Question

2.
By bow much will the susface of mercusy be
depressed in a glass tube of radius 0.01cm if the
angle of contact of mercury is 135º and its surface
tension is 0.487 Nm^-1? Cos 135 : -0.707.​

Answer 1

Answer:

$\boxed{\mathfrak{Height \ by \ which \ surface \ of \ mercury \ is \ depressed \approx 5.2 \ cm}}$

Given:

Radius of glass tube (r) = 0.01 cm = $\sf 1 \times 10^{-4}$

Angle of contact (θ) = 135°

(Cos 135° = -0.707)

Surface tension (S) = 0.487 N/m

Density of mercury (ρ) = 13,546 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

To Find:

Height by which surface of mercury will be depressed in a glass tube (h)

Explanation:

Formula:

$\boxed{ \bold{h = \frac{2S cos \theta}{r \rho g} }}$

Substituting value of S, cosθ, r, ρ & g in the equation:

$\sf \implies h = \frac{2 \times 0.487 \times ( - 0.707)}{1 \times {10}^{ - 4} \times 13546 \times 9.8} \\ \\ \sf \implies h = - \frac{0.688618}{132750.8 \times {10}^{ - 4} } \\ \\ \sf \implies h = - \frac{0.688618}{13.27508} \\ \\ \sf \implies h \approx - 0.052 \: m \\ \\ \sf \implies h \approx - 5.2 \: cm$

Negative sign denotes that surface is depressed in the glass tube.

$\therefore$

Height by which surface of mercury is depressed in a glass tube (h) ≈ 5.2 cm

Answer 2

Answer:

r = 0.01 cm = 10^-4 m

θ = 135°

(Cos 135° = -0.707)

S = 0.487 N/m

ρ = 13546 kg/m³

g = 10 m/s²

Equation:

h = 2Scosθ/rρg

Substituting values in the equation we get:

h = 0.05 m