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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
By how much will the surface of a liquid be depressed in a glass tube of radius 0.2 mm if the angle of contact of the liquid is `135^(@)` and the surface tension is `0.547 Nm^(1)`? Density of the liquid is `13.5xx10^(3) kg m^(-3)`. [ Hint: Neglecting the liquid in the meniscus we have for equilibrium `pi r_(2)h rho g=2pirT cos theta,h=(2T cos theta)/(rho g r)`] |
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Answer» Correct Answer - `2.92xx10^(-2)m` |
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| 52. |
When liquid medicine of density `rho` 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 te 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 surfacetension 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 `r=5xx10^(-4)`m `overset(~)(n)=10^(3)kgm^(-3)`, `g=10ms^(-2)`, `T=0.11Nm^(-1)` the radius of the drop when it detaches from the dropper is approximatelyA. `1.4 xx 10^(-3)m`B. `3.3 xx 10^(-3)m`C. `2.0 xx 10^(-3)m`D. `4.1 xx 10^(-3)m` |
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Answer» Correct Answer - A Equating force on the drop : `T_(2pir^(2))/(R) = rho(4)/(3)piR^(3)g` (Assume drop as a complete sphere `R = ((3Tr^(2))/(2rhog))^(1//4) = ((3 xx 0.11 xx 25 xx 10^(-8))/(2 xx 10^(3) xx 10))^(1//4)` `= 14.25 xx 10^(-4)m = 1.425 xx 10^(-3)jm` |
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| 53. |
When liquid medicine of density `rho` is to be put in the eye, it is done with the help of a dropper. As the bylb on the top of the dropper is pressed. A drop frons 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=0.11 when the radius of the drop is R=1.4mm. when this force become smaller than the weight of the drop the drop gets detached from the dropper. After the drop detaches, its surface energy isA. `1.4 xx 10^(-6)J`B. `2.7 xx 10^(-6)J`C. `5.4 xx 10^(-6)J`D. `8.1 xx 10^(-6)J` |
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Answer» Correct Answer - B Surface energy of the drop `U = TA` `= 0.11 xx 4pi (1.4 xx 10^(-3))^(2)` `= 2.7 xx 10^(-6) J` |
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| 54. |
If a capillary tube of radius ` 2 xx 10^(-4)` m is is dipped vertically in a water of surface tension ` 7 xx 10^(-2)` N/m and density `10^(3) "kg/m"^(3)` then the height to which water rises will beA. ` 7 xx 10^(-2)` cmB. ` 0.7 xx 10^(-2)` cmC. `14 xx 10^(-2)` cmD. ` 1.4 xx 10^(-2)` cm |
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Answer» Correct Answer - A ` h = (2T cos theta )/(rho gr) = (2xx7xx10^(-2)xx1)/(10^(3)xx10xx2xx10^(-4))=7xx10^(-2)` m |
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| 55. |
Water rise in capillary tube when its one end is dipped vertically in it, is 3 cm. If the surface tension of water is `75xx10^-3(N)/(m)`, then the diameter of capillary will be (Take angle of contact`=0^@C`)A. 0.1 mmB. 0.5 mmC. 1.0 mmD. 2.0 mm |
| Answer» Correct Answer - C | |
| 56. |
Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by `75xx10^-4` newton force due to the weight of the liquid. If the surface tension of water is `6xx`10^-2` newton/metre the inner circumference of the capillary must be:A. `1.25xx10^(-2) m`B. `0.50 xx10^(-2) m`C. `6.5xx10^(-2) m`D. `12.5xx10^(-2) m` |
| Answer» Correct Answer - D | |
| 57. |
An astronaut tries to fill ink in an ink pen in an artificial satellite . The inkA. will be filled into the penB. will not be filledC. filling will depend upon the quality of inkD. will fill slowly |
| Answer» Correct Answer - B | |
| 58. |
It is not possible to write directly on blotting paper or newspaper with ink penA. Because of viscosityB. Because of inertiaC. Because of frictionD. Because of capillarity |
| Answer» Correct Answer - D | |
| 59. |
A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?A. B. C. D. |
| Answer» Correct Answer - B | |
| 60. |
The temperature at which the surface tension of water is zero (1) `370^(@)C ` (2) `0^(@)C` (3) Slightly less than `647 K` (4) `277 K`A. `0^(@)C`B. 277 KC. `370^(@)C`D. Slightly less than 647 K |
| Answer» Correct Answer - C::D | |
| 61. |
The work done to get `n` smaller equal size spherical drops from a bigger size spherical drop of wate is proportional to :A. `(1/(n^(2//3)))-1`B. `(1/(n^(1//3)))-1`C. `n^(1//3)-1`D. `n^(4//3)-1` |
| Answer» Correct Answer - C | |
| 62. |
Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one?A. `2^(1//2) : 1`B. `1:1`C. `2^(2//3) :1`D. None of these |
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Answer» Correct Answer - C Volume remains constant after coalescing. Thus, `(4)/(3)pi R^(3) = 2 xx (4)/(3)pi r^(3)` where, R is radius of bigger drop and r is radius of each smaller drop. `:. R = 2^(1//3) r` Surface energy per unit surface area is the surface tension. So, surface energy `E = Delta DA = 4pi R^(2)S` for bigger drop, `E_(1) = 4pi (2^(1//3)r)^(2) S = (2^(1//3)) 4pi r^(2)S` For smaller drop, `E_(2) = 4pi r^(2)S` Hence, required ratio, `(S_(1))/(S_(2)) = 2^(2//3) : 1` |
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| 63. |
Water rises to a height of 2 cm in a capillary tube . If its diameter is made `(2//3)^(rd)` , the water will now rise to a height ofA. 6 cmB. 3 cmC. 2/3 cmD. 3/2 cm |
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Answer» Correct Answer - B `(h_(2))/(h_(1)) = (d_(1))/(d_(2))` `(d_(1))/(2/3d_(1))=3/2 ` ` h _(2) = 3/2 xx 2 = 3 cm ` |
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| 64. |
Water rises in a capillary tube a height h. Choose false statement regarding capillary rise from the following.A. On the surface of jupiter, height will be less than hB. In a lift moving up with contact acceleration height is less than hC. On the surface of moon the height is more than hD. In a lift moving down with constant acceleration height is less than h |
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Answer» Correct Answer - D The height (h) to which water rises in a capillary tube is given by `h = (2T cos theta)/(r rho g)` When lift moves down with constant acceleration, height is more than h, because effective value of acceleration due to gravity decreases, hence h increases. |
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| 65. |
Calculate the heat evolved for the rise of water when one end of the capillary tube of radius r is immeresed vartically into water. Asssume surface tension =T and density of water to be `rho`A. `(2pi S)/(rho g)`B. `(pi S^(2))/(rho g)`C. `(2pi S^(2))/(rho g)`D. `(4pi S^(2))/(rho g)` |
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Answer» Correct Answer - C Potential energy of water column, `U = (mgh)/(2) = (2pi S^(2))/(rho g)` The work performed by force of surface tension is `W = 2pi rSh = (4pi S^(2))/(rho g)` From conservation of energy the heat evolved, `Q = W-U = (2pi S^(2))/(rho g)` |
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| 66. |
If a capillary tube is `45^(@) and 60^(@)` from the vertical then the ratoio of lengths `l_(1) and l_(2)` liquid columns in it will beA. ` 1: sqrt(2)`B. ` sqrt(2):1`C. `2:1`D. `1:4` |
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Answer» Correct Answer - A `(l_(2))/(l_(2))= (costheta_(2))/(costheta_(1))= (1xxsqrt(2))/(sqrt(2)xxsqrt(2))= 1:sqrt(2)` |
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| 67. |
A soap bubble encloses air inside itA. the soap film consists of two surface layers of molecules back to backB. the bubble encloses air inside itC. the pressure inside the bubble is less than the atmospheric pressure has compressed if equally from all sides it a spherical shapeD. because of the elastic property of the film , it will tend to shrink to have as small as surface is possible for the volume it has enclosed |
| Answer» Correct Answer - C | |
| 68. |
A glass capillary of radius 0.4 mm is inclined at `60^(@)` with the vertical in water. Find the length of water in the capillary tube. (Given, surface tension of water `= 7 xx 10^(-2) N^(-1)`).A. 7.1 cmB. 3.6 cmC. 1.8 cmD. 0.9 cm |
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Answer» Correct Answer - A Use, `h = l cos alpha` `rArr l = (h)/(cos alpha)` where, `h = (2S cos theta)/(rho rg) " "(theta_("water" = 0^(@)))` `:.` The length of water in capillary tube `l = (2S cos theta)/(rho rg cos alpha) = (2 xx 7 xx 10^(-2) cos 0^(@))/(10^(3) xx 0.4 xx 10^(-3) xx 9.8 cos 60^(@))` `= 7.1 cm` |
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| 69. |
If a single drop of liquid is splited into large number of droplets all of the same size , then the energy in this process will beA. liberatedB. absorbedC. neither liberated nor absorbedD. some mass is converted into energy |
| Answer» Correct Answer - B | |
| 70. |
The angle of contanct between glass and mercury isA. `0^(@)`B. `30^(@)`C. `90^(@)`D. `135^(@)` |
| Answer» Correct Answer - D | |
| 71. |
A liquid drop having surface energy E is spread into 512 droplets of same size. The final surface energy of the droplets isA. 2EB. 4RC. 8ED. 12 |
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Answer» Correct Answer - C According to question, the surface area of the liquid drop is splits into 512 droplets, the surface area becomes `A_(2) = 512 xx 4pi r^(2)` [r = radius of smaller drop] r can be calculated by comparing the total volume of bigger and all smaller droplets. i.e., `(4)/(3)pi R^(3) = 512 xx (4)/(3) pi r^(3) rArr r = (R)/(8)` Hence, total area of smaller droplets is given by `A_(1) = 512 xx 4 xx pi xx ((R)/(8))^(2) = 8A` Change in surface area is given by `A_(2) - A_(1)` `= 4pi ((512 xx R^(2))/(64) - R^(2))` `4pi (8R^(2) - R^(2)) = 7R^(2)` Surface energy, E = AT [ A area, T - tension] So, `(E_(n))/(E_(o)) = (A_(1) xx T)/(A xx T) = (8A)/(A) = 8` `:. E_(n) = 8E` |
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| 72. |
When mercury is in contact with glass m then the surface of mercury isA. concaveB. convexC. planeD. irregular shape |
| Answer» Correct Answer - B | |
| 73. |
The surface tension for pure water in a capillary tube experiment isA. `(pg)/(2hr)`B. `(2)/(hrpg)`C. `(rpg)/(2h)`D. `(hrpg)/(2)` |
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Answer» Correct Answer - D Surface tension, `S = (rh rho g)/(2 cos theta)` `= (rh rho g)/(2 cos 0^(@))` (`because` for pure water `theta = 0^(@)`) `= (rh rho g)/(2)` |
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| 74. |
A mercury drop does not spread on a glass plate because the angle of contact between glass and mercury isA. AcuteB. ObtuseC. zeroD. `90^(@)` |
| Answer» Correct Answer - B | |
| 75. |
The surface tension of pure water isA. `0.065 `N/mB. `0.072`N/mC. `0.045` N/mD. `0.045` dyne/cm |
| Answer» Correct Answer - B | |
| 76. |
The potential energy per unit change in area of the liquid surface , isA. surface energyB. work doneC. surface tensionD. total energy |
| Answer» Correct Answer - A | |
| 77. |
The relation between surface tension and surface energy isA. W=A/TB. W=T/AC. T=W/AD. T=W.A |
| Answer» Correct Answer - C | |
| 78. |
A liquid is coming out from a vertical tube. The relation between the weight of the drop W , surface tension of the liquid T and radius of the tube r is given by, if the angle of contact is zeroA. `W=pir^(2)T`B. `W=2pirT`C. `W=2r^(2)piT`D. `W=3/4 pir^(3) T` |
| Answer» Correct Answer - B | |
| 79. |
The correct relation isA. `r=(2T cos theta)/(hdg)`B. `r=(hdg)/(2T cos theta)`C. `r=(2T dgh)/(cos theta)`D. `r=(T cos theta)/(2hdg)` |
| Answer» Correct Answer - A | |
| 80. |
The work done to split a big drop into n number of identical droplets all of the same size will beA. `(n-n^(2//3))`B. `(n^(1//3)-n^(2//3))`C. `(n^(1//3-1))`D. `(n^(3)-1)` |
| Answer» Correct Answer - C | |
| 81. |
The action of a nib split at the top is explained byA. gravity flowB. Diffusion of fluidC. Capillar actionD. Osmosis of liquid |
| Answer» Correct Answer - C | |
| 82. |
A vessel whose bottom has round holes with a diameter of `d=0.1mm` is filled with water. The maximum height of the water level h at which the water does not flow out, will be (The water does not wet the bottom of the vessel). `[ST of water=70 "dyne"//cm]`A. `h = 24.0 cm`B. `h = 25.0 cm`C. `h = 26.0 cm`D. `h = 28.0 cm` |
| Answer» Correct Answer - D | |
| 83. |
A vessel, whose bottom has round holes with diameter of 0.1 mm , is filled with water. The maximum height to which the water can be filled without leakage is (S.T. of water =75 dyne/cm , g=1000 cm/s)A. 100 cmB. 75 cmC. 60 cmD. 30 cm |
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Answer» Correct Answer - D Balancing axcess pressure we have `h rho g=(2S)/r` `implies h=(2S)/(r rho g)=(2xx75)/(0.005xx1xx1000)=30` cm |
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| 84. |
Two drops of water of radius `2xx10^(-7)` m coalesce. What is the resultant rise in temperature? (S.T. of water `=74xx10^(-3) N m^(-1)`, sp. Heat capacity of water =4200 J/kg/K). |
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Answer» Correct Answer - `0.05^(@)C` |
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| 85. |
To what height can mercury be filled in a vessel without any leakage if there is a pin hole of diameter 0.1 mm at the bottom of the vessel? ( Density of mercury `=13.6xx10^(3) kg m^(-3)`, surface tension of mercury `=550xx10^(-3) N m^(-1)`, Neglect angle of contact. |
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Answer» Correct Answer - `16.49xx10^(-2)`m |
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| 86. |
Excess pressure inside a drop of water of radius 2 mm , is 70 `"N/m"^(2)` . The pressure in a drop of radius 4 mm isA. `55 N//m^(2)`B. `35 N//m^(2)`C. `45n//m^(2)`D. `25 N//m^(2)` |
| Answer» Correct Answer - B | |
| 87. |
The bubbles have radii in the ratio 3 : 4. the ratio of excess pressure inside them isA. `4:3`B. `3:4`C. `2:1`D. `4:1` |
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Answer» Correct Answer - A `(P_(1))/(P_(2))= (4T//r_(1))/(4T//r_(2)) = (r_(2))/(r_(1)) = 4/3` |
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| 88. |
Two liquid drop have diameters of 1 cm and 1.5 cm. The ratio of excess pressures inside them isA. `1:1`B. `5:3`C. `2:3`D. `3:2` |
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Answer» Correct Answer - D Excess pressure inside a liquid drop, `Delta p = (2S)/(R)` where, S is surface tension and R is radius of liquid drop. `:. (Delta p_(1))/(Delta p_(2)) = (R_(2))/(R_(1)) = (0.75)/(0.50) = (3)/(2)` |
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| 89. |
The ratio of radii of two bubbles is `2 : 1`. What is the ratio of excess pressures inside them ?A. `1:2`B. `1:4`C. `2:1`D. `4:1` |
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Answer» Correct Answer - A Excess pressure inside the bubble , `p = (4S)/(r)` For two different situation, `(p_(1))/(p_(2)) = (4S//r_(1))/(4S//r_(2)) = (r_(2))/(r_(1)) = (1)/(2)` |
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| 90. |
A mercury drop of radius 1 cm is broken into `10^6` droplets of equal size. The work done is `(T=35xx10^-2(N)/(m)`)A. `4.4xx10^(-3) J`B. `2.2xx10^(-4) J`C. `8.8xx10^(-4) J`D. `10^(4) J` |
| Answer» Correct Answer - A | |
| 91. |
Under a pressure head, the rate of orderly volume of liquid flowing through a capillary tube is Q. If the length of capaillary tube were doubled and diameter of the bore is halved, the rate of flow would becomeA. `(Q)/(4)`B. 16 QC. `(Q)/(8)`D. `(Q)/(32)` |
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Answer» Correct Answer - D As, `Q = (pi p r^(4))/(8 eta l)` and `Q_(1) = (pi r(r//2)^(4))/(8 eta (2l)) = (Q)/(32)` |
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| 92. |
A mercury drop of radius 1 cm is broken into `10^6` droplets of equal size. The work done is `(T=35xx10^-2(N)/(m)`)A. `4.35 xx 10^(-2) J`B. `4.35 xx 10^(-3) J`C. `4.35 xx 10^(-6) J`D. `4.35 xx 10^(-8) J` |
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Answer» Correct Answer - A If r is the radius of smaller droplet and R is the radius of bigger drop, then according to question, `(4)/(3) pi R^(3) = 10^(6) xx (4)/(3) pi r^(3) rArr r = (R)/(100) = 0.01 R` `0.01 xx 10^(-2) m = 10^(-4) m` `:.` Work done = Surface tension `xx` increase in area `= 35 xx 10^(-2) xx [(10^(6) xx 4pi xx (10^(-4))^(2) - 4pi xx (10^(-3))^(2)]` `= 4.35 xx 10^(-2) J` |
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| 93. |
A mercury drop of radius 1 cm is sprayed into `10^(6)` drops of equal size. The energy expended in joule is (surface tension of mercury is `(460xx10^(-3)N//m)`A. `5.7`B. `5.7 xx 10^(-4)`C. `0.057`D. `5.7 xx 10^(-6)` |
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Answer» Correct Answer - C dE `= 4piR^(2)T (n^(1//3)-1)` ` = 12 .56 xx 10^(-4) xx 46 xx10^(-2) xx 99 = 0.057` |
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| 94. |
A mercury drop of radius `1.0 cm` is sprayed into `10^(8)` droplets of equal size. Calculate the energy expanded. (Surface tension of mercury `= 32 xx 10^(-2) N//m`). |
| Answer» Correct Answer - `3.98 xx 10^(-2) J` | |
| 95. |
A mercury drop of radius 1.0 cm is sprayed into `10^(6)` droplets of equal sizes. The energy expended in this process is (Given, surface tension of mercury is `32 xx 10^(-2) Nm^(-1)`)A. `3.98 xx 10^(-4) J`B. `8.46 xx 10^(-4) J`C. `3.98 xx 10^(-2) J`D. `8.46 xx 10^(-2) J` |
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Answer» Correct Answer - C Let r be the radius of one droplet. Now, `(4)/(3)pi R^(3) = 10^(6) xx (4)/(3)pi r^(3)` `rArr r = (R)/(100) = (1)/(100) cm = 10^(-4) m` `A_(i) = 4pi R^(2)` and `A_(f) = 10^(6) xx 4pi r^(2)` Change in area, `Delta A = A_(f) = A_(i) = 4pi xx 99 xx 10^(-4) m^(2)` Increases in surface energy `= S Delta A = 32 xx 10^(-2) xx 4pi xx 99 xx 10^(-4) J = 3.98 xx 10^(-4) m^(2)` Increase in surface energy `= S Delta A = 32 xx 10^(-2) xx 4pi xx 99 xx 10^(-4) J = 3.98 xx 10^(-2) J` The increase in surface energy is on the expense of internal energy, so expended `= 3.98 xx 10^(-2) J` |
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| 96. |
A mercury drop of radius 1 cm is sprayed into `10^(6)` drops of equal size. The energy expended in joule is (surface tension of mercury is `(460xx10^(-3)N//m)`A. 0.057B. 5.7C. `5.7xx10^(-4)`D. `5.7xx10^(-6)` |
| Answer» Correct Answer - A | |
| 97. |
The oil is sprinkled on sea waves to calm them down. Why ?A. surface tension of water decreases so that oil spreads over waterB. surface tension of water increases so that water spreads overC. does not affect the surface tensionD. only water surface increases |
| Answer» Correct Answer - A | |
| 98. |
If more aire is pushed in a soap bubble the pressure in itA. decreasesB. increasesC. remains the sameD. is zero |
| Answer» Correct Answer - A | |
| 99. |
The hot soup taste better than cold soup , surface tension of hot soup isA. greater than surface tension of cold soupB. less than surface tension of cold soup so thatC. less than surface tension of cold soup so that hot soup spreads over larger area than cold soup.D. equal to surface tension of cold soup so that hot soup spreads over larger area than cold sup. |
| Answer» Correct Answer - C | |
| 100. |
Statement-1: hot soup tastes better than the cold soup. Statement-2: Hot soup spread properly on our tongue due to lower surface tension.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false |
| Answer» Correct Answer - C | |