JEE Mains · Physics · STD 11 - 3.2 motion in plane
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\(......\)
- A \(0.41\)
- B \(0.50\)
- C \(0.37\)
- D \(0.73\)
Answer & Solution
Correct Answer
(D) \(0.73\)
Step-by-step Solution
Detailed explanation
Rain is falling vertically downwards. \(\vec{v}_{r / m}=\vec{v}_{r}-\vec{v}_{m}\) \(\tan 60^{\circ}=\frac{ V _{ r }}{ v _{ m }}=\sqrt{3}\) \(v _{ r }= v _{ m } \sqrt{3}= v \sqrt{3}\) Now, \(v _{ m }=(1+ B ) v\) and \(\theta=45^{\circ}\) \(\tan 45=\frac{ v _{ r }}{ v _{ m }}=1\)…
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