Two narrow bores of diameters 3.0 mm and 6.0 ...

Mechanical Properties of Fluids

Question : 30 of 31

Marks: +1, -0

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 \times 10^{-2} \text{ N m}^{-1}

. Take the angle of contact to be zero and density of water to be

1.0 \times 103 \text{ kg m}^{-3} (g = 9.8 \text{ m s}^{-2})

Solution:

Here,

T=7.3\times 10^{-2} \text{ N m}^{-1};

\rho=1.0\times 10^{3} \text{ kg m}^{-3}; \theta=0^{\circ}

For narrow tube

2 r_{1}=3.00 \text{ mm}=3\times 10^{-3} \text{ m}

r_{1}=1.5 \times 10^{-3} \text{ m}

For wider tube,

2 r_{2}=6.00 \text{ mm}=6 \times 10^{-3} \text{ m}

r_{2}=3 \times 10^{-3} \text{ m}

Let

h_1, h_2

be the heights to which water rises in narrow tube and wider tube respectively

Then,

h_{1}= \frac{2 T \cos \theta}{r_{1} \rho g}

and

h_{2}= \frac{2 T \cos \theta}{r_{2} \rho g}

∴ Difference in levels of water in two limbs of U tube is,

h_{1}-h_{2}= \frac{2 T \cos \theta}{\rho g} \left[ \frac{1}{r_{1}} - \frac{1}{r_{2}} \right]

= \frac{2 \times 7.3 \times 10^{-2} \times \cos 0^{\circ}}{10^{3} \times 9.8} \times \left[ \frac{1}{1.5 \times 10^{-3}} - \frac{1}{3 \times 10^{-3}} \right]

=4.97 \times 10^{-3} \text{ m}

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