Q: Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm3 of oleic per cm3 of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm2 by considering 100 spherical drops of radius (40π3)31×10−3 cm . Then the thickness of oleic acid layer will be x×10−14 m . Where x is..............

4tT=100×34πr3=100×=100×34π×40π3×10−9=10−8 cm3tT=25×10−10 cm=25×10−12 mt0=0.01 tT=25×10−14 mx=25

Q: When two soap bubbles of radii a and b(b > a) coalesce, the radius of curvature of common surface is :

aba4T=R4T+b4T;R1=a1−b1;R=b−aba

Q: How much work will be done in increasing the diameter of a soap bubble from 2 cm to 5 cm ? Surface tension of soap solution is 3.0×10−2 N/m.

Soap bubble has two surfaces. Hence,W=TΔA(T=ΔAΔW)Here,ΔA=2[4π{(2.5×10−2)2−(1.0×10−2)2}]=1.32×10−2 m2W=(3.0×10−2)(1.32×10−2)J=3.96×10−4 J

Q: Two separate air bubbles (radii 0.004 m and 0.002 m ) formed of the same liquid (surface tension 0.07 N/m ) come together to form a double bubble. Find the radius and the sense of curvature of the internal film surface common to both the bubbles.

p1=p0+r14T⇒p2=p0+r24T∴p2>p1p=p2−p1=4T(r1r2r1−r2) ...(i)This excess pressure acts from concave to convex side, the interface will be concave towards smaller bubble and convex towards larger bubble. Let R be the radius of interface then,p=R4T ...(ii)From Eqs. (i) and (ii), we getR=r1−r2r1r2=(0.004−0.002)(0.004)(0.002)=0.004 m

Q: When liquid medicine of density ρ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop.We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r≪R ) is

Vertical force due to surface tension,Fv=Fsinθ=(T)(2πr)(r/R)=R2πr2T∴ Correct option is (c).

Question 1

At critical temperature the surface tension of liquid:

Accepted Answer

At critical temperature, surface tension of liquid is zero as there is no interface present.

Question 2

Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm3 of oleic per cm3 of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm2 by considering 100 spherical drops of radius (40π3​)31​×10−3 cm . Then the thickness of oleic acid layer will be x×10−14 m . Where x is..............

Accepted Answer

​4tT​=100×34​πr3=100×=100×34π​×40π3​×10−9=10−8 cm3tT​=25×10−10 cm=25×10−12 mt0​=0.01 tT​=25×10−14 mx=25​

Question 3

When two soap bubbles of radii a and b(b > a) coalesce, the radius of curvature of common surface is :

Accepted Answer

aba4T​=R4T​+b4T​;R1​=a1​−b1​;R=b−aba​​

Question 4

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible?The surface tension is T, density of liquid is ρ and L is its latent heat of vaporisation.(a) ρL/T(b) T/ρL​(c) T/ρL(d) 2T/ρL

Accepted Answer

Exp. (d)If a layer of thickness dr is evaporated, then Change in surface energy= Change in surface area xT=[4π(r+dr)2−4πr2]T=[4πr2+4πdr2+8πrdr−4πr2]As dr is very small, so dr2 neglected.So, change in surface energy =8πrdrTEnergy required to evaporate layer of thicknessdr=(4πr2dr)ρLThe process of evaporation just starts, if change in surface energy is just sufficient to evaporate the water layer.∴(4πr2dr)ρL=(8πrdr)T⇒ r=ρL2T​

Question 5

A soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume, with time, at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by

Accepted Answer

Exp. (b)When soap bubble is being inflated and its temperature remains constant, then it follows Boyle's law, sopV=constant(k)⇒p=Vk​Differentiating above equation with time, we getdtdp​=k⋅dtd​(V1​)⇒dtdp​=k(V2−1​)⋅dtdV​It is given that, dtdV​=c (a constant)So, dtdp​=V2−kc​........(i)Now, from dtdV​=c, we getdV=cdtor ∫dV=∫cdt or V=ct...................(ii)From Eqs. (i) and (ii), we getdtdp​=c2t2−kc​or dtdp​=−(ck​)t−2⇒dp=−ck​⋅t−2dtIntegrating both sides, we get∫dp=−ck​∫t−2dtp=−ck​⋅(−2+1t−2+1​)=−ck​⋅t−1​=ctk​or p∝t1​Hence, p versus t1​ graph is a straight line, which is correctly represented in option (b).

Question 6

On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r≪R andthe surface tension of water is T, value of r just before bubbles detatch is (density of water is ρ )(a) R23Tρw​g​​(b) R26Tρw​g​​(c) R2Tρw​g​​(d) R2T3ρw​g​​

Accepted Answer

Exp. (*)The bubble will detach, if Buoyant force ≥ Surface tension forcei.e., 34​πR3ρw​g≥∫T×dlsinθρw​)(34​πR3)g≥(T)(2πr)sinθsinθ=Rr​Solving, r=3T2ρw​R4g​​=R23T2ρw​g​​ No option matches with the correct answer.

Question 7

How much work will be done in increasing the diameter of a soap bubble from 2 cm to 5 cm ? Surface tension of soap solution is 3.0×10−2 N/m.

Accepted Answer

Soap bubble has two surfaces. Hence,W=TΔA(T=ΔAΔW​)Here,ΔA=2[4π{(2.5×10−2)2−(1.0×10−2)2}]=1.32×10−2 m2W=(3.0×10−2)(1.32×10−2)J=3.96×10−4 J

Question 8

Two separate air bubbles (radii 0.004 m and 0.002 m ) formed of the same liquid (surface tension 0.07 N/m ) come together to form a double bubble. Find the radius and the sense of curvature of the internal film surface common to both the bubbles.

Accepted Answer

p1​=p0​+r1​4T​⇒p2​=p0​+r2​4T​∴p2​>p1​p=p2​−p1​=4T(r1​r2​r1​−r2​​) ...(i)This excess pressure acts from concave to convex side, the interface will be concave towards smaller bubble and convex towards larger bubble. Let R be the radius of interface then,p=R4T​ ...(ii)From Eqs. (i) and (ii), we getR=r1​−r2​r1​r2​​=(0.004−0.002)(0.004)(0.002)​=0.004 m

Question 9

When liquid medicine of density ρ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop.We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r≪R ) is

Accepted Answer

Vertical force due to surface tension,Fv​=Fsinθ=(T)(2πr)(r/R)=R2πr2T​∴ Correct option is (c).

Question 10

Explain:Small mercury droplets are spherical and larger ones tend to flatten.

Accepted Answer

Reason Small mercury droplets are spherical because the forces of surface tension tend to reduce their area to a minimum value and a sphere has minimum surface area for a given volume.Larger drops of mercury are flattened due to the large gravitational force acting on them. Here the shape is such that the sum of the gravitational potential energy and the surface potential energy must be minimum. Hence the centre of gravity moves down as low as possible. This explains flattening of the larger drops.

Question 11

Give reason:

Accepted Answer

Reason: This is because the water films formed on them tend to contract to minimum area.

Question 12

Small mercury droplets are spherical and larger ones tend to flatten.

Accepted Answer

Reason Small mercury droplets are spherical because the forces of surface tension tend to reduce their area to a minimum value and a sphere has minimum surface area for aFig. 16.68 given volume.Larger drops of mercury are flattened due to the large gravitational force acting on them. Here the shape is such that the sum of the gravitational potential energy and the surface potential energy must be minimum. Hence the centre of gravity moves down as low as possible. This explains flattening of the larger drops.

Question 13

The hair of a painting brush clings together when taken out of water.

Accepted Answer

This is because the water films formed on them tend to contract to minimum area.

Question 14

How much work will be done in increasing the diameter of a soap bubble from 2 cm to 5 cm ? Surface tension of soap is 3.0×10−2 N/m

Accepted Answer

Soap bubble has two surfaces. Hence,W=TΔA(T=ΔAΔW​)Here,ΔAW​=2[4π{(2.5×10−2)2−(1.0×10−2)2}]=1.32×10−2 m2=(3.0×10−2)(1.32×10−2)J=3.96×10−4 J​

Question 15

Calculate the energy released when 1000 small water drops each of same radius 10−7 m coalesce to form one large drop. The surface tension of water is 7.0×10−2 N/m.

Accepted Answer

Let r be the radius of smaller drops and R of bigger one. Equating the initial and final volumes, we have34​πR3=(1000)(34​πr3)R=10r=(10)(10−7)mR=10−6 m​Further, the water drops have only one free surface. Therefore,ΔA=4πR2−(1000)(4πr2)=4π[(10−6)2−(103)(10−7)2]=−36π(10−12)m2Here, negative sign implies that surface area is decreasing. Hence, energy released in the process.U=T∣ΔA∣​=(7×10−2)(36π×10−12)J=7.9×10−12 J​

Question 16

What should be the pressure inside a small air bubble of 0.1 mm radius situated just below the water surface. Surface tension of water =7.2×10−2 N/m and atmospheric pressure =1.013×105 N/m2.

Accepted Answer

Surface tension of water T=7.2×10−2 N/mRadius of air bubble R=0.1 mm=10−4 mThe excess pressure inside the air bubble is given by,p2​−p1​=R2T​∴ Pressure inside the air bubble,p2​=p1​+R2T​Substituting the values, we havep2​​=(1.013×105)+10−4(2×7.2×10−2)​=1.027×105 N/m2​

Question 17

Two separate air bubbles (radii 0.004 m and 0.002 m ) formed of the same liquid (surface tension 0.07 N/m ) come together to form a double bubble. Find the radius and the sense of curvature of the internal film surface common to both the bubbles.

Accepted Answer

p1​=p0​+r1​4T​⇒p2​r2​​=p0​+r2​4T​<r1​p2​>p1​​ i.e. pressure inside the smaller bubble will be more. The excess pressure p=p2​−p1​=4T(r1​r2​r1​−r2​​) ...(i)This excess pressure acts from concave to convex side, the interface will be concave towards smaller bubble and convex towards larger bubble. Let R be the radius of interface then,p=R4T​ ...(ii)From Eqs. (i) and (ii), we getR​=r1​−r2​r1​r2​​=(0.004−0.002)(0.004)(0.002)​=0.004 m​

Question 18

Under isothermal condition two soap bubbles of radii r1​ and r2​ coalesce to form a single bubble of radius r. The external pressure is p0​. Find the surface tension of the soap in terms of the given parameters.

Accepted Answer

As mass of the air is conserved,∴∴​n1​+n2​RT1​p1​V1​​+RT2​p2​V2​​​=n=RTpV​​As temperature is constant,T1​=T2​=T∴p1​V1​+p2​V2​=pV∴(p0​+r1​4S​)(34​πr13​)+(p0​+r2​4S​)(34​πr23​)=(p0​+r4S​)(34​πr3)Solving, this we getS=4(r12​+r22​−r2)p0​(r3−r13​−r23​)​

Question 19

Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced, what will be the new velocity?

Accepted Answer

∵ Volume of bigger drop=n× volume of smaller drop34​πR3R​=2×34​πr3=21/3r​Terminal velocity ∝r2∴v′=22/3v

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