10.30 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)

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Vishal Baghel | Contributor-Level 10

5 months ago

Diameter of the 1st bore, $d_{1}$ = 3 mm = 3 $\times 10^{- 3}$ m

Radius of the first bore, $r_{1}$ = 1.5 mm = 1.5 $\times 10^{- 3}$ m

Diameter of the 2nd bore, $d_{2}$ = 6 mm = 6 $\times 10^{- 3}$ m

Radius of the 2nd bore, $r_{2}$ = 3.0 mm = 3 $\times 10^{- 3}$ m

Surface tension of water, s= 7.3 $\times 10^{- 2}$ N/m

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

Density of water, $ρ = 1.0 \times 10^{3}$ 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 1st and 2nd tubes respectively. These heights are given by the relations:

$h_{1} = {(2 s c o s ? θ) / ρ g r}_{1}$ …(i)

$h_{2} = {(2 s c o s ? θ) / ρ g r}_{2}$ …(ii)

The difference in level of water i

...more

Diameter of the 1st bore, $d_{1}$ = 3 mm = 3 $\times 10^{- 3}$ m

Radius of the first bore, $r_{1}$ = 1.5 mm = 1.5 $\times 10^{- 3}$ m

Diameter of the 2nd bore, $d_{2}$ = 6 mm = 6 $\times 10^{- 3}$ m

Radius of the 2nd bore, $r_{2}$ = 3.0 mm = 3 $\times 10^{- 3}$ m

Surface tension of water, s= 7.3 $\times 10^{- 2}$ N/m

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

Density of water, $ρ = 1.0 \times 10^{3}$ 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 1st and 2nd tubes respectively. These heights are given by the relations:

$h_{1} = {(2 s c o s ? θ) / ρ g r}_{1}$ …(i)

$h_{2} = {(2 s c o s ? θ) / ρ g r}_{2}$ …(ii)

The difference in level of water in the 2 limbs

$h_{1} - h_{2}$ = $\frac{2 s c o s ? θ}{ρ g}$ ( $\frac{1}{r_{1}}$ - $\frac{1}{r_{2}})$

= $\frac{2 \times 7.3 \times 10^{- 2} \times c o s ? 0}{1.0 \times 10^{3} \times 9.8}$ ( $\frac{1}{1.5 \times 10^{- 3}}$ - $\frac{1}{3 \times 10^{- 3}})$ = 4.965 $\times 10^{- 3}$ m = 4.965 mm

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