A capiIIary tube of radius ' r ' is immersed in water and water rises to a height of ' h '. Mass of water in the capillary tube is 5 × 10-3 kg. The same capillary tube is now immersed in a liquid whose surface tension is √2 times the surface tension of water. The angle of contact between the capillary tube and this liquid is 45° . The mass of liquid which rises into the capillary tube now is, (in kg )

Question

A capiIIary tube of radius ' r ' is immersed in water and water rises to a height of ' h '. Mass of water in the capillary tube is 5 × 10-3 kg. The same capillary tube is now immersed in a liquid whose surface tension is √2 times the surface tension of water. The angle of contact between the capillary tube and this liquid is 45° . The mass of liquid which rises into the capillary tube now is, (in kg ) (A) 5 × 10-3 (B) 2.5 × 10-3 (C) 5 √2 × 10-3 (D) 3.5 × 10-3

Tardigrade Admin · Accepted Answer

We knows height of water rise in a capillary tube

h = \frac{2 T c o s θ}{r d g}

h_{1} = \frac{2 T _{1} c o s θ _{1}}{r d g}, h_{2} = \frac{2 T _{2} c o s θ _{2}}{r d g}

Given,

h_{1} = h, T_{1} - T, θ_{1} = 0

∴ h = \frac{2 T}{r d g}

...(i)
Given,

T_{2} = 2 T, θ = 4 5^{\circ}, cos 4 5^{\circ} = \frac{1}{2}

∴ h_{2} = \frac{2 2 T \times \frac{1}{2}}{r d g}

...(ii)
From Eqs. (i) and (ii), we observe

h_{2} = h .

Hence, same mass of liquid rises into the capillary as before

5 \times 1 0^{- 3} k g

.

$5 \times 1 0^{- 3}$

$2.5 \times 1 0^{- 3}$

$52 \times 1 0^{- 3}$

$3.5 \times 1 0^{- 3}$

5×10−3

2.5×10−3

52​×10−3

3.5×10−3

$5 \times 1 0^{- 3}$

$2.5 \times 1 0^{- 3}$

$52 \times 1 0^{- 3}$

$3.5 \times 1 0^{- 3}$