+\beta )v,$

$\tan {45}^{\circ }=\frac{v_r}{(\beta +1)v}...\left(ii\right)$

Dividing $\left(i\right)$ by $\left(ii\right)$ we get,

$\sqrt{3}=(\beta +1)$

$\Rightarrow \beta =\sqrt{3}-1=\mathrm{0.732.}$

Hence, $\left(D\right)$ is the correct answer.