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Standard XII

Physics

Young's Modulus of Elasticity

A glass U-tub...

Question

A glass U-tube is such that the diameter of one limb is 3.0 mm and that of the other is 6.00 mm. The tube is inverted vertically with the open ends below the surface of the water in a beaker. What is the difference between the heights to which water rises in the two limbs? The surface tension of water is 0.07 $n m^{- 1}$ . Assume that the angle of contact between water and glass is $0^{\circ}$

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Solution

Suppose pressures at the points A, B, C and D be $P_{A}, P_{B}, P_{C}$ and $P_{D}$ respectively
The pressure on the concave side of the liquid surface is greater than that on the other side by 2T/R
An angle of contact $θ$ is given to be $0^{\circ}$ hence R cos $0^{\circ}$ = r or R = r
$∴$ $P_{A} = P_{B} + 2 T / r_{1}$ and $P_{C} = P_{D} + 2 T / r_{2}$
where $r_{1}$ and $r_{2}$ are the radii of the two limbs
But $P_{A} = P_{C}$ $∴$ $P_{B} + \frac{2 T}{r_{1}} = P_{D} + \frac{2 T}{r_{2}}$
or $P_{D} - P_{B} = 2 T (\frac{1}{r_{1}} - \frac{1}{r_{2}})$
where h is the difference in water levels in the two limbs
Now $h = \frac{2 T}{ρ g} (\frac{1}{r_{1}} - \frac{1}{r_{2}})$ Given that T = $0.07 N m^{- 1}, ρ = 1000 k g m^{- 3}$
$r_{1} = \frac{3}{2} m m = \frac{3}{20} c m = \frac{3}{20 \times 100} m = 1.5 \times 10^{- 3} m, r_{2} = 3 \times 10^{- 3} m$
$∴$ $h = \frac{2 \times 0.07}{1000 \times 9.8} (\frac{1}{1.5 \times 10^{- 3}} - \frac{1}{3 \times 10^{- 3}}) m = 4.76 \times 10^{- 3} m = 4.76 m m$

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