MHT CET · Physics · Motion In Two Dimensions
A particle rotates in horizontal circle of radius 'R' in a conical funnel, with 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 \((\mathrm{g}=\) acceleration due to gravity\()\)
- A \(\frac{\mathrm{V}^{2}}{2 \mathrm{~g}}\)
- B \(\frac{\mathrm{V}}{\mathrm{g}}\)
- C \(\frac{\mathrm{V}^{2}}{\mathrm{~g}}\)
- D \(\frac{V}{2 g}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{V}^{2}}{\mathrm{~g}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{mg}=\mathrm{R} \sin \theta\)
\(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{R} \cos \theta\)
\(\tan \theta=\frac{\mathrm{rg}}{\mathrm{v}^{2}}\)
\(\tan \theta=\frac{\mathrm{r}}{\mathrm{h}}\)
\(\mathrm{h}=\frac{\mathrm{v}^{2}}{\mathrm{~g}}\)
\(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{R} \cos \theta\)
\(\tan \theta=\frac{\mathrm{rg}}{\mathrm{v}^{2}}\)
\(\tan \theta=\frac{\mathrm{r}}{\mathrm{h}}\)
\(\mathrm{h}=\frac{\mathrm{v}^{2}}{\mathrm{~g}}\)
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