Kinematics Question 803

Question: A particle describes a horizontal circle in a conical funnel whose inner surface it’s smooth with speed of 0.5 m/s. What is the height of the plane of circle from vertex of the funnel ? [J&K CET 2005]

Options:

A)0.25 cm

B) 2 cm

C) 4 cm

D) 2.5 cm

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Answer:

Correct Answer: D

Solution:

The particle is moving in circular path , $ mg=R\sin \theta $ -(i) $ \ \frac{mv^{2}}{r}=R\cos \theta $ -(ii)

From equation (i) and (ii) we get $ \tan \theta =\frac{rg}{v^{2}} $ but $ \tan \theta =\frac{r}{h} $ $ h=\frac{v^{2}}{g}=\frac{{{(0.5)}^{2}}}{10}=0.025m=2.5cm $



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