Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = $0.075 N m^{- 1}$ .

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Solution

Given , T = 0.075 N/m,

d = 1 mm = $10^{- 3} m$ ,

$ρ = 10^{3} k g / m^{3}$

Now, T(2L) = $[1 \times (10)^{3} \times h] ρ g$

$\Rightarrow h = \frac{2 \times (0.075)}{10^{- 3} \times 10^{3} \times 10}$

= 0.015 m = 1.5 cm

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Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water =0.075Nm−1.

Given , T = 0.075 N/m, d = 1 mm = 10−3m, ρ=103kg/m3 Now, T(2L) = [1×(10)3×h]ρg ⇒h=2×(0.075)10−3×103×10 = 0.015 m = 1.5 cm

Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = $0.075 N m^{- 1}$ .

Given , T = 0.075 N/m,

d = 1 mm = $10^{- 3} m$ ,

$ρ = 10^{3} k g / m^{3}$

Now, T(2L) = $[1 \times (10)^{3} \times h] ρ g$

$\Rightarrow h = \frac{2 \times (0.075)}{10^{- 3} \times 10^{3} \times 10}$

= 0.015 m = 1.5 cm