Question:
When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed \(v\), he sees that rain drops 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 :
When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed \(v\), he sees that rain drops 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 :
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The rain falls vertically at speed \(u\); relative to the moving car it has horizontal component equal to the car speed, so \(\tan\theta=u/v_{car}\).
Updated On: Nov 5, 2026
- 0.50
- 0.73
- 0.37
- 0.41
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The Correct Option is B
Solution and Explanation
Let the rain speed be \(u\) (vertical). Relative to the car, the rain has a vertical component \(u\) and a horizontal component equal to the car speed.
Car speed \(v\): \(\tan60^\circ=\dfrac{u}{v}\Rightarrow u=\sqrt3\,v\).
Car speed \((1+\beta)v\): \(\tan45^\circ=\dfrac{u}{(1+\beta)v}=1\Rightarrow(1+\beta)v=u=\sqrt3\,v\).
\(1+\beta=\sqrt3=1.732\Rightarrow\beta=0.732\approx0.73\).
Answer: (2) 0.73
Car speed \(v\): \(\tan60^\circ=\dfrac{u}{v}\Rightarrow u=\sqrt3\,v\).
Car speed \((1+\beta)v\): \(\tan45^\circ=\dfrac{u}{(1+\beta)v}=1\Rightarrow(1+\beta)v=u=\sqrt3\,v\).
\(1+\beta=\sqrt3=1.732\Rightarrow\beta=0.732\approx0.73\).
Answer: (2) 0.73
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