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 f revolutions made by the fan during the time interval between switch of and rest are (Angular retardation is uniform)

Options

A. `(4n)/(15)`
B. `(8n)/(15)`
C. `(16n)/(15)`
D. `(32n)/(15)`

Answer

C

Explanation

Given, `omega_(0) = omega = (omega)/(4), theta =2pin`
From the relation, `omega^(2) = omega_(0)^(2) - 2alpha theta`
Substituting the given values, we get
`(omega/4)^(2) = omega^(2) - 2 alphan(2pi)`
`2alphan(2pi) = omega^(2) - (omega^(2))/(16) rArr 2 pin =(15)/(16) ((omega^(2))/(2alpha))`
Now, when the fan has been switched off,
`omega = 0, omega_(0) = omega, theta = 2pin'`
`rArr 0 = omega^(2 ) - 2 alphan' (2pi)`
`:. 2pin' = (omega^(2))/(2alpha)`
`:. n' = (16)/(15) n`