A particle describes a circular motion in horizontal plane in a conical funnel with speed v, the height of plane of circle from bottom is 40 cm, find v.

4 m / s

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8 m / s

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1 m / s

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2 m / s

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Solution

The correct option is D $2 m / s$

$N c o s θ = \frac{m v^{2}}{r}$
$N s i n θ = m g$
$t a n θ = \frac{r g}{v^{2}} = \frac{/ r g}{v^{2}} = \frac{/ r}{h}$
$v^{2} = g h = \sqrt{g h} = 2 m / s$

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A particle describes a circular motion in horizontal plane in a conical funnel with speed v, the height of plane of circle from bottom is 40 cm, find v.

The correct option is D 2 m/s N cos θ=mv2r N sin θ=mg tan θ=rgv2=/rgv2=/rh v2=gh=√gh=2 m/s

The correct option is D $2 m / s$

$N c o s θ = \frac{m v^{2}}{r}$
$N s i n θ = m g$
$t a n θ = \frac{r g}{v^{2}} = \frac{/ r g}{v^{2}} = \frac{/ r}{h}$
$v^{2} = g h = \sqrt{g h} = 2 m / s$