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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Motion in Two Dimensions

NTA Abhyas 2020 · Physics · Motion in Two Dimensions

To a stationary man, rain appears to be falling at his back at an...

To a stationary man, rain appears to be falling at his back at an angle 30o{30}^{o} with the vertical. As he starts moving forward with a speed of 0.5ms−1,0.5 ms^{- 1}, he finds that rain falls vertically. The speed of rain with respect to stationary man isTo a stationary man, rain appears to be falling at his back at an angle 30 ^ o with the vertical. As he starts movingTo a stationary man, rain appears to be falling at his back at an angle 30 ^ o with the vertical. As he starts moving

Options

  1. A0.5ms−10.5 ms^{- 1}
  2. B1ms−11ms^{- 1}
  3. C32ms−1\dfrac{3}{2 }ms^{- 1}
  4. D13ms−1\dfrac{1}{3}ms^{- 1}

Correct answer

B. 1ms−11ms^{- 1}

Step-by-step solution

For stationary person
To a stationary man, rain appears to be falling at his back at an angle 30 ^ o with the vertical. As he starts moving
For running person
To a stationary man, rain appears to be falling at his back at an angle 30 ^ o with the vertical. As he starts moving
Speed of rain with respect to a stationary man =ν= ν
\overset{→}{v} = vsin{30}^{o}\overset{^}{i} - vcos{30}^{o}\overset{^}{j}
The velocity of rain with respect to man,
v→RM=v→R−v→M\overset{→}{v}RM = \overset{→}{v}R - \overset{→}{v}M
= ( vsin{30}^{o}\overset{^}{i} - vcos{30}^{o}\overset{^}{j} ) - ( 0.5\overset{^}{i} )
Since V→RM\overset{→}{V}RM is vertically downward, its xx-component will become zero.
∴vsin30o−0.5=0∴ vsin{30}^{o} - 0.5 = 0
⇒v=0.5sin30o=1ms−1⇒ v = \dfrac{0.5}{sin{30}^{o}} = 1ms^{- 1}
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