# Physics Questions and Answers – Fluids Mechanical Properties – Surface Tension

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Fluids Mechanical Properties – Surface Tension”.

1. Surface energy is ________________
a) kinetic energy of the surface molecules
b) the force per unit length acting on surface particles
c) the energy of the molecules inside the beaker
d) the extra energy that the molecules at the surface have relative to molecules inside the liquid

Explanation: Molecules much below the surface are attracted by equal forces in all directions, while molecules on the surface have a net force acting on them as they are exposed to air from one side and water on another. This net force acting on the surface molecules has a corresponding energy called surface energy.

2. Consider two lines on the surface of water kept in a beaker. The length of one line is twice the others. A force will be acting on the lines from either side. What is the ratio of magnitude of F1 & F2? a) 2:1
b) 1:1
c) 1:2
d) 2:3

Explanation: Surface tension, a surface phenomenon, is defined as force per unit length. It is a property of a liquid.
Let surface tension be ‘S’.
We can say S = F1/2l and S = F2/l.
Therefore F1/2l = F2/l,
which implies F1/F2 = 2/1.

3. The unit of surface tension is same as that of _________
a) surface energy per unit volume
b) force per unit area
c) surface energy per unit area
d) surface energy per unit length

Explanation: Surface tension is defined as force per unit length or surface energy per unit area.
The unit is therefore N/m.
Another unit people use is J/m2 (where J refers to joule, the unit of energy).
But obviously it is equal to Nm/m2 = N/m.
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4. A U-shaped frame consists of a sliding rod connecting its two arms. It is dipped in a soap solution and then placed horizontally. The weight of the sliding rod is 100gm. Surface tension (S) = 0.05 & length of sliding wire = 20cm. If the sliding rod starts from rest at t=0, find its velocity at t=5s. Assume that there are no resistive forces. a) 0.5 m/s
b) 3.6 km/hr
c) 1 km/hr
d) 2 m/s

Explanation: The sliding rod will be pulled by two surfaces of the soap film.
One upper layer and one bottommost layer as both are surface layers exposed to air and soap on either side.
So, the net force on the sliding rod is 2SL.
Acceleration = 2SL/m = 2*0.05*0.2 / 0.1 = 0.2m/s2.
Velocity at any time = at.
Therefore, at t = 5s, velocity = 0.2*5
= 1m/s = 3.6km/hr.

5. What will be the change in surface energy when a drop of liquid (S = 0.08N/m) is divided into 10 equal droplets? Radius of initial single drop (R)= 5cm.
a) 0.1259 J
b) 0.1141 J
c) 0.2356 J
d) 0.1765 J

Explanation: Volume of drop = 4/3πR2.
Let the radius of droplets be ‘r’. Volume of big drop will be equal to volume of 10 droplets.
10*4/3πr3 = 4/3πR3 ⇒ r = 101/3 *R.
Surface area of big drop = 4πR2 & surface area of 10 droplets = 10 * 4πr2 = 40πR2(102/3).
Surface energy of big drop = 4πR2 *S = 4π(0.0025)*0.08 = 0.0025
Surface energy of droplets = 40πR2(102/3) *S = 40π(0.0025)(102/3)*0.08 = 0.1166.
Change in surface energy = 0.1166 – 0.0025 = 0.1141 J.

6. Consider a spherical drop of liquid having radius ‘R’. Assume a diametric cross-section which divides the drop into two hemispheres. The hemispheres pull each other due to surface tension along the circumference of the diametrical cross-section. What is the value of this force?
a) 2πRS
b) πR2 S
c) 2RS
d) 0

Explanation: The diametrical cross section of a sphere will be a circle of radius R. One hemisphere will exert a force of Sdl (dl is a small element of length along periphery) on dl length.
Total length = perimeter of circle = 2&7pi;R. Therefore, total force = 2πRS.

7. What is the value of pressure inside the bubble of radius 1cm, if the bubble is in water at a depth (H) of 1m? Assume atmospheric pressure to be 1atm, surface tension = 0.075N/m & density of water = 1000kg/m3.
a) 110015 Pa
b) 109985 Pa
c) 26 atm
d) 4 atm

Explanation: The pressure in the bubble will be more than the pressure just outside it by an amount 2S/R.
The pressure just outside the bubble is P0 + ρgh.
Therefore, pressure inside bubble = P0 + ρgh + 2S/R
= 105 + (1000*10*1) + (2*0.075/0.01)
= 110015 Pa.

8. A soap bubble contains two surface layers. What is the excess pressure just inside the first layer when seen from outside the bubble?
a) 2S/R
b) 4S/R
c) -2S/R
d) -4S/R

Explanation: Let the pressure outside the bubble be P1, between the two layers be P2 and pressure inside the bubble be P3.
P2 – P1 = 2S/R
P3 – P2 = 2S/R
∴ P3 – P1 = 4S/R
This shows that excess pressure just inside the first layer is 2S/R while excess pressure completely inside the bubble is 4S/R.

9. The contact angle determines whether liquid will rise or get depressed along a solid surface. Select the correct statement regarding the same.
a) If contact angle is found to be less than 90°, the liquid will have raise along the solid surface
b) If contact angle is found to be greater than 90°, then liquid will have depressed along the solid surface
c) Liquid always rises along solid surface, irrespective of contact angle
d) Liquid always gets depressed along solid surface, irrespective of contact angle

Explanation: A liquid rises or gets depressed along a solid surface such that its surface at that place is perpendicular to the net force. Near the solid and water interface the particles of fluid are acted upon by forces from solid & liquid particles and also gravity. If the resultant force vector on those fluid particles passes through the solid surface, the surface of fluid has to rise to ensure that net force is perpendicular to its surface. When liquid rises the contact, angle becomes less than 90°. Refer to the diagram given. 10. Consider a capillary tube in which water has risen. The contact angle is θ. The radius of the capillary tube is ‘r’. The surface tension is ‘S’. And the density of water is ‘ρ’. What is the expression for the height of water risen in the tube? Assume the radius of meniscus to be ‘R’. Let the height of water risen in the tube be ‘h’.
a) 2Ssinθ / Rρg
b) 2S / Rρg
c) 2Scosθ / Rρg
d) 2S / rρg

Explanation: In the given diagram the surface tension force will be acting along the red tangent. Therefore, the force is 2πrcosθ*S. This will be balanced by the weight of water risen in the tube.
∴ πr2 hρg = 2πrcosθ*S
∴ h = 2Scosθ / rρg
= 2S / Rρg. (∵cosθ = r/R) 11. If a glass tube is of insufficient length, water rises and overflows.
a) True
b) False

Explanation: If the tube is of insufficient length then the contact angle changes and doesn’t let water overflow. The surface tension force balances the weight of water risen.
2πrcosθ*S = πr2 hρg.
Here, theta will be different from the known value of contact angle for water and glass tube.

Sanfoundry Global Education & Learning Series – Physics – Class 11.

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