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Asked by labdhisangoi101 | 30 Mar, 2024, 02:51: PM

When the capillary tube of radius 2 mm is dipped inside water to a depth 8 mm , there is a capillary rise of water to a height h .

Height h of capillary rise is

where T is surface tension of water , θ is angle of contact , ρ is density of water, g is acceleration due to gravity

and r is radius of capillary tube .

Angle θ of contact with water on glass is zero . Surface tension T of water  is 0.072 N/m .

Hence height h of capillary rise is

h = 7.3 × 10-3 m = 7.3 mm

If the bubble is at the end of capillary tube inside water as shown in figure,

then pressure inside the bubble is  greater by (4T)/r than pressure outside bubble.

Pressure P inside the bubble is

P = Pa + [ ρ g (h+8) × 10-3 ] + [ (4T)/r ]

where Pa is atmospheric pressure .

If Pg is gauge pressure inside the bubble , then

Pg = P - Pa = [ ρ × g × (h+8) × 10-3 ] + [ (4T)/r ]

Pg = [ 1000 × 9.8 ×  (7.3 + 8.0 ) ] + [ ( 4 × 0.072 ) / 0.002 ]  \ Pa

Pg = 294  Pa

Answered by Thiyagarajan K | 30 Mar, 2024, 06:07: PM
NEET neet - Physics
Asked by labdhisangoi101 | 30 Mar, 2024, 02:51: PM