1aree bhaiya....nobody interested in thermo dynamics or what???
A thin tube of uniform cross section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal ends containing air at the same pressure "P". When tube is held at an angle 60° with the vertical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. What is the value of P in cm of mercury?(Temperature of system is kept at 30°C
a) 76 cm of Hg
b) 74.6 cm of Hg
c) 75.4 cm of Hg
d) 72.8 cm of Hg
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5 Answers
1
length of each part when tube is kept horizontal will be l1+l22 = 45.25 cm
let p1 and p2 be pressures in the upper and lower parts when tube is kept inclined
applying boyles law , for the upper part
p1l1A = pl0A
→ p1 = pl0l1 .................(1)
similarly for lower part,
p2 = pl0l2 .................(2)
LET mass of mercury column is m
forces along length of tube are p1A and p2A down the tube and up the tube respectively
and mgcos 60°
balancing forces for equilibrium
p2 = p1 + mgcos60°A
substituting (1) and (2)
pl0l2 = pl0l1 + mg2A
→ pl_{0}\left(\frac{1}{l_{2}}-\frac{1}{l_{1}} \right)=\frac{mg}{2A}
→ p=\frac{mg}{2Al_{0}\left(\frac{1}{l_{2}}-\frac{1}{l_{1}} \right)}
p = h\rho g
and also m = (5cm)AÏ
hence
h\rho g=\frac{(5cm)A\rho g}{2Al_{0}\left(\frac{1}{l_{2}}-\frac{1}{l_{1}} \right)}
substituting the values we get h = 75.4 cm
hence pressure ' p ' is equal to height of mercury column = 75.4 cm