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).

Asked by Pragya Singh | 1 year ago |  134

1 Answer

Solution :-

Diameter of the first bore, d1 = 3.0 mm = 3 × 10-3 m

Radius of the first bore, r\( \dfrac{3}{2}\) = 1.5 x 10-3 m.

Diameter of the second bore, d2 =6mm

Radius of the second bore, r=\( \dfrac{6}{2}\) = 3 x 10-3 mm

Surface tension of water, s = 7.3 × 10-2 N /m

Angle of contact between the bore surface and water, θ= 0

Density of water, ρ =1.0 × 103 kg/m-3

Acceleration due to gravity, g = 9.8 m/s2

Let, h1 and h2 be the heights to which water rises in the first and second tubes respectively.

Thus, the difference in the height:

h1 – h2 =\( \dfrac{2sCos\Theta }{r_{1}\rho g} -\dfrac{2sCos\Theta }{r_{2}\rho g}\)

Since, h = \( \dfrac{2sCos\Theta }{r\rho g}\)

h1 – h2 = \( \dfrac{2sCos\Theta }{\rho g}[\dfrac{1}{r_{1}} -\dfrac{1}{r_{2}}]\)

=\( \dfrac{2\times 7.3\times 10^{-2}\times 1 }{10^{3}\times 9.8} \times\)

\( \left [ \dfrac{1}{1.5\times 10^{-3}} -\dfrac{1}{3 \times 10^{-3}} \right]\)

= −3×10−31​] = 4.97 mm

Therefore, the difference in the water levels of the two arms =2.482 mm.

Answered by Sudhanshu | 1 year ago

Related Questions

Answer the following questions:-

(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].

Class 11 Physics Mechanical Properties of Fluids View Answer

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

Class 11 Physics Mechanical Properties of Fluids View Answer

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.

Class 11 Physics Mechanical Properties of Fluids View Answer

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 = \( 9.8 m/s^2\)

Class 11 Physics Mechanical Properties of Fluids View Answer

Answe the following question:-

(a) What is the largest average velocity of blood flow in an artery of radius 2×10–3m if the flow must remain lanimar? 

(b) What is the corresponding flow rate ? (Take viscosity of blood to be 2.084 × 10–3 Pa s).

Class 11 Physics Mechanical Properties of Fluids View Answer