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A leaf is caught in a whirlpool. At a given instant, the leaf is at a distance of 120 m from the centre of the whirlpool. The whirlpool can be described by the following velocity distribution:
Vr = - 60 × 103 m/s and Vθ = 300 × 103 m/s 2πr 2πr
where r (in meters) is the distance from the centre of the whirlpool. What will be the distance of the leaf from the centre when it has moved through half a revolution?
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- 48 m
- 64 m
- 120 m
- 142 m
- 48 m
Correct Option: B
Radial distance = 120 m
Vr = - | m /s | |
2πr |
& Vθ = | m /s | |
2πr |
& θ = π
= | |||
Vθ | 5 |
we know that
Vr = | ...(i) | |
dt |
& Vθ = rw = r | ||
dt |
-5Vr = r | ||
dt |
Vr = | × | ...(ii) | ||
5 | dt |
By equating (i) & (ii), we get
= | . | ||||
dt | 5 | dt |
Integrating both sides, we get
![](http://images.interviewmania.com/wp-content/uploads/2019/10/as-75.jpg)
ln | ![]() | ![]() | = | |||
120 | 5 |
By solving above, we get
r = 64 m