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

In transitional motion, we have `v^(2)=u^(2)+2as`
In rotational motion, we have `omega^(2)=omega_(0)^(2)+2alphatheta`
In the first case, `omega=(omega_(0))/(4)`
`therefore" "((omega_(0))/(4))^(2)=omega_(0)^(2)+2alph theta`
Where `theta=2pin` and n is the no. of rotations in time t
`therefore" "(omega_(0)^(2))/(16)-omega_(0)^(2)=2alpha theta`
`therefore" "-(15)/(16)omega_(0)^(2)=2alpha theta" ...(1)"`
`alpha` is `-ve` (retardation)
In the second case, `omega=0`
and in time 't' it rotates through an angle `theta'=2pin`'
where n' is the no. of rotations.
`therefore 0=omega_(0)^(2)+2alpha theta'`
`therefore -omega_(0)^(2)=2alpha theta'" ...(2)"`
`therefore (omega_(0)^(2))/((15)/(16)omega_(0)^(2))=(2alpha theta')/(2alpha theta)`
`therefore" "(16)/(15)=(theta')/(theta)" "therefore theta'=(16)/(15)theta`
`therefore 2pin'=(16)/(15)(2pin)`
`therefore n'=(16)/(15)n`