Q177 / 180 · KCET 2020 · Physics
Rain is falling vertically with a speed of $12 \mathrm{~ms}^{-1}$. A woman rides a bicycles with a speed of $12 \mathrm{~ms}^{-1}$ in east to west direction. What is the direction in which she should hold her umbrella?
- A$30^{\circ}$, towards east
- Correct answerB$45^{\circ}$, towards east
- C$30^{\circ}$, towards west
- D$45^{\circ}$, towards west
Solution
The given situation is shown below
Here, velocity of rain, $v_{r}=12 \mathrm{~ms}^{-1}$ and velocity of woman, $v_{w}=12 \mathrm{~ms}^{-1}$ The relative velocity of rain w.r.t. woman is $\begin{aligned} v_{r w} &=\sqrt{v_{w}^{2}+v_{r}^{2}} \\ &=\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1} \end{aligned}$ The direction in which the woman should hold her umbrella is $\begin{aligned} \sin \theta &=\frac{v_{w}}{v_{r w}}=\frac{12}{12 \sqrt{2}} \\ \Rightarrow \quad \theta &=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}, \text { towards east } \end{aligned}$
Here, velocity of rain, $v_{r}=12 \mathrm{~ms}^{-1}$ and velocity of woman, $v_{w}=12 \mathrm{~ms}^{-1}$ The relative velocity of rain w.r.t. woman is $\begin{aligned} v_{r w} &=\sqrt{v_{w}^{2}+v_{r}^{2}} \\ &=\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1} \end{aligned}$ The direction in which the woman should hold her umbrella is $\begin{aligned} \sin \theta &=\frac{v_{w}}{v_{r w}}=\frac{12}{12 \sqrt{2}} \\ \Rightarrow \quad \theta &=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}, \text { towards east } \end{aligned}$
Hence, the correct answer is option: 'B'.