Q 11-03-145JEE MainJEE Main 2020 (6 Sep, Shift 2)Easy
When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed $v$, he sees the rain drops 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:
Answer: (D) $0.73$
Let the rain fall vertically with speed $u$. Relative to the car, the rain has a horizontal component equal to the car's speed, so the angle $\theta$ with the horizontal satisfies $\tan\theta = u/v_{\text{car}}$.
$\tan60^\circ = \dfrac uv \Rightarrow u = \sqrt3v$
$\tan45^\circ = \dfrac{u}{(1+\beta)v} \Rightarrow 1+\beta = \sqrt3 \Rightarrow \beta \approx 0.73$
Solution by Sreeraj P, M.Sc Physics