1.

The figureshows a modelof perfume atomizer . Whenthe bulbA is compressed , air ("density "rho_(a))flows throughthe verticaltubeand emerges throughthe end . If excesspressure appliedto the bulbins progress beDelta p thenthe minimu speed ofair in the tubeto liftperfume is :

Answer»

` sqrt((2 [ Delta p +o gh ])/(rho_(a)))`
` sqrt((2[ Delta p - rhogh ])/(rho_(a)))`
` sqrt((Delta RHO + rho gh)/(rho_(a)))`
None of these

SOLUTION :ApplyingBernoull.sprincipleat 1 and 2 , we obtain
`P_(1) +1/2 rho_(a)v_(1)^(2) = P_(2) + 1/2 rho_(a) v_(2)^(2)`
Putting` v_(1) = 0 `since of the air is at REST inside the BULB ,
`P_(1) = P_(10) + DeltaP_(r) P_(2) = P_(0) - rho gh ` whre `rho = ` densityof liquid( perfume)
` rho_(a)` densityof air
`v_(a)`= minimumvelocityof airto lift the liquidthrough the heighth = v (say)
` P_(0)` = atmosphericpressure
Hence ( A) is CORRECT


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