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Question

The figure shows a model of perfume atomizer. When the bulb A is compressed, air having density ρa flows through the narrow tube. Consequently, pressure at the position of the vertical tube reduces. The liquid (perfume) of density ρ rises through the vertical tube and emerges through the end. If the excess pressure applied to the bulb in this process be Δp, the minimum speed of air in the tube to lift the perfume is

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Solution

The correct option is **A** √2(Δp+ρgh)ρa

Applying Bernoulli’s principle at 1 and 2, we obtain

P1+12ρa v21=P2+12 ρ v22

putting, v1=0 since the air is at rest inside the bulb,

Also, P1=P0+Δp

And, P2=P0−ρgh

where ρ = density of liquid (perfume)

ρa = density of air

v2 = minimum velocity of air to lift the liquid through the height h

p0 = atmospheric pressure

⇒P0+Δp+0=P0−ρgh+12 ρ v22

⇒v2=√2(Δp+ρgh)ρa

Hence, option (A) is correct.

Applying Bernoulli’s principle at 1 and 2, we obtain

P1+12ρa v21=P2+12 ρ v22

putting, v1=0 since the air is at rest inside the bulb,

Also, P1=P0+Δp

And, P2=P0−ρgh

where ρ = density of liquid (perfume)

ρa = density of air

v2 = minimum velocity of air to lift the liquid through the height h

p0 = atmospheric pressure

⇒P0+Δp+0=P0−ρgh+12 ρ v22

⇒v2=√2(Δp+ρgh)ρa

Hence, option (A) is correct.

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