What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

Asked by Pragya Singh | 1 year ago |  64

##### Solution :-

Given:

Surface tension of mercury, S = 4.65 × 10-1 N m-1

Radius of the mercury drop, r = 3.00 mm = 3 × 10-3 m

Atmospheric pressure, P0 = 1.01 × 105 Pa

We know:

Total pressure inside the mercury drop

= Excess pressure inside mercury + Atmospheric pressure

=$$\dfrac{2S}{r}$$ + P$$\left (\dfrac{ 2 \times 4.65 \times 10^{ -1 } }{ 3 \times 10^{ -3 } } \right ) + 1.01 \times 10^{ 5 }$$

$$= 1.0131 \times 105 Pa$$

= 1.01 x 105 Pa

Excess pressure = $$\dfrac{ 2S }{ r }$$

=$$\left (\dfrac{ 2 \times 4.65 \times 10^{ -1 } }{ 3 \times 10^{ -3 } } \right )= 310 Pa$$

Answered by Abhisek | 1 year ago

### Related Questions

#### According to the law of atmospheres density of air decreases with increase in height y

(a) According to the law of atmospheres density of air decreases with increase in height y as $$ρ=\rho _{0}e^{\dfrac{-y}{y_{0}}}$$. Where ρ0 =1.25 kg m-3 is the density of air at sea level and yis a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

(b) A zeppelin of volume 1500 m3 is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y0 = 8000 m and ρHe =0.18 kg m-3].

#### Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends.

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 m s–2).

#### Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius

Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m–1. Density of mercury = 13.6 × 103 kg m–3

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius $$2.0 × 10^{–5} m$$ and density $$1.2 × 10^3 kg m^{–3}$$. Take the viscosity of air at the temperature of the experiment to be $$1.8 × 10^{–5} Pa$$. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
A plane is in level flight at constant speed and each of its wings has an area of $$25 m^2$$. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be $$1 kg/m^3$$), g = $$9.8 m/s^2$$