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 × 10^{3} kg m^{–3} (g = 9.8 m s^{–2}).

Asked by Pragya Singh | 1 year ago | 134

Diameter of the first bore, d_{1} = 3.0 mm = 3 × 10^{-3} m

Radius of the first bore, r_{1 }= \( \dfrac{3}{2}\) = 1.5 x 10^{-3} m.

Diameter of the second bore, d_{2} =6mm

Radius of the second bore, r_{2 }=\( \dfrac{6}{2}\) = 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 × 10^{3} kg/m^{-3}

Acceleration due to gravity, g = 9.8 m/s^{2}

Let, h_{1} and h_{2} be the heights to which water rises in the first and second tubes respectively.

Thus, the difference in the height:

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

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

**h _{1} – h_{2} =** \( \dfrac{2sCos\Theta }{\rho g}[\dfrac{1}{r_{1}} -\dfrac{1}{r_{2}}]\)

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

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

= −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**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 ρ_{0} =1.25 kg m^{-3} is the density of air at sea level and y_{0 }is a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

**(b)** A zeppelin of volume 1500 m^{3} 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?[ y_{0} = 8000 m and ρH_{e} =0.18 kg m^{-3}].

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 × 10^{3} 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 = \( 9.8 m/s^2\)

Answe the following question:-

**(a) **What is the largest average velocity of blood flow in an artery of radius 2×10^{–3}m 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).