MHT CET · Physics · Motion In Two Dimensions
A particle rotates in a horizontal circle of radius ' R ' in a conical funnel with constant speed ' V '. The inner surface of the funnel is smooth. The height of the plane of the circle from the vertex of the funnel is (g-acceleration due to gravity)
- A \(\frac{\mathrm{V}}{\mathrm{g}}\)
- B \(\frac{\mathrm{V}}{2 \mathrm{~g}}\)
- C \(\frac{\mathrm{V}^2}{2 \mathrm{~g}}\)
- D \(\frac{\mathrm{V}^2}{\mathrm{~g}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{V}^2}{\mathrm{~g}}\)
Step-by-step Solution
Detailed explanation
\(\text { From figure, } \)
\( N \sin \theta=m g \)
\( N \cos \theta=\frac{m V^2}{R} \)
\( \therefore \tan \theta=\frac{R g}{V^2} \)
\( \therefore \frac{R}{h}=\frac{R g}{V^2} \)
\( \therefore h=\frac{V^2}{g}\)

\( N \sin \theta=m g \)
\( N \cos \theta=\frac{m V^2}{R} \)
\( \therefore \tan \theta=\frac{R g}{V^2} \)
\( \therefore \frac{R}{h}=\frac{R g}{V^2} \)
\( \therefore h=\frac{V^2}{g}\)

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