306 + Interview Questions in SURFACE-TENSION IN GENERAL AWARENESS Page 5 InterviewSolution

201.	A glass rod of diameter `d_(1)=1.5 mm` in inserted symmetrically into a glass capillary with inside diameter `d_(2)=2.0` mm. Then the whole arrangement is vertically oriented and brought in contact with the surface of water. To what height will the liquid rise in the capillary? Surface tension of water `=73xx10^(-3)N//m`
Answer» Correct Answer - `6`

Discussion

202.	There is a howizontal film of soap solution. On it a thread is placed in the form of a loop. The film is pierced inside the loop and the thread becomes a circular loop of radius `R`. If the surface tension of the loop be `T`, then what will be the tension in the thread?A. `pi R^(2)//T`B. `pi R^(2)T`C. `2pi RT`D. `2RT`
Answer» Correct Answer - D

Discussion

203.	The work done in forming a soap film of size ` 10 cm xx 10 ` cm will be , it if the surface tension of soap solution is ` 3xx 10^(-2)` N/mA. ` 3 xx 0^(-4)`JB. ` 3xx 10^(-2)`JC. `6xx10^(-4)`JD. ` 6xx 10^(-3)`J
Answer» Correct Answer - C

Discussion

204.	A rectangular plate of dimension `6 cm xx 4cm` and thickness `2mm` is placed with its largest face flat on the surface of water. Find the downward force on the plate due to surface tension. Surface tension of water is `7.0 xx 10^(-2)Nm^(-1)`.A. `1.8 xx 10^(-2)N`B. `1.4 xx 10^(-2)N`C. `2 xx 10^(-2) N`D. `2.5 xx 10^(-2) N`
Answer» Correct Answer - B Given, `l = 6 xx 10^(-2)m`, `b = 4 xx 10^(-2)m`, `d = 2 xx 10^(-3) m`, and `S = 7.0 xx 10^(-2) Nm^(-1)` Force on the plate due to surface tension is `F = 2(l+b)S` `= 2(6 xx 10^(-2) + 4 xx 10^(-2)) xx 7.0 xx 10^(-2)` `= 1.4 xx 10^(-2) N`

Discussion

205.	An air bubble of radius 0.1 mm is ready to leave the surface of a lake. Then the pressure inside in `N//m^(2)` is (Surface tension of water : `70 xx 10^(-3) ` N/m, 1 atm = `1 xx 10^(5) N//m^(2)`)A. `1400 N//m^(2)`B. `1.014 xx10^(5) N//m^(2)`C. `14 N//m^(2)`D. `1.014 xx10^(4) N//m^(2)`
Answer» Correct Answer - D ` Delta P = (2T)/R , P_(0) + Delta P = P_("inside")` ` :. Delta P = (2T)/R = (2 xx 70 xx 10^(-3)N//m)/(0.1xx10^(-3)m)` ` = 1400 " N/m"^(2)` ` = 0.014 xx 10^(5) " N/m"^(2)` ` P_("inside")= P_(0) +Delta P` ` = 1.014 xx 10^(5) " N/m"^(2)`

Discussion

206.	A hollow sphere has a small hole. When the sphere is taken to a depth of 0.5 m inside water the air bubbles started coming out from the hole. If the surface tension of water is the radius of the hole isA. `2.8 xx10^(5)` mB. ` 1.35 xx10^(-5)`mC. `0.67 xx10^(-5)`mD. `0.28 xx10^(-5)`m
Answer» Correct Answer - A ` p = (2T)/R and P = hdg rArr hdg = (2T)/R` ` R = (2T)/("hdg")` ` = 2.8 xx 10^(-5)` m

Discussion

207.	A bubble of radius 10 cm is formed with a solution that has surface tention .What is the excess pressure inside the bubble inA. `0.16`B. 16C. `1.6`D. 10
Answer» Correct Answer - C ` P = (4T)/R = (4 xx 40 xx 10^(-3))/(0.1)` ` = 1600 xx 10^(-3)` ` = 1.6 " N/m"^(2)`

Discussion

208.	The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. What is the ratio between the volume of the first and the second bubble?A. `1:3`B. `1:9`C. `27:1`D. `1:27`
Answer» Correct Answer - D

Discussion

209.	A soap film of surface tension `3 xx 10^(-2)N//m` formed in a rectangular frame can support a straw as shown in Fig. If `g = 10 ms^(-2)`, the mass of the straw is A. `0.06 gm`B. `0.6 gm`C. `6 gm`D. `60 gm`
Answer» Correct Answer - B

Discussion

210.	The height of liquid column in capillary tube is h the radius of capillary tube is r.`rho` is the density of liquid , g is acceleration due to gravity and `theta ` is angle of contact . The surface tension of the liquid is ,A. `t = (r h rho g)/(2 cos theta)`B. `T= (2 cos theta )/(h rho g)r`C. ` T = (2 cos theta )/(r rho h g)`D. all of these
Answer» Correct Answer - A

Discussion

211.	A liquid rises in a capillary tube of radius r upto height h and tha mass of liquid is 20 gm . If a the radius of capillary tube is half and height is twice , then the new mass of a liquid will be ,A. 10 gmB. 80 gmC. 20 gmD. 40 gm
Answer» Correct Answer - A `2pir_(1) = m_(1)g and 2pir_(2)= m_(2)g`

Discussion

212.	The force of surface tension on a ring situated on the surface of water is ` 14 pi xx 10^(-4)` N . Then the diameter of the ring isA. 5 mmB. 1 cmC. 2 cmD. 4 cm
Answer» Correct Answer - B `T =F/(4pir)= F/(2pid)` ` :. D = (14 pi xx 10^(-4))/(2xxpixx70xx10^(-3)) = 0.01 m = 1 cm`

Discussion

213.	A liquid rises to a height of 9 cm in a glass liquid column in a glass capillary of radius 0.03 cm isA. `13.5` cmB. 9 cmC. 6 cmD. 12 cm
Answer» Correct Answer - C `h_(1)r_(1)=h_(2)r_(2)`

Discussion

214.	A circular wire of length 0.1 m is touching the surface of the liquid of surface tension ` 3 xx 10^(-2)` N/m The mass of a wire is `10^(-4)` kg . If g = 10 `m//s^(2)` then the force needed to lift the circular wire isA. `7xx10^(-3)`NB. `0.7 xx 10^(-3)`NC. `0.07 xx 10^(-3)`ND. `70 xx 10^(-3)` N
Answer» Correct Answer - A `F = mg + T xx 2l = 10^(-4) xx 10 + 3 xx 10^(-2) xx 0.2 ` ` = 1 xx 10^(-3) + 6 xx 10^(-3) = 7 xx 10^(-3)` N

Discussion

215.	A metal ring of internal and external radii 15 mm and 20 mm respectively is placed horizontally on the surface of a liquid of surface tension 0.03 N/m To detach the ring from the liquid surface the force required isA. `28 xx 10^(-5)`NB. `66xx10^(-4)`NC. `6.6 xx10^(-4)`ND. `425 xx 10^(-4)`N
Answer» Correct Answer - B `2pi (r+R)T = =F` ` F = 2pi (15xx 10^(-3) + 20 xx 10^(-3))xx3 xx 10^(-2)` ` 2pi (35) xx 10^(-3) xx 3xx 10^(-2)` ` = 659.7 xx10^(-5)` ` = 66 xx 10^(-4)` N

Discussion

216.	A thin disc of radius `R`, just touching the liquid surface forms one arm of a balance. The palte is balanced by some weight on the other side of the balance. How much extra weight should be added on the other side, so that the disc can just come cut of water ?
Answer» Surface tension force on the disc is `(T)(2piR)` For balance `(T)(2piR) = (Deltam)g rArr Deltam = ((T)(2piR))/(g)`

Discussion

217.	A circular loop of thin wire of radius 7 cm is lifted from the surface of a liquid . An additional force of ` 44 xx 10^(-2)` N is required to detach the loop from the surface of the liquid . The surface tension of the liquid in N/m isA. `0.5`B. 1C. `2.86`D. `6.826`
Answer» Correct Answer - A `F = 2 l T` l is the circumference of the loop = `2pir` ` :. 2(2pir) T = F` ` T = F/(4pir) = (44xx10^(-2)N)/(4xx7xx10^(-2)xxpim)` ` = 0.5 ` N/m

Discussion

218.	A circular loop of a thin wire of radius `(7//pi)` cm is suspended from one arm of a balance . The plane of the loop is in contact with the surface of soap solution . If the pull on the loop due to surface tension is found to be ` 0.6 xx 10^(-3)` kg wt , then the surface tension of soap solution will be ,A. `11xx10^(-3)`N/mB. `21xx10^(-5)`N/mC. `21xx10^(-3)`N/mD. `41 xx 10^(-3)` N/m
Answer» Correct Answer - C ` T = F/(2xx2pir)=(6xx10^(-4)xxpixx10)/(4xxpixx7xx10^(2))=21 xx10^(-3) ` N/m

Discussion

219.	The force required to take away a flat circular plate of radius 4 cm from the surface of water isA. 560 `pi` dyneB. 560 dyneC. `5.6 pi`D. `56 pi` dyne
Answer» Correct Answer - A `F = T xx 2 pir = 70 xx 2 xx pi xx 4 = 560 pi ` dyne

Discussion

220.	The surface tension of a liquid is `10^(8)` dyne `cm^(-1)`. It is equivalent toA. `10^(7)`N/mB. `10^(6)`N/mC. `10^(5)`N/mD. `10^(4)`N/m
Answer» Correct Answer - C `T = 10^(8) dy//cm = (10^(8)xx10^(-5))/(10^(-2))` ` = 10^(5) N//m^(2)`

Discussion

221.	The surface tension of a liquid is `70 "dyne" // cm` . In MKS system its value isA. 70 N/mB. `70 xx 10^(-2)`N/mC. `7xx10^(2)` N/mD. `7xx10^(3)` N/m
Answer» Correct Answer - B `T_(w) = 70 dy//cm = (70 xx 10^(-5))/(10^(-2))` ` = 70 xx 10^(-3) = 70 xx 10^(-2)` N/m

Discussion

222.	The surface tension of a liquid is `10^(8)` dyne `cm^(-1)`. It is equivalent toA. `10^(7)` N/mB. `10^(6)`N/mC. `10^(5)`N/mD. `10^(4)`N/m
Answer» Correct Answer - A ` T = 10^(8) dy//cm^(2) = (10^(8) xx10^(-5))/(10^(-4)) = 10^(-7)`

Discussion

223.	In order to float a ring of area 0.04 m in a liquid of surface tension 75 N/m, the required surface energy will beA. 3JB. 6.5 JC. 1.5 JD. 4J
Answer» Correct Answer - A

Discussion

224.	A film is held on a ring of area 0.05 `m^(2)` of liquid of surface tension 0.05 N/m then the surface energy isA. 25 JB. `0.01` JC. `0.5` JD. `25xx10^(-4)`J
Answer» Correct Answer - D Surface energy , `= 4piR^(2)T` ` = 0.05 xx 0.05 ` N/m ` = 25 xx10^(-4)` J

Discussion

225.	If the maximum force in addition to the weight required to pull a wire frame 5.0 cm long from water surface at temperature of `20^(@)` C is 720 dyne the surface tension of water will beA. `72.0`dyne/cmB. 145 dyne/cmC. `7.28` dyne/cmD. `7.28` N/m
Answer» Correct Answer - A S.T = `F/(2l) = 720/10 = 72 ` dyne/cm

Discussion

226.	A ring of radius r, and weight W is lying on a liquid surface . If the surface tension of the liquid is T , then the minimum force required to be applied in order to lift the ring upA. WB. 2WC. `W+4 pi rT`D. `W+2 pi rT`
Answer» Correct Answer - C `F = W + 2 TL ` `= w + 2pir xx 2 = W + 4 pi rT `

Discussion

227.	In the last question let the length of the tube be L and its outer radius be R. Water rises in it to a height h. Calculate the vertical force needed to hold the tube in this position. Mass of empty tube is M.
Answer» Correct Answer - `Mg+pi P_(o)[R^(2)-(Lr^(2))/(L-h)]+2pi(R+r)T`

Discussion

228.	A straight piece of wire 4 cm long is placed horizontally on the surface of water annd is gently pulled up . It is found that a force of 560 dyne in addition to the weight of the wire is required for this purpose . Then surface tension of the water isA. 70 dyne/cmB. 44.8 dyne/cmC. 700 dyne/cmD. none
Answer» Correct Answer - A `T = F/(2l) = 560/(2xx4) = 70 ` dyne/cm

Discussion

229.	In the previous question, in place of disc a ring is used whose inner radius is `R_(1)` and outer radius is `R_(2)` Now how much extra weight should be added on the other side, so that the ring can just come out of water ?
Answer» Surface tension force on the disc is `(T)2pi(R_(1)+R_(2))` For balance `(T)2pi(R_(1) + R_(2)) = (Deltam)g rArr ((T)2pi(R_(1)) - R_(2))/(g)`

Discussion

230.	If `sigma ` be the surface tension, the work done in breaking a big drop of radius R in n drops of equal radius isA. `Rn^(2//3) sigma`B. `(n^(2//3)-1)Rsigma`C. `(n^(1//3)-1) R sigma`D. `4piR^(2)(n^(1//3)-1) sigma`
Answer» Correct Answer - D

Discussion

231.	A long thin walled capillary tube of mass M and radius r is partially immersed in a liquid of surface tension T. The angle of contact for the liquid and the tube wall is `30^(@)`. How much force is needed to hold the tube vertically? Neglect buoyancy force on the tube.
Answer» Correct Answer - `2sqrt(3)pi rT+Mg`

Discussion

232.	`1000` small water drops, each of radius `r`, combine and form a big drop. In this process, find decrease in surface energy.
Answer» Suppose radius of big drop is `R`. During this process, mass will be conserved, so volume will also be conserved. `("Volume")_("initial") = ("Volume")_("final")` `((4)/(3)pir^(3)) xx 1000 = ((4)/(3)piR^(3)) rArr R = 10r` loss in surface energy `DeltaU_("loss") = TDeltaA_("loss") = T((4pir^(2)) xx 1000 - 4pi(10r)^(2))` `DeltaU_("loss") = (T)(900 xx 4pir^(2))` this energy loss will be converted into heat. So increase in temperature of the drop can be found from `T(900 xx 4pir^(2)) = msDeltaT`, From this get the incerase in temperature `DeltaT`.

Discussion

233.	One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy isA. `1000:1`B. `1:1000`C. `10:1`D. `1:10`
Answer» Correct Answer - D

Discussion

234.	(i) One end of a uniform glass capillary tube of radius `r = 0.025` cm is immersed vertically in water to a depth `h = 1cm`. Contact angle is `0^(@)`, surface tension of water is `7.5 × 10–2 N//m`, density of water is `rho = 10^(3) kg//m^(3)` and atmospheric pressure is `P_(o) = 10^(5) N // m^(2)` Find the excess pressure to be applied on the water in the capillary tube so that - (a) The water level in the tube becomes same as that in the vessel. (b) Is it possible to blow out an air bubble out of the tube by increasing the pressure? (ii) A container contains two immiscible liquids of density `rho_(1)` and `rho_(2) (rho_(2) gt rho_(1))`. A capillary of radius r is inserted in the liquid so that its bottom reaches up to denser liquid and lighter liquid does not enter into the capillary. Denser liquid rises in capillary and attain height equal to h which is also equal to column length of lighter liquid. Assuming zero contact angle find surface tension of the heavier liquid.
Answer» Correct Answer - (1) (a) `600 P_(a)` (b) Yes (3) `T=r/2(rho_(2)-rho_(1))gh`

Discussion

235.	8000 identical water drops combine together to form a big drop. Then the ratio of the final surface energy of all the initial surface energy of all the drops together isA. `1:10`B. `1:15`C. `1:20`D. `1:25`
Answer» Correct Answer - C

Discussion

236.	A capillary tube when immersed vertically into a liquid records a rise of 3 cm . If the tube is held immersed in the liquid at an angle of `30^(@)` with the vertical , the liquid column along the tube will beA. `3 . 464` cmB. 3 cmC. `4.5 ` cmD. `7.5` cm
Answer» Correct Answer - A `h = l cos theta` , ` :. l = h/(costheta) = 3/(cos30^(@))= (3xx2)/(sqrt(2))=3.464` cm

Discussion

237.	A capillary tube when immersed vertically in a liquid records a rise of 3cm.if the tube is immersed in the liquid at an angle of `60^(@)` with the vertical, then find the length of the liqiud column along the tube.A. 3 cmB. `4.5` cmC. 6 cmD. ` 7.5 ` cm
Answer» Correct Answer - C ` h = l sin theta " " :. L = h/(sin theta)`

Discussion

238.	A capillary tube when immersed vertically in a liquid records a rise of 3cm.if the tube is immersed in the liquid at an angle of `60^(@)` with the vertical, then find the length of the liqiud column along the tube.A. 9 cmB. 6 cmC. 3 cmD. 2 cm
Answer» Correct Answer - B Length, `l = (h)/(cos theta) = (3)/(cos 60^(@)) = 6 cm`

Discussion

239.	Find the work done in blowing a soap bubble of radius R if pressure outside in `p_(0)` and the surface tension of the soap solution is T.
Answer» Correct Answer - `W=8piR^(2)T+p_(f)V_(f) " In " (pf)/(p_(0)) " where " p_(f)=p_(0)+(4T)/(R ) " and " V_(f)=(4 pi)/(3) R^(3)`

Discussion

240.	Work done in increasing the size of a soap bubble from a radius of `3 cm` to `5 cm` is nearly. (surface tension of soap solution `= 0.3 Nm^(-1))`A. `4pimJ`B. `0.2pimJ`C. `2pimJ`D. `0.4 pi mJ`
Answer» Correct Answer - D `W = TDeltaA` `= 0.03 (2 xx 4pi xx (5^(2) - 3^(2))10^(-4)` `= 24pi(16)xx 10^(-6)` `= 0.384 pi xx 10^(-3)` Joule `-= 0.4 pimJ`

Discussion

241.	A drop of liquid of diameter 2.8 mm breaks up into 125 identical drops. The change in energy is nearly (S.T. of liquid =75 dynes / cm )A. ZeroB. 19 ergC. 46 ergD. 74 erg
Answer» Correct Answer - D

Discussion

242.	Find the difference of air pressure between the inside and outside of a soap bubble is 5mm in diameter, if the surface tension is `1.6 Nm^(-1)`.A. `2560 Nm^(-2)`B. `3720 Nm^(-2)`C. `1208 Nm^(-2)`D. `10132 Nm^(-2)`
Answer» Correct Answer - A The excess pressure p of bubble in air is given by `p = (4S)/(R) = (4 xx 1.6)/(2.5 xx 10^(-3))` `= 2560 Nm^(-2)`

Discussion

243.	Two water droplets merge with each other to from a larger droplet. In this procesA. Energy is liberatedB. Energy is absorbedC. Neither liberated nor absorbedD. Some mass is converted into energy
Answer» Correct Answer - A

Discussion

244.	Surface tension of a soap solution is `1.9xx10^(-2) N//m`. Work done in blowing a bubble of 2.0 cm diameter will beA. `45.54 pi xx 10^(6)` JB. `15.2 pi xx 10^(-6)`JC. `1.9 pi xx10^(-6)`JD. `1 pi xx 10^(-4)`J
Answer» Correct Answer - B `dw = S.T xx "surface area "` ` = 1.9 xx10^(-2) xx (4piR^(2)) xx 2 ` `= 1.9 xx 10^(-2) 4xx pi (1 xx 10^(-2))xx2` ` = 15.2 xx10^(-6) pi ` J

Discussion

245.	What would be the excess pressure in side a small air bubble of 0.2 mm diameter situated just below the surface of water ? (S.T of water = 0.072 N/m)A. `1 . 44 xx 10^(2)` PaB. ` 1..44 xx10^(3)`PaC. ` 1.44 xx 10^(4)`PaD. `1.44 xx10^(5)`Pa
Answer» Correct Answer - B ` P = (2T)/r = (2 xx 72 xx 10^(-3))/(10^(-4))= 1440 = 1.44 xx 10^(3)` pa

Discussion

246.	P is the excess pressure inside a water drop . If that drop is divided into 8 indentical droplets , excess pressure inside smaller droplet isA. PB. P/2C. 2PD. P/8
Answer» Correct Answer - C ` v_(1)= 8v_(2)` `(4pi)/3 r_(1)^(3) = 8(4pi)/3 r_(2)^(3)` ` r_(1) = 2r_(2)" " :. r_(2) = r_(1)/2` ` P_(1)/(P_(2)) = (r_(2))/(r_(1)) = (r/2)/r=1/2` `P_(2)= 2P`

Discussion

247.	A soap bubble in air (two surfaces) has surface tension `0.03 Nm^(-1)`. Find the excess pressure inside a bubble of diameter 30 mm.A. 2 PaB. 4 PaC. 16 PaD. 10 Pa
Answer» Correct Answer - D Excess pressure is given by , `p = (4S)/(r)` Given, `S = 0.03 Nm^(-1)` `r = (d)/(2) = (30)/(2) mm = (30)/(2) xx 10^(-3) m` `rArr p = (4 xx 0.03)/(15 xx 10^(-3)) = 8 Pa`

Discussion

248.	A soap bubble of radius `10^(-2)` m is formed . The surface tension of the soap bubble is 0.04 N/m The excess of pressure inside the bubble isA. `160 N//m^(2)`B. `16 N//m^(2)`C. `1.6 N//m^(2)`D. `0.16 N//m^(2)`
Answer» Correct Answer - B ` P = (4T)/r = (4 xx 0.04 )/(10^(-2)) = 16 " N/m"^(2)`

Discussion

249.	If two soap bubbles of different radii are connected by a tubeA. air flows from the bigger bubble to the smaller bubble till the sizes become equalB. air flows from bigger bubble to the smaller bubble till the sizes are interchangedC. air flows from the smaller bubble to the smaller bubble till the sizes are interchangedD. there is no flow of air
Answer» Correct Answer - C The excess pressure inside the soap bubble is inversely proportional to radius of soap bubble i.e., `p prop 1//r`, where r being radius of soap bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one.

Discussion

250.	The surface tension of soap solution is ` 2.1 xx 10^(-2)` N/m . Then the work done in blowing a soap bubble of diameter 3.0 isA. `4.4 xx10^(-5)`JB. `11.9 xx 10^(-5)`JC. `8.95 xx 10^(-5)`JD. `23.10^(-5)`J
Answer» Correct Answer - B `W = Txx2A` ` = 2.1 xx 10^(-2) xx 2xx 4 xx3.14 xx2.25 xx10^(-4)` ` = 11.9 xx 10^(-5)` J

Discussion

Explore topic-wise InterviewSolutions in .

The work done in forming a soap film of size ` 10 cm xx 10 ` cm will be , it if the surface tension of soap solution is ` 3xx 10^(-2)` N/mA. ` 3 xx 0^(-4)`JB. ` 3xx 10^(-2)`JC. `6xx10^(-4)`JD. ` 6xx 10^(-3)`J

An air bubble of radius 0.1 mm is ready to leave the surface of a lake. Then the pressure inside in `N//m^(2)` is (Surface tension of water : `70 xx 10^(-3) ` N/m, 1 atm = `1 xx 10^(5) N//m^(2)`)A. `1400 N//m^(2)`B. `1.014 xx10^(5) N//m^(2)`C. `14 N//m^(2)`D. `1.014 xx10^(4) N//m^(2)`

A hollow sphere has a small hole. When the sphere is taken to a depth of 0.5 m inside water the air bubbles started coming out from the hole. If the surface tension of water is the radius of the hole isA. `2.8 xx10^(5)` mB. ` 1.35 xx10^(-5)`mC. `0.67 xx10^(-5)`mD. `0.28 xx10^(-5)`m

A bubble of radius 10 cm is formed with a solution that has surface tention .What is the excess pressure inside the bubble inA. `0.16`B. 16C. `1.6`D. 10

The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. What is the ratio between the volume of the first and the second bubble?A. `1:3`B. `1:9`C. `27:1`D. `1:27`

A soap film of surface tension `3 xx 10^(-2)N//m` formed in a rectangular frame can support a straw as shown in Fig. If `g = 10 ms^(-2)`, the mass of the straw is A. `0.06 gm`B. `0.6 gm`C. `6 gm`D. `60 gm`

A liquid rises in a capillary tube of radius r upto height h and tha mass of liquid is 20 gm . If a the radius of capillary tube is half and height is twice , then the new mass of a liquid will be ,A. 10 gmB. 80 gmC. 20 gmD. 40 gm

The force of surface tension on a ring situated on the surface of water is ` 14 pi xx 10^(-4)` N . Then the diameter of the ring isA. 5 mmB. 1 cmC. 2 cmD. 4 cm

A liquid rises to a height of 9 cm in a glass liquid column in a glass capillary of radius 0.03 cm isA. `13.5` cmB. 9 cmC. 6 cmD. 12 cm

A circular wire of length 0.1 m is touching the surface of the liquid of surface tension ` 3 xx 10^(-2)` N/m The mass of a wire is `10^(-4)` kg . If g = 10 `m//s^(2)` then the force needed to lift the circular wire isA. `7xx10^(-3)`NB. `0.7 xx 10^(-3)`NC. `0.07 xx 10^(-3)`ND. `70 xx 10^(-3)` N

A metal ring of internal and external radii 15 mm and 20 mm respectively is placed horizontally on the surface of a liquid of surface tension 0.03 N/m To detach the ring from the liquid surface the force required isA. `28 xx 10^(-5)`NB. `66xx10^(-4)`NC. `6.6 xx10^(-4)`ND. `425 xx 10^(-4)`N

A thin disc of radius `R`, just touching the liquid surface forms one arm of a balance. The palte is balanced by some weight on the other side of the balance. How much extra weight should be added on the other side, so that the disc can just come cut of water ?

A circular loop of thin wire of radius 7 cm is lifted from the surface of a liquid . An additional force of ` 44 xx 10^(-2)` N is required to detach the loop from the surface of the liquid . The surface tension of the liquid in N/m isA. `0.5`B. 1C. `2.86`D. `6.826`

The force required to take away a flat circular plate of radius 4 cm from the surface of water isA. 560 `pi` dyneB. 560 dyneC. `5.6 pi`D. `56 pi` dyne

The surface tension of a liquid is `10^(8)` dyne `cm^(-1)`. It is equivalent toA. `10^(7)`N/mB. `10^(6)`N/mC. `10^(5)`N/mD. `10^(4)`N/m

The surface tension of a liquid is `70 "dyne" // cm` . In MKS system its value isA. 70 N/mB. `70 xx 10^(-2)`N/mC. `7xx10^(2)` N/mD. `7xx10^(3)` N/m

The surface tension of a liquid is `10^(8)` dyne `cm^(-1)`. It is equivalent toA. `10^(7)` N/mB. `10^(6)`N/mC. `10^(5)`N/mD. `10^(4)`N/m

In order to float a ring of area 0.04 m in a liquid of surface tension 75 N/m, the required surface energy will beA. 3JB. 6.5 JC. 1.5 JD. 4J

A film is held on a ring of area 0.05 `m^(2)` of liquid of surface tension 0.05 N/m then the surface energy isA. 25 JB. `0.01` JC. `0.5` JD. `25xx10^(-4)`J

If the maximum force in addition to the weight required to pull a wire frame 5.0 cm long from water surface at temperature of `20^(@)` C is 720 dyne the surface tension of water will beA. `72.0`dyne/cmB. 145 dyne/cmC. `7.28` dyne/cmD. `7.28` N/m

A ring of radius r, and weight W is lying on a liquid surface . If the surface tension of the liquid is T , then the minimum force required to be applied in order to lift the ring upA. WB. 2WC. `W+4 pi rT`D. `W+2 pi rT`

In the last question let the length of the tube be L and its outer radius be R. Water rises in it to a height h. Calculate the vertical force needed to hold the tube in this position. Mass of empty tube is M.

In the previous question, in place of disc a ring is used whose inner radius is `R_(1)` and outer radius is `R_(2)` Now how much extra weight should be added on the other side, so that the ring can just come out of water ?

If `sigma ` be the surface tension, the work done in breaking a big drop of radius R in n drops of equal radius isA. `Rn^(2//3) sigma`B. `(n^(2//3)-1)Rsigma`C. `(n^(1//3)-1) R sigma`D. `4piR^(2)(n^(1//3)-1) sigma`

A long thin walled capillary tube of mass M and radius r is partially immersed in a liquid of surface tension T. The angle of contact for the liquid and the tube wall is `30^(@)`. How much force is needed to hold the tube vertically? Neglect buoyancy force on the tube.

`1000` small water drops, each of radius `r`, combine and form a big drop. In this process, find decrease in surface energy.

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy isA. `1000:1`B. `1:1000`C. `10:1`D. `1:10`

8000 identical water drops combine together to form a big drop. Then the ratio of the final surface energy of all the initial surface energy of all the drops together isA. `1:10`B. `1:15`C. `1:20`D. `1:25`

A capillary tube when immersed vertically into a liquid records a rise of 3 cm . If the tube is held immersed in the liquid at an angle of `30^(@)` with the vertical , the liquid column along the tube will beA. `3 . 464` cmB. 3 cmC. `4.5 ` cmD. `7.5` cm

A capillary tube when immersed vertically in a liquid records a rise of 3cm.if the tube is immersed in the liquid at an angle of `60^(@)` with the vertical, then find the length of the liqiud column along the tube.A. 3 cmB. `4.5` cmC. 6 cmD. ` 7.5 ` cm

A capillary tube when immersed vertically in a liquid records a rise of 3cm.if the tube is immersed in the liquid at an angle of `60^(@)` with the vertical, then find the length of the liqiud column along the tube.A. 9 cmB. 6 cmC. 3 cmD. 2 cm

Find the work done in blowing a soap bubble of radius R if pressure outside in `p_(0)` and the surface tension of the soap solution is T.

Work done in increasing the size of a soap bubble from a radius of `3 cm` to `5 cm` is nearly. (surface tension of soap solution `= 0.3 Nm^(-1))`A. `4pimJ`B. `0.2pimJ`C. `2pimJ`D. `0.4 pi mJ`

A drop of liquid of diameter 2.8 mm breaks up into 125 identical drops. The change in energy is nearly (S.T. of liquid =75 dynes / cm )A. ZeroB. 19 ergC. 46 ergD. 74 erg

Find the difference of air pressure between the inside and outside of a soap bubble is 5mm in diameter, if the surface tension is `1.6 Nm^(-1)`.A. `2560 Nm^(-2)`B. `3720 Nm^(-2)`C. `1208 Nm^(-2)`D. `10132 Nm^(-2)`

Two water droplets merge with each other to from a larger droplet. In this procesA. Energy is liberatedB. Energy is absorbedC. Neither liberated nor absorbedD. Some mass is converted into energy

Surface tension of a soap solution is `1.9xx10^(-2) N//m`. Work done in blowing a bubble of 2.0 cm diameter will beA. `45.54 pi xx 10^(6)` JB. `15.2 pi xx 10^(-6)`JC. `1.9 pi xx10^(-6)`JD. `1 pi xx 10^(-4)`J

What would be the excess pressure in side a small air bubble of 0.2 mm diameter situated just below the surface of water ? (S.T of water = 0.072 N/m)A. `1 . 44 xx 10^(2)` PaB. ` 1..44 xx10^(3)`PaC. ` 1.44 xx 10^(4)`PaD. `1.44 xx10^(5)`Pa

P is the excess pressure inside a water drop . If that drop is divided into 8 indentical droplets , excess pressure inside smaller droplet isA. PB. P/2C. 2PD. P/8

A soap bubble in air (two surfaces) has surface tension `0.03 Nm^(-1)`. Find the excess pressure inside a bubble of diameter 30 mm.A. 2 PaB. 4 PaC. 16 PaD. 10 Pa

A soap bubble of radius `10^(-2)` m is formed . The surface tension of the soap bubble is 0.04 N/m The excess of pressure inside the bubble isA. `160 N//m^(2)`B. `16 N//m^(2)`C. `1.6 N//m^(2)`D. `0.16 N//m^(2)`

The surface tension of soap solution is ` 2.1 xx 10^(-2)` N/m . Then the work done in blowing a soap bubble of diameter 3.0 isA. `4.4 xx10^(-5)`JB. `11.9 xx 10^(-5)`JC. `8.95 xx 10^(-5)`JD. `23.10^(-5)`J