Grade 11Mechanics

A column of mercury of 10 cm length is contained in the middle of a narrow horizontal 1 m long tube which is closed at both the ends. Both the halves of the tube contain air at a pressure of 76 cm of mercury. By what distance will the column of mercury be displaced if the tube is held vertically

Radhika Batra

12 Years agoGrade 11

2 Answers

Kevin Nash

12 Years ago

. M is the mid-point of tube AB.

At equilibrium

p₁ x A + mg = p₂ x A

p₁ x A + 10 x A x d_Hg = p₂ x A

⇒ p₁ + 10d_Hg x g = p₂

For air present in column AP

p x 45 x A = p₁ x (45 + x ) x A

⇒ p₁ = 45 / 45 + x x 76 d_Hg x g …..(ii)

For air present in column QB

p x 45 x A = p₂ x (45 – x ) x A

⇒ p₂ = 45 / 45 - x x 76 d_Hg x g (iii)

From (i), (ii) and (iii)

45 x 76 x d_Hgg / 45 + x + 10 d_Hg x g = 45 / 45 – x x 76 x d_Hg x g

⇒ 45 x 76 / 45 + x + 10 = 45 x 76 / 45 – x

x = 2.95 cm.

Rishi Sharma

6 Years ago

Dear Student,
Please find below the solution to your problem.

M is the mid-point of tube AB.
At equilibrium
p1 x A + mg = p2 x A
p1 x A + 10 x A x dHg = p2 x A
⇒ p1 + 10dHg x g = p2
For air present in column AP
p x 45 x A = p1 x (45 + x ) x A
⇒ p1 = 45 / 45 + x x 76 dHg x g …..(ii)
For air present in column QB
p x 45 x A = p2 x (45 – x ) x A
⇒ p2 = 45 / 45 - x x 76 dHg x g (iii)
From (i), (ii) and (iii)
45 x 76 x dHgg / 45 + x + 10 dHg x g = 45 / 45 – x x 76 x dHg x g
⇒ 45 x 76 / 45 + x + 10
= 45 x 76 / 45 – x
x = 2.95 cm.

Thanks and Regards