(i) One end of a uniform glass capillary tube of radius r = 0.025 cm is immersed vertically in water to a depth h = 1cm. Contact angle is 0°, surface tension of water is 7.5 × 10–2 N/m,density of water is ρ = 103 kg/m3 and atmospheric pressure is Po = 105 N / m2Find the excess pressure to be applied on the water in the capillary tube so that -The water level in the tube becomes same as that in the vessel.

(i) (a) P1 = P2P0+ΔP−2TR=P0[R = radius of curvature of the curved surface. Pressure on convex side is lesser by 2TR than pressure on concave side]For θ=0∘;R=rΔP=2Tr=2×7.5×10−22.5×10−4=600Pa

(i) One end of a uniform glass capillary tube of radius r = 0.025 cm is immersed vertically in water to a depth h = 1cm. Contact angle is 0°, surface tension of water is 7.5 × 10–2 N/m,density of water is ρ = 103 kg/m3 and atmospheric pressure is Po = 105 N / m2Find the excess pressure to be applied on the water in the capillary tube so that -The water level in the tube becomes same as that in the vessel.

(i) (a) P1 = P2P0+ΔP−2TR=P0[R = radius of curvature of the curved surface. Pressure on convex side is lesser by 2TR than pressure on concave side]For θ=0∘;R=rΔP=2Tr=2×7.5×10−22.5×10−4=600Pa

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(i) One end of a uniform glass capillary tube of radius r = 0.025 cm is immersed vertically in water to a depth h = 1cm. Contact angle is 0°, surface tension of water is 7.5 × 10^–2 N/m,density of water is $ρ$ = 10³ kg/m³ and atmospheric pressure is P_o = 10⁵ N / m²
Find the excess pressure to be applied on the water in the capillary tube so that -
The water level in the tube becomes same as that in the vessel.

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200kPa

300 Pa

150 kPa

600 Pa

answer is B.

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Detailed Solution

(i) (a) P₁ = P₂
$P_{0} + ΔP - \frac{2 T}{R} = P_{0}$
[R = radius of curvature of the curved surface. Pressure on convex side is lesser by $\frac{2 T}{R}$ than pressure on concave side]
For $θ = 0^{\circ}; R = r$
$Δ P = \frac{2 T}{r} = \frac{2 \times 7.5 \times 10^{- 2}}{2.5 \times 10^{- 4}} = 600 Pa$