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 :

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

Answer 1

Answer: (D)

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 $\theta=45^{\circ}$

$\tan 45=\frac{ v _{ r }}{ v _{ m }}=1$
$v _{ r }= v _{ m }$
$v \sqrt{3}=(1+\beta) v$
$\sqrt{3}=1+\beta$
$\Rightarrow \beta=\sqrt{3}-1$
$=0.73$