# 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° from the horizontal. On further increasing the speed of the car to $(1+\beta )v,$ this angle changes to 45°. The value of $\beta$ is close to : Option: 1 0.50 Option: 2 0.41 Option: 3 0.37   Option: 4 0.73

\begin{aligned} &\text { Rain is falling vertically downwards. }\\ &\overrightarrow{\mathrm{v}}_{r / \mathrm{m}}=\overrightarrow{\mathrm{v}}_{\mathrm{r}}-\overrightarrow{\mathrm{v}} \end{aligned}

$\begin{array}{l} \tan 60^{\circ}=\frac{\mathrm{v}_{\mathrm{r}}}{\mathrm{v}_{\mathrm{m}}}=\sqrt{3} \\ \mathrm{v}_{\mathrm{r}}=\mathrm{v}_{\mathrm{m}} \sqrt{3}=\mathrm{v} \sqrt{3} \\ \text { Now, } \mathrm{v}_{\mathrm{m}}=(1+\mathrm{B}) \mathrm{v} \\ \text { and } \theta=45^{\circ} \\ \tan 45=\frac{\mathrm{v}_{\mathrm{r}}}{\mathrm{v}_{\mathrm{m}}}=1 \\ \mathrm{v}_{\mathrm{r}}=\mathrm{v}_{\mathrm{m}} \\ \mathrm{v} \sqrt{3}=(1+\beta) \mathrm{v} \\ \sqrt{3}=1+\beta \\ \Rightarrow \beta=\sqrt{3}-1=0.73 \end{array}$

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