Air is blown through a pipe AB at a rate of `10 "Lmin"^(-1)`. The cross-sectional area of the broad portion of the pipe AB is `2 cm^(2)` and that of the narrow portion is `0.5 cm^(2)`. The difference in water level h is (density of air `=1.32 kg m^(-3))`

Options

A. 16 mm
B. 1.5 mm
C. 10 mm
D. 3.2mm

Answer

B

Explanation

`Q=A_(1)v_(1)=A_(2)v_(2)=15L "min"^(-1)=(15xx10^(-3))/(60)=2.5xx10^(-4) m^(3) s^(-1)`
`v_(1)=(2.5xx10^(-4))/(0.5xx10^(-4))=5 ms^(-1)`
Now difference in pressure = difference in kinetic energy
`rho_(w)gh=(1)/(2)rho_(a)(v_(2)^(2)-v_(1)^(2))`
`therefore " " h=(rho_(a)(v_(2)^(2)-v_(1)^(2)))/(2rho_(w)g)=1.5xx10^(-3) m ~~1.5 mm`
or `hrho_(w)g=(1)/(2)rho_("air")(v_(2)^(2)-v_(1)^(2))`
or `h=(rho_("air"))/(2rho_(w)g)(v_(2)^(2)-v_(1)^(2))=(1.32)/(2xx10^(3)xx10)[(5)^(2)-(1.25)^(2)]`
`=1.5xx10^(-3) m ~~1.5 mm`