A J-tube shown in the figure contains a volume V of dry air trapped in arm A of the tube. The atmospheric pressure is H cm of mercury. When more mercury is poured in arm B, the volume of the enclosed air and its pressure undergo a change. What should be the difference in mercury levels in the two arms so as to reduce the volume of air to V/2?

H cm

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\frac{H}{2}

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2 H cm

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\frac{H}{30}

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Solution

The correct option is C H cm
$G i v e n V v o l u m e o f a i r t r a p p e d w i t h p r e s s u r e = P a t m (a t m o s h p h e r i c p r e s s u r e) w h e n i t s v o l u m e i s V / 2, s i n c e T i s k e p t c o n s t a n t h e n c e, P 1 V 1 = P 2 V 2 P V = V / 2 \times P n e w P n e w = 2 P P = ρ g H L e t t h e h e i g h t o f m e r c u r y b e X p d o w n x = P o + ρ g X = ρ g (H + X) 2 P = ρ g (H + X) 2 ρ g H = ρ g (H + X) X = H a n s$

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Similar questions

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The correct option is C H cmGivenVvolumeofairtrappedwithpressure=Patm(atmoshphericpressure)whenitsvolumeisV/2,sinceTiskeptconstanthence,P1V1=P2V2PV=V/2×PnewPnew=2PP=ρgHLettheheightofmercurybeXpdownx=Po+ρgX=ρg(H+X)2P=ρg(H+X)2ρgH=ρg(H+X)X=Hans