Question

.
Find the difference in levels of mercury in the two limbs of a
U tube, if the diameter of bore of one limbis 1 mm and of the
other 4 mm. The surface tension of mercury is 544 x 103 and angle of contact is 130°.​

Answer 1

Answer:

0.00787 m

Explanation:

Given :

r₁ = 1 mm = 10⁻³ m

r₂ = 4 mm = 4 × 10⁻³ m

S = 544 × 10⁻³ N / m

Ф = 130

ρ = 13.6 × 10³ kg / m³

Let h₁ and h₂ be the height of mercury rises.

We know :

h₁ = 2 S cos Ф / r₁  ρ g   and  h₂ = 2 S cos Ф / r₂  ρ g

Difference in level of mercury in two limbs of U-tube :

= > h₂ - h₁ = 2 S cos Ф / ρ g [ 1 / r₂ - 1 / r₁ ]

Putting values here we get :

= > h₂ - h₁ = 2 × 544 × 10⁻³  × cos 130 / 13.6 × 10³ × 9.8 [ 1 / 4 × 10⁻³ - 1 / 10⁻³ ]

= >  h₂ - h₁ = 2 × 544 × cos 130 / 13.6 × 10³ × 9.8 [ 1 / 4  - 1  ] m

= >  h₂ - h₁ = 7.87 × 10⁻⁴ m

= >  h₂ - h₁ = 0.00787 m

Therefore , we get required answer.