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```				   A man is running along a circular track has a speed of 10 kmph. A source of light is at the center of the circular track. A wall along the tangent from which he starts. what is the speed of the shadow of the man on the wall when he covers 1/8 th of the track.
```

6 years ago

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```										Dear ANAM

from the figur when man is at point P shadow will be at point B
tanΘ =x/r
x =r tanΘ
differentiate
dx/dt = r sec2Θ dΘ/dt
=r sec2Θ * W      where w is angular velocity
=v sec2Θ
at 1/8 th of the track it will cover an angel  π/4
so dx/dt = 10* 2 =20 kmph
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
6 years ago
```										This means when particle move from point A along P..the shadow move along AB...if we assume AP=10..then shadow must travel AB in 1 hr ....then 2*pi*r=80...subtituting r in tan formula we dont get x=20...can u tel me wt wrong i did in this?
```
10 months ago

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