Numerical Problems on Surface Tension

In this article, we shall study numerical problems to find surface tension of liquid.

Science > Physics > Surface Tension > Numerical Problems on Surface Tension

Example – 1:

A needle 5 cm long can just rest on the surface of the water without wetting. What is its weight? Surface tension of water = 0.07 N/m.

Given: Length of needle = l = 5 cm = 5 × 10^-2 m, Surface tension = T = 0.07 N/m

To Find: Weight of needle = F =?

Solution:

In case of a needle, the water wets it from two sides, Hence, the effective length = 2 l

We have, T = F / effective length

∴   F = T × effective length = T × 2 l

∴   F = 0.07 × 2 × 5 × 10^-2 = 0.007 N

Ans: The weight of needle is 0.007 N

Example – 2:

A light square wireframe each side of which is 10 cm long hangs vertically in the water with one side just touching the water surface. Find the additional force necessary to pull the frame clear of the water (T=0.074 N/m)

Given: Length of frame in contact with water = l = 10 cm = 10 × 10^-2 m, Surface tension = T = 0.074 N/m

To Find: Pull required = F =?

Solution:

In case of wire, the water wets it from two sides,

Hence, the effective length = 2 l

We have, T = F / effective length

∴   F = T × effective length = T × 2 l

∴   F = 0.074 × 2 × 10 × 10^-2 = 0.0148 N

Ans: The pull required is 0.0148 N

Example – 3:

A rectangular film of liquid is formed in a frame of wire and a movable rod of length 4 cm. What force must be applied to the rod to keep it in equilibrium if the surface tension of the liquid is 40 × 10^-3 N/m?

Given: Length of the rod = l = 4 cm = 4 × 10^-2 m, Surface tension = T = 40 × 10^-3 N/m

To Find: Force required = F =?

Solution:

In case of rod, the liquid wets it from two sides,

Hence, the effective length = 2 l = 2 × 4 × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = 40 × 10^-3 × 2 × 4 × 10^-2 = 3.2 × 10^-3 N

Ans: Force required is 3.2 × 10^-3 N

Example – 4:

A thin and light ring of the material of radius 3 cm is rested flat on the liquid surface. When slowly raised, it is found that the pull required is 0.03 N more before the film breaks than after. Find the surface tension of the liquid.

Given: Radius of the ring = l = 3 cm = 3 × 10^-2 m, Pull required = F = 0.03 N

To Find: Surface tension = T =?

Solution:

In case of ring, the liquid wets it from two sides (inside and outside),

Hence, the effective length = 2 × 2πr

We have, T = F / effective length

∴   T = F / (2 × 2πr) =    0.03 / (2 × 2 × 3.142 × 3 × 10^-2) = 0.08 N/m

Ans: The surface tension of liquid is 0.08 N/m.

Example – 5:

A horizontal circular loop of wire of radius 0.02 m is lowered into crude oil form film. The force due to the surface tension of the liquid is 0.0113 N. Calculate the surface tension of crude oil.

Given: Radius of the loop = l = 0.02 m, Force due to surface tension = F = 0.0113 N

To Find: Surface tension = T =?

Solution:

In case of loop, the liquid wets it from two sides (inside and outside),

Hence, the effective length = 2 × 2πr

We have, T = F / effective length

∴   T = F / (2 × 2πr) =    0.0113 / (2 × 2 × 3.142 × 0.02) =  0.04496 N/m

Ans: The surface tension of liquid is 0.04496 N/m.

Problem – 6:

Calculate the force required to pull away a horizontal circular loop of wire of radius 0.02 m from the surface of the water. The surface tension of water is 0.075 N/m.

Given: Radius of the ring = l = 0.02 m, Surface tension = T = 0.075 N/m.

To Find: Pull required = F =?

Solution:

In case of a ring, the liquid wets it from two sides (inside and outside),

Hence, the effective length = 2 × 2πr

We have, T = F / effective length

∴ F = T × (2 × 2πr) =    0.075 × 4 × 3.142 × 0.02 =  0.0188 N

Ans: The pull required is 0.0188 N

Example – 7:

Calculate the force required to take away a flat circular plate of radius 0.01 m from the surface of the water. The surface tension of water is 0.075 N/m.

Given: Radius of disc = r = 0.01 m, Surface tension = T = 0.075 N/m.

To Find: Force required = F =?

Solution:

In case of a circular plate, the water wets only its outer edges,

Hence, the effective length = circumference = 2 π r = 2 × 3.142 × 0.01 m

We have, T = F / effective length

∴   F = T × effective length

∴   F = 0.075 × 2 × 3.142 × 0.01 = 4.713 × 10^-3 N

Ans: Force required is 4.713 × 10^-3 N

Example – 8:

A thin wire is bent in the form of a rectangle of length 4 cm and breadth 3 cm. What force due to the surface tension does the side experience when a soap film is formed in the frame? S.T. of soap solution = 0.030 N/m.

Solution:

Given: Length of frame = l = 4 cm = 4 × 10^-2 m, Breadth of frame = b = 3 cm = 3 × 10^-2 m, Surface tension = T = 0.030 N/m

To Find: surface tension experienced = F =?

In a case of a rectangular frame, the water wets it from two sides (inside and outside),

Hence, the effective length = 2 × perimeter = 2 × 2 (l + b)

∴ The effective length = 2 × perimeter = 2 × 2 (4 + 3) = 28 cm = 28 × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = T × 2 l = 0.030 × 28 × 10^-2 = 8.4 × 10^-3 N

Ans: The surface tension experienced is 8.4 × 10^-3 N

Example – 9:

A rectangular glass plane (height 10 cm, breadth 8 cm, and thickness = 0.2 cm) rests with the largest face flat on the surface of the water. Calculate the additional force required to pull the plane clear of the water. What is the force due to surface tension acting on the plane it is held vertically with the edge of the longest side just touching the water surface? Surface Tension of water = 0.070 N/m.

Part – I

Given: Length of face touching water = l = 10 cm = 10 × 10^-2 m, Breadth of face touching water = b = 8 cm = 8 × 10^-2 m, Surface tension = T = 0.070 N/m

To Find: surface tension experienced = F =?

Solution:

In case of a rectangular plate, the water wets only its outer edges,

Hence, the effective length = perimeter = 2 (l + b)

Effective length =   2 (10 + 8) = 36 cm = 36 × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = 0.070 × 36 × 10^-2 = 0.0252 N

Ans:  Force due to surface tension is 0.0252 N

Part – II

Given: Length of face touching water = l = 10 cm = 10 × 10^-2 m, Breadth of face touching water = b = 0.2 cm = 0.2 × 10^-2 m, Surface tension = T = 0.070 N/m

To Find: surface tension experienced = F =?

Solution:

In case of a rectangular plate, the water wets only its outer edges,

Hence, the effective length = perimeter = 2 (l + b)

Effective length =   2 (10 + 0.2) = 20.4 cm = 20.4 × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = 0.070 × 20.4 × 10^-2 = 0.0143 N

∴ Force due to surface tension is0.0143 N

Example – 10:

A square glass plate 10 cm long and 1 mm in thickness is suspended vertically with the lower edge horizontal from one arm of a balance and counterpoised so that the beam is horizontal. When the glass plate is allowed to touch the surface of a soap solution, an additional mass of 0.72g has to be added to the other pan to keep the beam horizontal. Find the S.T. of the liquid. g = 9.8 m/s².

Given: Length of face touching water = l = 10 cm = 10 × 10^-2 m, Breadth of face touching water = b = 1 mm = 0.1 cm = 0.1 × 10^-2 m, Pull required = F = 0.74 gf = 0.74 × 9.8 × 10^-3 N.

To Find: surface tension = T =?

Solution:

In case of rectangular plate, the water wets only its outer edges,

Hence, the effective length = perimeter = 2 (l + b)

Effective length =   2 (10 + 0.1) = 20.2 cm = 20.2 × 10^-2 m

We have, T = F / effective length

∴   T = (0.74 × 9.8 × 10^-3) / (20.2 × 10^-2) =  0.074 N/m

Ans: The surface tension of liquid is 0.074 N/m

Example – 11:

By dipping a U-shaped wire in a soap solution, a film is formed between it and a light sliding wire resting on it. The sliding wire supports a weight of 0.01 N when its length is 20 cm. Find the surface tension of the liquid.

Given: Length of movable wire = l = 20 cm = 20 × 10^-2 m, Supporting weight = F = 0.01 N

To Find: Surface tension = T =?

Solution:

In case of wire, the liquid wets it from two sides,

Hence, the effective length = 2 l

We have, T = F / effective length

∴   T = F /2 l =    0.01 / (2 × 20 × 10^-2) = 0.025 N/m

Ans: The surface tension of liquid is 0.025 N/m.

Example – 12:

A glass tube of internal diameter 3.5 cm and thickness 0.5 cm is held vertically with its lower end immersed in water. Find the downward pull on the tube due to surface tension. The surface tension of water is 0.074 N/m.

Given: Internal diameter of the tube = 3.5 cm, Internal radius of tube = r_i = 3.5/2 = 1.75 cm  = 1.75 × 10^-2 m, thickness of tube = 0.5 cm, external radius of the tube = r_o =1.75 cm + 0.5 cm = 2.25 cm = 2.25 × 10^-2 m, Surface tension = T = 0.074 N/m.

To Find: Downward pull = F =?

Solution:

In case of a tube, the liquid wets it from two sides (inside and outside),

Hence, the effective length = (2πr_i + 2πr_o) = 2π (r_i + r_o)

∴ The effective length = 2π (1.75 + 2.25) = 2π × 4 = 8 π cm = 8 π × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = 0.074 × 8 π × 10^-2 = 0.074 × 8 × 3.142 × 10^-2 = 0.0186 N

Ans: Downward pull is 0.0186 N

Example – 13:

A ring of glass is cut from a tube 7.4 internal and 7.8 cm external diameter. This ring with its lower side horizontal is suspended from one arm of a balance so that the lower edge is just immersed in a vessel of water. It is found that an additional weight of 3.62 g must be placed in the other scale pan to compensate for the pull of surface tension on the ring. Calculate the S.T. of water. g = 9.8 m/s².

Given: Internal diameter of the tube = 7.4 cm, Internal radius of tube = r_i = 7.4/2 = 3.7 cm = 3.7 × 10^-2 m, external diameter of the tube = 7.8 cm, external radius of tube = r_o = 7.8/2 = 3.9 cm = 3.9 × 10^-2 m, Pull required = F = 3.62 gf = 3.62 × 9.8 × 10^-3 N.

To Find: Surface tension = T =?

Solution:

In case of tube, the liquid wets it from two sides (inside and outside),

Hence, the effective length = (2πr_i + 2πr_o) = 2π (r_i + r_o)

∴ The effective length = 2π (3.7 + 3.9) = 2π × 7.6 = 15.2 π cm = 15.2 π × 10^-2 m

We have, T = F / effective length

∴   T = (3.62 × 9.8 × 10^-3) / (15.2 π × 10^-2) = 0.074 N/m

Ans: The surface tension of liquid is 0.074 N/m.

Example – 14:

The surface tension of water at 0 °C is 70 dynes/cm. Find the surface tension of water at 25 °C. Given α for water = 0.0027/°C.

Given: Initial temperature θ₁ = 0 °C, Final temperature θ₂ = 25 °C, α for water = 0.0027/°C, Initial surface tension = T₁ = 70 dynes/cm

To Find: Final surface tension = T₂ =?

Solution:

T₂ = T₁ (1 – αθ) = T₁ (1 – α (θ₂ – θ₁))

∴ T₂ = 70 × (1 – 0.0027 × (25 – 0))

∴ T₂ = 70 × (1 – 0.0027 ×25)

∴ T₂ = 70 × (1 – 0.0675)

∴ T₂ = 70 × 0.9325

∴ T₂ = 65.28 dyne/cm

Ans: The surface tension of water at 25 °C is 65.28 dyne/cm

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