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Standard VII

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Mughal attack on Shivaji

A capillary t...

Question

A capillary tube of radius $0.20 m m$ is dipped vertically in water. The height (in $c m$ ) of the water column raised in the tube, will be (surface tension of water $= 0.075 N / m$ and density of water $= 1000 k g / m^{3}$ . Take $g = 10 m / s^{2}$ and contact angle $0^{\circ}$ ).

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Solution

We have,

$h = \frac{2 T cos θ}{r ρ g}$

Here, $T = 0.075 N / m$ and $r = 0.20 m m = 0.2 \times 10^{- 3} m, ρ = 1000 k g / m^{3}, θ = 0^{\circ}, g = 10 m / s^{2}$

$h = \frac{2 \times 0.075 \times 1}{(0.20 \times 10^{- 3}) \times (1000) (10)}$

$h = 0.075 m = 7.5 c m$

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Mughal attack on Shivaji

Standard VII History

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A capillary tube of radius 0.20 mm is dipped vertically in water. The height (in cm) of the water column raised in the tube, will be (surface tension of water =0.075 N/m and density of water =1000 kg/m3. Take g=10 m/s2 and contact angle 0∘).

We have, h=2Tcosθrρg Here, T=0.075 N/m and r=0.20 mm=0.2×10−3m,ρ=1000 kg/m3,θ=0∘,g=10 m/s2 h=2×0.075×1(0.20×10−3)×(1000)(10) h=0.075 m=7.5 cm

A capillary tube of radius $0.20 m m$ is dipped vertically in water. The height (in $c m$ ) of the water column raised in the tube, will be (surface tension of water $= 0.075 N / m$ and density of water $= 1000 k g / m^{3}$ . Take $g = 10 m / s^{2}$ and contact angle $0^{\circ}$ ).

We have,

$h = \frac{2 T cos θ}{r ρ g}$

Here, $T = 0.075 N / m$ and $r = 0.20 m m = 0.2 \times 10^{- 3} m, ρ = 1000 k g / m^{3}, θ = 0^{\circ}, g = 10 m / s^{2}$

$h = \frac{2 \times 0.075 \times 1}{(0.20 \times 10^{- 3}) \times (1000) (10)}$

$h = 0.075 m = 7.5 c m$