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

##### 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

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