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Question

# A thin tube of uniform cross section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the parts on its two sides containing air at the same pressure P. When the tube is held at an angle of 60∘ with the vertical, the length of the air column above and below the mercury pellet are 46cm and 44.5cm respectively. Calculate the pressure P in centimeters of mercury. The temperature of the system is kept at 30∘C.

A

68.45 cms Hg

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B

72.00 cms Hg

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C

75.39 cms Hg

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D

82.21 cms Hg

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Solution

## The correct option is Bb 75.39 cms Hg.When the tube is kept inclined to the vertical, the length of the upper part is l1=46 cm, and that of the lower part is l2=44.5 cm. When the tube lies horizontally, the length on each side is - l0=[l1+l22]=[46 cm+44.5 cm2]=45.25 cm. Let P1 and P2 be the pressures in the upper and the lower parts when the tube is kept inclined. As the temperature is constant throughout, we can apply Boyle's law. For the upper part, P1l1A=Pl0A P1=(Pl0l1).................(i) Similarly, for the lower part, P2=(Pl0l2)................(ii) Thus, P2=P1+mgAcos 60∘. Putting from (i) and (ii), (Pl0l2)=(Pl0l1)+(mg2A) ⇒Pl0(1l2−1l1)=mg2A ⇒P=⎡⎢⎣mg2Al0(ll2−ll1)⎤⎥⎦. If the pressure P is equal to a height h of mercury, P=ρgh. Also, m = (5 cm) Aρ ⇒ρgh=⎡⎢⎣(5 cm)Aρg2Al0(1l2−1l1)⎤⎥⎦ ⇒h=[(5 cm)2×(45.25 cm)×(144.5−146 cm)]=75.39 cm. The pressure P is equal to 75.39 cm of Hg.

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