Surface Tension

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   i. An imaginary sphere with a molecule at its centre and radius equal to the molecular range is called the sphere of influence of the molecule. ii. The angle of contact (θ) between a liquid and a solid surface is defined as the angle between the tangents drawn to the free surface of the liquid and surface of the solid at the point of contact, measured within the liquid.

2. 2.↑ back to the question

   Surface tension: Surface tension is defined as the tangential force acting per unit length on both sides of an imaginary line drawn on the free surface of liquid. Effect of impurities on surface tension: The effect of impurities on surface tension of liquid can be either due to soluble impurities or insoluble impurities on the surface of liquid. i. Effect of soluble impurities: When soluble substance such as common salt (i.e., sodium chloride) is dissolved in water, the surface tension of water increases. ii. Effect of partly soluble impurities: When a sparingly soluble substance such as phenol or a detergent is mixed with water, surface tension of water decreases. Example: a detergent powder is mixed with water to wash clothes. Due to this, the surface tension of water decreases and water makes good contact with the fabric and is able to remove tough stains. iii. Effect of insoluble impurities: When insoluble impurity is added into water, surface tension of water decreases. When impurity gets added to any liquid, the cohesive force of that liquid decreases which affects the meniscus. If mercury gathers dust then its surface tension is reduced. It does not form spherical droplets unless the dust is completely removed.

3. 3.↑ back to the question

   Surface tension: [From 2.4, Q2 (Definition only)] Surface tension: Surface tension is defined as the tangential force acting per unit length on both sides of an imaginary line drawn on the free surface of liquid. Effect of impurities on surface tension: The effect of impurities on surface tension of liquid can be either due to soluble impurities or insoluble impurities on the surface of liquid. i. Effect of soluble impurities: When soluble substance such as common salt (i.e., sodium chloride) is dissolved in water, the surface tension of water increases. ii. Effect of partly soluble impurities: When a sparingly soluble substance such as phenol or a detergent is mixed with water, surface tension of water decreases. Example: a detergent powder is mixed with water to wash clothes. Due to this, the surface tension of water decreases and water makes good contact with the fabric and is able to remove tough stains. iii. Effect of insoluble impurities: When insoluble impurity is added into water, surface tension of water decreases. When impurity gets added to any liquid, the cohesive force of that liquid decreases which affects the meniscus. If mercury gathers dust then its surface tension is reduced. It does not form spherical droplets unless the dust is completely removed. i. Unit: N/m or J/m$^2$ in SI system. ii. Dimensions: $[L^0M^1T^{-2}]$

4. 4.↑ back to the question

   Explanation of capillary action: i. Suppose a capillary tube is dipped into water. ii. Consider the situation before the movement of water inside the capillary. The shape of the surface of water in the capillary is concave. iii. Let us consider four points as shown in the figure. Point A is just above the concave surface inside the capillary. B is just below the concave surface inside the capillary. C is just above the plane surface outside the capillary. D is just below the plane surface outside the capillary and is at the same horizontal level as that of B. iv. Let $P_A$, $P_B$, $P_C$ and $P_D$ be the pressures at points A, B, C and D respectively. v. Since pressure on concave side of liquid surface is greater than that on the convex side. $\therefore P_A>P_B$ vi. As the pressure is same on both sides of a plane surface, $\therefore P_C=P_D$ vii. The points A and C are at the same level, the pressure at both these points is the same. $P_A=P_C=P_D=$ atmospheric pressure $\therefore P_D>P_B$ viii. But points B and D are at same horizontal level in the liquid. Therefore, in order to maintain the same pressure, the liquid out of the capillary flows into the capillary tube and rises above the point B, till the pressure at B becomes the same as that at D. Because of this, there is a rise in the level of liquid inside the capillary tube.

5. 5.↑ back to the question

   The question is a real board question. Its answer and working are ours — this paper was published without an official answer key.

   The diagram to draw

   Draw a glass capillary tube of inner radius $r$ held vertically with its lower end dipped in water in a beaker, and label these parts:

   • The water inside the tube standing at a height $h$ above the free surface of the water in the beaker.
   • The concave meniscus inside the tube, with the angle of contact $\theta$ marked between the tangent to the meniscus and the inner wall of the tube, measured within the liquid.
   • At the point of contact of the meniscus with the wall, the force of surface tension per unit length, $T$, drawn along the tangent to the meniscus — outward and upward, making the angle $\theta$ with the wall.
   • Its two rectangular components at that point, drawn as dotted arrows: the vertical component $T\cos\theta$, along the wall, directed upwards, and the horizontal component $T\sin\theta$, perpendicular to the wall.
   • The same pair of components drawn at the diametrically opposite point of contact, to show that the two horizontal components are equal and opposite.
   • The weight of the raised liquid column, $\pi r^{2}h\rho g$, marked as a downward arrow at the centre of the column.

   What the diagram shows

   • The liquid is in contact with the tube along the circumference $2\pi r$, so the total force due to surface tension is $f_{T} = T\times 2\pi r$, acting along the tangent to the meniscus.
   • The horizontal components $T\sin\theta$ at diametrically opposite points are equal in magnitude and opposite in direction, so they cancel all round the circle and play no part in the rise.
   • Only the vertical components are effective. Their resultant is

   $$(f_{T})_{v} = 2\pi r\,T\cos\theta \quad \text{(upwards)}$$

   • This supports the weight of the liquid column of height $h$, which is $w = \pi r^{2}h\rho g$ (downwards). At equilibrium,

   $$2\pi r\,T\cos\theta = \pi r^{2}h\rho g$$

   $$h = \dfrac{2T\cos\theta}{r\rho g}$$

6. 6.↑ back to the question

   The question is a real board question. Its answer and working are ours — this paper was published without an official answer key.

   The diagram to draw

   Draw the vertical wall of a solid container on the left and the liquid to its right, with the liquid surface curving upward to meet the wall — a concave meniscus. Mark a molecule A on the liquid surface, close to the wall, and show these labelled arrows and lines starting at A:

   • $AP$ — the net adhesive force on A, drawn horizontal (pointing from the liquid into the solid wall). It is horizontal because the wall is vertical.
   • $AC$ — the net cohesive force on A, drawn at nearly $45^\circ$ to either surface, directed into the body of the liquid.
   • $AR$ — the resultant of $AP$ and $AC$, completed by the parallelogram law. Since the adhesive force is the larger here, $AR$ is directed inside the solid.
   • $AT$ — the tangent to the liquid surface at A, drawn perpendicular to $AR$.
   • $\theta$ — the angle of contact, marked between the tangent $AT$ and the solid surface, measured within the liquid. It is acute.

   What the diagram shows

   • For the liquid surface to be stable, the resultant force $AR$ on a surface molecule must be normal to the liquid surface at every point.
   • For $AR$ to be normal to the surface, the liquid near the wall piles up against the solid boundary so that the tangent $AT$ is perpendicular to $AR$. The meniscus is therefore concave.
   • The angle of contact between $AT$ and the wall, measured inside the liquid, comes out acute, and such a liquid wets the solid. Kerosene or water in a glass bottle is the standard example.
   • The condition on the magnitudes for an acute angle of contact is

   $$AP > \dfrac{AC}{\sqrt{2}}$$

   since $AP = \dfrac{AC}{\sqrt{2}}$ would give exactly $\theta = 90^\circ$.

7. 7.↑ back to the question

   i. Surface tension: [From 2.4, Q2 (Definition only)] Surface tension: Surface tension is defined as the tangential force acting per unit length on both sides of an imaginary line drawn on the free surface of liquid. Effect of impurities on surface tension: The effect of impurities on surface tension of liquid can be either due to soluble impurities or insoluble impurities on the surface of liquid. i. Effect of soluble impurities: When soluble substance such as common salt (i.e., sodium chloride) is dissolved in water, the surface tension of water increases. ii. Effect of partly soluble impurities: When a sparingly soluble substance such as phenol or a detergent is mixed with water, surface tension of water decreases. Example: a detergent powder is mixed with water to wash clothes. Due to this, the surface tension of water decreases and water makes good contact with the fabric and is able to remove tough stains. iii. Effect of insoluble impurities: When insoluble impurity is added into water, surface tension of water decreases. When impurity gets added to any liquid, the cohesive force of that liquid decreases which affects the meniscus. If mercury gathers dust then its surface tension is reduced. It does not form spherical droplets unless the dust is completely removed. ii. Surface energy: The extra energy of the molecules on the surface layer of a liquid is called surface energy of the liquid.

8. 8.↑ back to the question

   Expression for excess pressure inside a drop: i. Free surface of drops or bubbles are spherical in shape. Let, $P_i$ = inside pressure of a drop or air bubble, $P_o$ = outside pressure of bubble, r = radius of drop or bubble. ii. As drop is spherical, $P_i>P_o$ $\therefore$ excess pressure inside drop $=P_i-P_o$ iii. Let the radius of drop increases from r to $r+\Delta r$ so that inside pressure remains constant. iv. Initial area of drop $A_1=4\pi r^2$, Final surface area of drop $A_2=4\pi(r+\Delta r)^2$ Increase in surface area, $\Delta A=A_2-A_1=4\pi[(r+\Delta r)^2-r^2]=4\pi[r^2+2r\Delta r+\Delta r^2-r^2]=8\pi r\Delta r+4\pi\Delta r^2$ v. As Δr is very small, the term containing $\Delta r^2$ can be neglected. $\therefore dA=8\pi r\Delta r$ vi. Work done by force of surface tension, $dW=TdA=(8\pi r\Delta r)T\qquad....(1)$ This work done is also equal to the product of the force F which causes increase in the area of the bubble and the displacement Δr which is the increase in the radius of the bubble. $\therefore dW=F\Delta r$ The excess force is given by, (Excess pressure) × (Surface area) $\therefore F=(P_i-P_o)\times4\pi r^2$ $\therefore dF=(P_i-P_o)A$ $dW=F\Delta r=(P_i-P_o)A\Delta r$ From equation (1), $(P_i-P_o)A\Delta r=(8\pi r\Delta r)T$ $$\therefore P_i-P_o=\dfrac{8\pi r\Delta rT}{4\pi r^2\Delta r}\qquad(\because A=4\pi r^2)$$ $$\therefore P_i-P_o=\dfrac{2T}{r}\qquad....(2)$$ Equation (2) represents excess pressure inside a drop or air bubble. It is also called Laplace's law of spherical membrane. vii. In case of soap bubble, there are two free surfaces in contact with air. $\therefore$ For a bubble, Total increase in the surface area of a soap bubble, while increasing its radius by Δr, is $2(8\pi r\Delta r)$ The work done by this excess pressure is $dW=(P_i-P_o)4\pi r^2\Delta r=T(16\pi r\Delta r)$ $$\therefore \text{Excess pressure, }P_i-P_o=\dfrac{4T}{r}$$

9. 9.↑ back to the question

   Expression for capillary rise: i. When glass capillary tube is dipped into a liquid, then the liquid rises in the capillary against gravity. Hence, the weight of the liquid column must be equal and opposite to the component of force due to surface tension at the point of contact. ii. The length of liquid in contact inside the capillary is the circumference 2πr. Let, r = radius of capillary tube, h = height of liquid level in the tube, T = surface tension of liquid, ρ = density of liquid, g = acceleration due to gravity iii. The force of magnitude $f_T$ acts tangentially on unit length of liquid surface which is in contact with wall of capillary tube and is given as, $f_T=T\times2\pi r$ This force can be resolved into two components: a. $f_T\cos\theta$ - vertically upward and b. $f_T\sin\theta$ - along horizontal iv. Vertical component is effective. Horizontal component is not responsible for capillary rise. v. Vertical component of force acting on liquid column $(f_T)_v=$ force per unit length × circumference $=T\cos\theta\times2\pi r$ vi. Upward force balances weight of liquid in the capillary. $W=mg=V\rho g=\pi r^2h\rho g$ where, V = volume of liquid rise in the tube (ignoring the liquid in the concave meniscus.) m = mass of the liquid in the capillary rise. vii. This must be equal and opposite to the vertical component of the force due to surface tension. If liquid in meniscus is neglected, then for equilibrium, $2\pi rT\cos\theta=\pi r^2h\rho g$ $$\therefore h=\dfrac{2T\cos\theta}{r\rho g}\qquad....(1)$$ This is the required expression for rise of liquid in capillary tube

10. 10.↑ back to the question

    Molecular theory of surface tension: i. Let PQRS = Surface film of liquid in a container containing liquid. PS is the free surface of liquid and QR is the inner layer parallel to PS at distance equal to the range of molecular force. ii. Now consider three molecules A, B and C in a liquid in a vessel such that the molecule A is well inside the liquid, the molecule B within surface film and molecule C is on the surface of liquid as shown in the figure. iii. The sphere of influence of the molecule A is entirely inside the liquid. As a result, molecule A is acted upon by equal cohesive forces in all directions. Thus, the net cohesive force acting on molecule A is zero. iv. For molecule B, a large part of its sphere of influence is inside the liquid and a smaller part is outside the surface (in air). The adhesive force acting on molecule B due to air molecules above it and within its sphere of influence is weak compared to the strong downward cohesive force acting on the molecule. As a result, the molecule B gets attracted inside the liquid. v. For molecule C, half of the sphere of influence is in air and half is in liquid. As density of air is much less than that of liquid, number of air molecules within the sphere of influence of the molecule C above the free surface of the liquid is much less than the numbers of liquid molecules within the sphere of influence that lies within the liquid. Thus, the adhesive force due to the air molecules acting on molecule C is weak compared to the cohesive force acting on the molecule. As a result, the molecule C also gets attracted inside the liquid. vi. Thus, all molecules in the surface film are acted upon by an unbalanced net cohesive force directed into the liquid. Therefore, the molecules in the surface film are pulled inside the liquid. This minimizes the total number of molecules in the surface film. As a result, the surface film remains under tension. The surface film of a liquid behaves like a stretched elastic membrane. This tension is known as surface tension and the force due to it acts tangential to the free surface of a liquid.

11. 11.↑ back to the question

    [From 2.4, Q1 (ii)]

    i. An imaginary sphere with a molecule at its centre and radius equal to the molecular range is called the sphere of influence of the molecule. ii. The angle of contact (θ) between a liquid and a solid surface is defined as the angle between the tangents drawn to the free surface of the liquid and surface of the solid at the point of contact, measured within the liquid. Properties of angle of contact: i. The angle of contact is constant for a given liquid-solid pair. ii. The value of angle of contact depends upon nature of liquid and solid in contact. iii. It depends upon the medium which exists above the free liquid surface. iv. The angle of contact changes due to impurity or temperature. v. A small contamination of the surface causes a large change in the angle of contact. [Any two properties]

12. 12.↑ back to the question

    Capillary action or capillarity: The phenomenon of rise or fall of liquid level inside a capillary tube when it is dipped in the liquid is called capillary action or capillarity. Uses of capillarity: i. Oil rises up the wick of a lamp. ii. Cloth rag sucks water. iii. Water rises up the crevices in rocks. iv. Sap and water rise up to the top most leaves in a tree. v. Blotting paper absorbs ink. [Any two uses]

13. 13.↑ back to the question

    Relation between surface tension and surface energy per unit area: i. Let ABCD be a rectangular frame of wire, fitted with a movable arm PQ. ii. The frame held in horizontal position is dipped into soap solution and taken out so that a soap film APQB is formed. Due to surface tension of soap solution, a force 'F' will act on each arm of the frame. Under the action of this force, the movable arm PQ moves towards AB. iii. Magnitude of force due to surface tension is, $F=2Tl\qquad....[\because T=F/l]$ (A factor of 2 appears because soap film has two surfaces which are in contact with wire.) iv. Let the wire PQ be pulled outwards through a small distance 'dx' to the position P'Q', by applying an external force F' isothermally, which is equal and opposite to F. Work done by this force, $dW=F'dx=2Tldx$. v. But, $2ldx=dA=$ increase in area of two surfaces of film. $\therefore dW=T\,dA$ vi. This work done in stretching the film is stored in the area dA in the form of potential energy (surface energy). $\therefore$ Surface energy, $E=T\,dA$ $\therefore \dfrac{E}{dA}=T$ Hence, surface tension = surface energy per unit area. vii. Thus, surface tension is equal to the mechanical work done per unit surface area of the liquid, which is also called as surface energy.

14. 14.↑ back to the question

    When detergent powder is mixed with water to wash clothes, the surface tension of water decreases and water makes good contact with the fabric and is able to remove tough stains.

15. 15.↑ back to the question

    [From 2.4, Q7 (Definition)]

    i. Surface tension: [From 2.4, Q2 (Definition only)] Surface tension: Surface tension is defined as the tangential force acting per unit length on both sides of an imaginary line drawn on the free surface of liquid. Effect of impurities on surface tension: The effect of impurities on surface tension of liquid can be either due to soluble impurities or insoluble impurities on the surface of liquid. i. Effect of soluble impurities: When soluble substance such as common salt (i.e., sodium chloride) is dissolved in water, the surface tension of water increases. ii. Effect of partly soluble impurities: When a sparingly soluble substance such as phenol or a detergent is mixed with water, surface tension of water decreases. Example: a detergent powder is mixed with water to wash clothes. Due to this, the surface tension of water decreases and water makes good contact with the fabric and is able to remove tough stains. iii. Effect of insoluble impurities: When insoluble impurity is added into water, surface tension of water decreases. When impurity gets added to any liquid, the cohesive force of that liquid decreases which affects the meniscus. If mercury gathers dust then its surface tension is reduced. It does not form spherical droplets unless the dust is completely removed. ii. Surface energy: The extra energy of the molecules on the surface layer of a liquid is called surface energy of the liquid.. [From 2.4, Q13] Relation between surface tension and surface energy per unit area: i. Let ABCD be a rectangular frame of wire, fitted with a movable arm PQ. ii. The frame held in horizontal position is dipped into soap solution and taken out so that a soap film APQB is formed. Due to surface tension of soap solution, a force 'F' will act on each arm of the frame. Under the action of this force, the movable arm PQ moves towards AB. iii. Magnitude of force due to surface tension is, $F=2Tl\qquad....[\because T=F/l]$ (A factor of 2 appears because soap film has two surfaces which are in contact with wire.) iv. Let the wire PQ be pulled outwards through a small distance 'dx' to the position P'Q', by applying an external force F' isothermally, which is equal and opposite to F. Work done by this force, $dW=F'dx=2Tldx$. v. But, $2ldx=dA=$ increase in area of two surfaces of film. $\therefore dW=T\,dA$ vi. This work done in stretching the film is stored in the area dA in the form of potential energy (surface energy). $\therefore$ Surface energy, $E=T\,dA$ $\therefore \dfrac{E}{dA}=T$ Hence, surface tension = surface energy per unit area. vii. Thus, surface tension is equal to the mechanical work done per unit surface area of the liquid, which is also called as surface energy.

16. 16.↑ back to the question

    Given: R = 2 mm $=2\times10^{-3}$ m, n = $10^6$, T = 0.5 N/m To find: Work done (W) Formula: $W=T\Delta A$ Calculation: Volume of a single drop $=\dfrac{4}{3}\pi R^3$ and Volume of a single droplet $=\dfrac{4}{3}\pi r^3$ $\therefore$ We have, $\dfrac{4}{3}\pi R^3=n\times\dfrac{4}{3}\pi r^3$ or $R^3=nr^3$ $\therefore r=\dfrac{R}{\sqrt[3]{n}}=\dfrac{2\times10^{-3}}{\sqrt[3]{10^6}}=2\times10^{-5}$ m From formula, $W=T(n\times4\pi r^2-4\pi R^2)=4\pi T(nr^2-R^2)$ $=4\times3.142\times0.5\times[10^6\times(2\times10^{-5})^2-(2\times10^{-3})^2]$ $=6.284\times[(20\times10^{-3})^2-(2\times10^{-3})^2]=6.284\times(400-4)\times10^{-6}=6.284\times396\times10^{-6}$ $\therefore W=2488.46\times10^{-6}=2.488\times10^{-3}$ J Ans: The work done when a spherical drop of mercury falls from some height and breaks into a million droplets is $2.488\times10^{-3}$ J.

17. 17.↑ back to the question

    Given: r = 12 cm, T = 30 dyne/cm To find: Work (W) Formula: $W=T\times\Delta A$ Calculation: Initial surface area of soap bubble = 0 Final surface area $A=2\times4\pi r^2$ $\therefore$ Increase in surface area $\Delta A=2\times4\pi r^2$ From formula, $W=T\times8\pi r^2=30\times8\times3.14\times(12)^2=108518.4$ erg $=108518.4\times10^{-7}$ J $\therefore W=1.085\times10^{-2}$ J Ans: The work done in blowing the soap bubble is $1.085\times10^{-2}$ J.

18. 18.↑ back to the question

    Given: d = 0.25 mm $\therefore r=\dfrac{d}{2}=\dfrac{0.25}{2}=0.125$ mm $=0.125\times10^{-3}$ m T = 0.0245 N/m, h = 4 cm = 0.04 m, θ = 28°, g = 9.8 m/s$^2$ To find: Density (ρ) Formula: $T=\dfrac{r h\rho g}{2\cos\theta}$ Calculation: From formula, $\rho=\dfrac{2T\cos\theta}{rhg}$ $$\therefore \rho=\dfrac{2\times0.0245\times\cos28°}{0.125\times10^{-3}\times0.04\times9.8}=\dfrac{2\times0.0245\times0.8829}{0.125\times10^{-3}\times0.04\times9.8}$$ ...(using log table, $\cos28°=0.8829$) $$=\dfrac{2\times245\times10^{-4}\times8829\times10^{-4}}{125\times10^{-3}\times10^{-3}\times4\times10^{-2}\times9.8}=\dfrac{490\times8829\times10^{-8}}{125\times4\times9.8\times10^{-8}}=\dfrac{490\times8829}{125\times4\times9.8}$$ = antilog {log(490)+log(8829) − [log(125)+log(4)+log(9.8)]} = antilog {2.6902+3.9459 − (2.0969+0.6021+0.9912)} = antilog{2.9459} $\therefore \rho=882.8$ kg/m$^3$ Ans: The density of paraffin oil is 882.8 kg/m$^3$.

19. 19.↑ back to the question

    Given: $h_1$ = 3.2 cm, $A_2=\dfrac{A_1}{2}$ To find: Height to which water will rise ($h_2$) Formula: $h\propto\dfrac{1}{r}$ Calculation: From formula, $h_1r_1=h_2r_2$ $\therefore \pi r_2^2=\dfrac{\pi r_1^2}{2}$ $\Rightarrow r_2=\dfrac{r_1}{\sqrt2}$ $\therefore h_2=\dfrac{3.2\times r_1}{r_1/\sqrt2}=3.2\times\sqrt2=4.525$ cm Ans: Water will rise to a height of 4.525 cm.

20. 20.↑ back to the question

    Given: T = 0.072 N/m, d = 4 mm $\therefore r=2\times10^{-3}$ m, $P_o=1.013\times10^5$ N/m$^2$ To find: Pressure inside the raindrop ($P_i$) Formula: $P_i=P_o+\dfrac{2T}{r}$ Calculation: From formula, $P_i=1.013\times10^5+\dfrac{2\times0.072}{2\times10^{-3}}$ $\therefore P_i=1.01372\times10^5$ Pa Ans: The pressure inside the raindrop is $1.01372\times10^5$ Pa.

21. 21.↑ back to the question

    Given: $E=5\pi T$ To find: Diameter of the drop (d) in C.G.S. system Formula: $E=T\Delta A$ Calculation: $\Delta A=4\pi r^2$ From formula, $E=4\pi r^2T$ $\therefore 5\pi T=4\pi r^2T$ $\therefore 4r^2=5$ $\therefore r=\sqrt{\dfrac{5}{4}}=\dfrac{\sqrt5}{2}$ $\therefore d=2r=\sqrt5=2.236$ cm Ans: The diameter of the drop is 2.236 cm.

22. 22.↑ back to the question

    Given: $E=\pi\sqrt2\,T$ To find: Diameter of drop (d) Formula: $E=T\Delta A$ Calculation: $\Delta A=4\pi r^2$ From formula, $E=4\pi r^2T$ $\therefore \pi\sqrt2\,T=4\pi r^2T$ $\therefore 4r^2=\sqrt2$ $\therefore r^2=\dfrac{\sqrt2}{4}=\dfrac{1.414}{4}$ $\therefore r^2=0.3535$ $\therefore r=\sqrt{0.3535}$ $r=0.5946$ m $=2r=2(0.5946)$ $\therefore d=1.1892$ m Ans: The diameter of the drop is 1.1892 m.

23. 23.↑ back to the question

    Given: $E=2\pi T$ To find: Diameter of drop (d) Formula: $E=T\Delta A$ Calculation: From formula, $\Delta A=\dfrac{E}{T}$ $\therefore \Delta A=\dfrac{2\pi T}{T}$ $\therefore \Delta A=2\pi$ We know, $\Delta A=4\pi r^2$ $\therefore 2\pi=4\pi r^2$ $\therefore 4r^2=2$ $\therefore r^2=\dfrac{2}{4}=\dfrac{1}{2}$ $\therefore r=\dfrac{1}{\sqrt2}$ m $d=2r=2\times\dfrac{1}{\sqrt2}=\sqrt2$ $\therefore d=1.414$ m Ans: The diameter of the drop is 1.414 m.

24. 24.↑ back to the question

    Given: $r_1:r_2=4:3$ To find: $W_1:W_2$ Formulae: $W=2TdA$ Calculation: From formula, Work done to blow both bubbles, $W_1=2T(4\pi r_1^2)$ and $W_2=2T(4\pi r_2^2)$ $\therefore \dfrac{W_1}{W_2}=\left(\dfrac{r_1}{r_2}\right)^2=\left(\dfrac{4}{3}\right)^2=16:9$ Ans: The ratio of work done to blow the bubbles is 16 : 9.

25. 25.↑ back to the question

    Given: $r_1$ = 1 cm, $r_2$ = 2 cm, T = 30 dyne/cm To find: Work (W) Formula: $W=T\Delta A$ Calculation: $A_1=4\pi r_1^2=4\pi\times1^2=4\pi$ cm$^2$ $A_2=4\pi r_2^2=4\pi\times2^2=16\pi$ cm$^2$ $\Delta A=A_2-A_1=(16\pi-4\pi)=12\pi$ cm$^2$ Since soap bubble has two surfaces From formula, $W=2T\times\Delta A=2\times30\times12\pi=2\times30\times12\times3.14$ $\therefore W=2.26\times10^3$ erg Ans: The work done in increasing the radius of the soap bubble is $2.26\times10^3$ erg.

26. 26.↑ back to the question

    Given: r = 0.02 m, F = 0.0113 N To find: Surface tension (T) Formula: $F=Tl$ Calculation: As there are two sides of a circular loop acted upon by the force, $l=2\times2\pi r$ From formula, $F=4\pi rT$ $\therefore T=\dfrac{F}{4\pi r}$ $\therefore T=\dfrac{0.0113}{4\times3.142\times0.02}=\dfrac{113\times10^{-4}}{4\times3.142\times2\times10^{-2}}=\dfrac{113\times10^{-2}}{4\times3.142\times2}$ = antilog {log(113) − [log(4)+log(3.142)+log(2)]}$\times10^{-2}$ = antilog{2.0531 − (0.6021+0.4972+0.3010)}$\times10^{-2}$ = antilog{0.6528}$\times10^{-2}=4.495\times10^{-2}$ $\therefore T=0.04495$ N/m Ans: The surface tension of crude oil is 0.04495 N/m.

27. 27.↑ back to the question

    Given: l = 4 cm = 0.04 m, $F=3.2\times10^{-3}$ N To find: Surface tension (T) Formula: $F=T\times$ length Calculation: In case of rod, liquid wets it from two sides. $\therefore$ Force acting on the rod due to surface tension, $F=T\times(2l)$ $\therefore$ From formula, $T=\dfrac{F}{2l}=\dfrac{3.2\times10^{-3}}{2\times0.04}=0.04$ N/m Ans: The surface tension of the liquid is 0.04 N/m.

28. 28.↑ back to the question

    Given: $T=30$ dyne/cm $=3\times10^{-2}$ N/m, Initial radius of bubble = 0 m, Final radius of bubble, r = 0.1 m To find: Work down (W) Formula: Work done $=T\times dA$ Calculation: Initial surface area of bubble = 0 A soap bubble has two surfaces, outer surface and inner surface. $\therefore$ Final surface area of soap bubble is, $A=2\times(4\pi r^2)=8\pi r^2$ $\therefore$ change in area $=dA=A-0=8\pi r^2=8\times3.142\times0.1^2=0.2514$ m$^2$ From formula, $W=3\times10^{-2}\times0.2514=7.542\times10^{-3}$ J Ans: Work done in blowing a soap bubble to a radius of 0.1 cm is $7.542\times10^{-3}$ J.

29. 29.↑ back to the question

    Given: r = 0.2 mm $=2\times10^{-4}$ m, n = 8 To find: Decrease in surface area (E) Formula: $dA=4\pi nr^2-4\pi R^2$ Calculation: Volume of a single drop $=\dfrac{4}{3}\pi R^3$ and Volume of a single droplet $=\dfrac{4}{3}\pi r^3$ $\therefore$ We have, $\dfrac{4}{3}\pi R^3=n\times\dfrac{4}{3}\pi r^3$ or $R^3=nr^3$ $\therefore R=\sqrt[3]{n}\,r=\sqrt[3]{8}\times2\times10^{-4}=2\times2\times10^{-4}=4\times10^{-4}$ m From formula, $dA=4\pi(nr^2-R^2)=4\times3.142\times[8\times(2\times10^{-4})^2-(4\times10^{-4})^2]$ $=12.568\times(32-16)\times10^{-8}=201.088\times10^{-8}=2.011\times10^{-6}$ m$^2$ Ans: Decrease in the surface area is $2.011\times10^{-6}$ m$^2$.

30. 30.↑ back to the question

    Given: r = 2 mm $=2\times10^{-3}$ m, $T_w=7.3\times10^{-2}$ N/m To find: Difference of pressure (ΔP) Formula: $\Delta P=\dfrac{2T}{r}$ Calculation: From formula, $\Delta P=\dfrac{2\times7.3\times10^{-2}}{2\times10^{-3}}=73$ N/m$^2$ Ans: The difference of pressure between inside and outside of a spherical water drop is 73 N/m$^2$.

31. 31.↑ back to the question

    Given: $T=2.5\times10^{-2}$ N/m, Initial radius of bubble = 0 cm, Final radius of bubble, r = 1 cm = 0.01 m To find: Work down (W) Formula: Work done $=T\times dA$ Calculation: Initial surface area of bubble = 0 A soap bubble has two surfaces, outer surface and inner surface. $\therefore$ Final surface area of soap bubble is, $A=2\times(4\pi r^2)=8\pi r^2$ $\therefore dA=A-0=8\pi r^2=8\times3.142\times0.01^2=0.00251$ m$^2=0.251\times10^{-2}$ m$^2$ From formula, $W=2.5\times10^{-2}\times0.251\times10^{-2}=6.275\times10^{-5}$ J Ans: Work done in blowing a soap bubble to a radius of 1 cm is $6.275\times10^{-5}$ J.

32. 32.↑ back to the question

    Given: $P_i-P_0=80$ N/m$^2$, $T=7.2\times10^{-2}$ N/m To find: Diameter of the water drop (D). Formula: $(P_i-P_0)=\dfrac{2T}{r}$ Calculation: From formula, $r=\dfrac{2T}{P_i-P_0}=\dfrac{2\times7.2\times10^{-2}}{80}=1.8\times10^{-3}$ m $\therefore D=2r=3.6$ mm Ans: Diameter of the water drop so that excess pressure inside it is 80 N/m$^2$ is 3.6 mm.