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Home » A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes (1/4)th of the original in time ′t′ and ′n′ revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest are (Angular retardation is uniform).
A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes (1/4)th of the original in time ′t′ and ′n′ revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest are (Angular retardation is uniform).
Physics | Physics
A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes (1/4)th of the original in time ′t′ and ′n′ revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest are (Angular retardation is uniform).
  1. 8n/15
  2. 4n/15
  3. 16n/15
  4. 32n/15

Solution: The correct answer is C. 16n/15

{{(\omega /4)}^{2}}={{\omega }^{2}}-2\alpha ~n(2\pi )

\therefore 2\alpha n(2\pi )={{\omega }^{2}}-\frac{{{\omega }^{2}}}{16}

\therefore 2\pi n=\frac{15}{16}\times (\frac{{{\omega }^{2}}}{2\alpha })

0={{\omega }^{2}}-2\alpha n\prime

2\pi n\prime =\frac{{{\omega }^{2}}}{2\alpha }

\therefore n\prime =\frac{16n}{15}

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