The angular frequency of a fan increases from $30 r p m$ to $60 r p m$ in $π s$ . A dust particle is present on the blades at a distance of $20 c m$ from axis of rotation. The net acceleration of dust particle at the end of $π s$ is

2.8 m s^{- 2}

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7.9 m s^{- 2}

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9.2 m s^{- 2}

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6.2 m s^{- 2}

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Solution

The correct option is B $7.9 m s^{- 2}$
Final velocity after $π s$ is,

$v = \frac{60 \times 2 π \times 0.2}{60}$

$v = 0.4 π$

Net acceleration after $π s$ is,

$a = \frac{v^{2}}{r}$

$a = \frac{{(0.4 π)}^{2}}{0.2}$

$a = \frac{9.87 \times 0.16}{0.2}$

$a = 7.9 m / s^{2}$

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The angular frequency of a fan increases from 30rpm to 60rpm in πs . A dust particle is present on the blades at a distance of 20cm from axis of rotation. The net acceleration of dust particle at the end of πs is

The correct option is B 7.9ms−2Final velocity after πs is, v=60×2π×0.260 v=0.4π Net acceleration after πs is, a=v2r a=(0.4π)20.2 a=9.87×0.160.2 a=7.9m/s2

The angular frequency of a fan increases from $30 r p m$ to $60 r p m$ in $π s$ . A dust particle is present on the blades at a distance of $20 c m$ from axis of rotation. The net acceleration of dust particle at the end of $π s$ is

The correct option is B $7.9 m s^{- 2}$
Final velocity after $π s$ is,

$v = \frac{60 \times 2 π \times 0.2}{60}$

$v = 0.4 π$

Net acceleration after $π s$ is,

$a = \frac{v^{2}}{r}$

$a = \frac{{(0.4 π)}^{2}}{0.2}$

$a = \frac{9.87 \times 0.16}{0.2}$

$a = 7.9 m / s^{2}$