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Question

The difference of two liquid levels in a manometer is 10cm and its density is 0.8gm/cm3. If the density of air is 1.3×103gm/cm3 then the velocity of air will be (in cm/s)

A
347
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B
34.7
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C
3470
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D
0.347
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Solution

The correct option is D 3470

The manometer here gives the stagnation pressure i.e. the velocity of liquid is zero at point B.Hence the pressure drop is given by ΔP=ρgh=PBPAρ=0.8gm/cm3=800kg/m3,h=0.1m,ρair=1.3×103gm/cm3=1.3kg/m3ΔP=800×9.81×0.1ΔP=784.8N/m2

From Bernoulli's equation for stagnation pressure

PA+ρairV22=PBV=2ΔPρairV=2×784.81.3V=34.74m/sV3470cm/s


171283_143612_ans.png

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