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A particle describes a horizontal circle on the smooth surface of an inverted cone. The plane of that circle is at a height of 9.8 cm above the vertex. Then the speed of the particle is

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0.98 ms^-1

3.92 ms^–1

0.49 ms^–1

1.96 ms^–1

answer is B.

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Detailed Solution

$Nsin θ = mg; Ncos θ = \frac{{mv}^{2}}{r}$
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$\begin{array}{l} \tan θ = rg / v^{2}; \frac{r}{h} = \frac{rg}{v^{2}} \\ V = \sqrt{gh} = \sqrt{9.8 \times 9.8 \times 10^{- 2}} = 9.8 \times 10^{- 2} = 0.98 \end{array}$

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