chalte chalte thermo. mercury in sealed tube

aree bhaiya....nobody interested in thermo dynamics or what???

arey use the concept of isothermal processes and an fbd

sayyad yeh Qs HCV se liya gaya hai....!!
Kya ans (c) hai?

haan but can u show the method!!!

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)Aρ
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