-
question_answer1)
Energy needed in breaking a drop of radius R into n drops of radii r is given by [CPMT 1982, 97]
A)
\[4\pi T(n{{r}^{2}}-{{R}^{2}})\] done
clear
B)
\[\frac{4}{3}\pi ({{r}^{3}}n-{{R}^{2}})\] done
clear
C)
\[4\pi T({{R}^{2}}-n{{r}^{2}})\] done
clear
D)
\[4\pi T(n{{r}^{2}}+{{R}^{2}})\] done
clear
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question_answer2)
The potential energy of a molecule on the surface of liquid compared to one inside the liquid is [MP PMT 1993]
A)
Zero done
clear
B)
Smaller done
clear
C)
The same done
clear
D)
Greater done
clear
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question_answer3)
Two droplets merge with each other and forms a large droplet. In this process [CBSE PMT 1993; RPMT 1997, 2000; CPMT 2001; BHU 2001; AFMC 2002]
A)
Energy is liberated done
clear
B)
Energy is absorbed done
clear
C)
Neither liberated nor absorbed done
clear
D)
Some mass is converted into energy done
clear
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question_answer4)
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) [CPMT 1989]
A)
Zero done
clear
B)
19 erg done
clear
C)
46 erg done
clear
D)
74 erg done
clear
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question_answer5)
Radius of a soap bubble is 'r', surface tension of soap solution is T. Then without increasing the temperature, how much energy will be needed to double its radius [CPMT 1991; Pb. PMT 2000; RPET 2001]
A)
\[4\pi {{r}^{2}}T\] done
clear
B)
\[2\pi {{r}^{2}}T\] done
clear
C)
\[12\pi {{r}^{2}}T\] done
clear
D)
\[24\pi {{r}^{2}}T\] done
clear
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question_answer6)
Work done in splitting a drop of water of 1 mm radius into 106 droplets is (Surface tension of water \[=72\times {{10}^{-3}}J/{{m}^{2}})\] [MP PET/PMT 1988; CPMT 1989; RPET 2001]
A)
\[9.58\times {{10}^{-5}}\,J\] done
clear
B)
\[8.95\times {{10}^{-5}}\,J\] done
clear
C)
\[5.89\times {{10}^{-5}}\,J\] done
clear
D)
\[5.98\times {{10}^{-6}}J\] done
clear
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question_answer7)
A spherical liquid drop of radius R is divided into eight equal droplets. If surface tension is T, then the work done in this process will be [CPMT 1990]
A)
2\[\pi {{R}^{2}}T\] done
clear
B)
3\[\pi {{R}^{2}}T\] done
clear
C)
4\[\pi {{R}^{2}}T\] done
clear
D)
2\[\pi R{{T}^{2}}\] done
clear
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question_answer8)
The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (T= surface tension of the solution) [MP PMT 1996]
A)
\[4\pi ({{D}^{2}}-{{d}^{2}})T\] done
clear
B)
\[8\pi ({{D}^{2}}-{{d}^{2}})T\] done
clear
C)
\[\pi ({{D}^{2}}-{{d}^{2}})T\] done
clear
D)
\[2\pi ({{D}^{2}}-{{d}^{2}})T\] done
clear
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question_answer9)
If T is the surface tension of soap solution, the amount of work done in blowing a soap bubble from a diameter D to 2D is [MP PMT 1990]
A)
2\[\pi {{D}^{2}}T\] done
clear
B)
4\[\pi {{D}^{2}}T\] done
clear
C)
6\[\pi {{D}^{2}}T\] done
clear
D)
8\[\pi {{D}^{2}}T\] done
clear
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question_answer10)
The radius of a soap bubble is increased from \[\frac{1}{\sqrt{\pi }}\]cm to \[\frac{2}{\sqrt{\pi }}\]cm. If the surface tension of water is 30 dynes per cm, then the work done will be [MP PMT 1986]
A)
180 ergs done
clear
B)
360 ergs done
clear
C)
720 ergs done
clear
D)
960 ergs done
clear
View Solution play_arrow
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question_answer11)
The surface tension of a liquid is 5 N/m. If a thin film of the area 0.02 m2 is formed on a loop, then its surface energy will be [CPMT 1977; MP PET 1989; BCECE 2005]
A)
\[5\times {{10}^{2}}\,J\] done
clear
B)
\[2.5\times {{10}^{-2}}\,J\] done
clear
C)
\[2\times {{10}^{-1}}\,J\] done
clear
D)
\[5\times {{10}^{-1}}\,J\] done
clear
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question_answer12)
If work W is done in blowing a bubble of radius R from a soap solution, then the work done in blowing a bubble of radius 2R from the same solution is [MP PET 1990]
A)
W/2 done
clear
B)
2W done
clear
C)
4W done
clear
D)
\[2\frac{1}{3}\]W done
clear
View Solution play_arrow
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question_answer13)
A spherical drop of oil of radius 1 cm is broken into 1000 droplets of equal radii. If the surface tension of oil is 50 dynes/cm, the work done is [MP PET 1990]
A)
18\[\pi \]ergs done
clear
B)
180\[\pi \]ergs done
clear
C)
1800\[\pi \]ergs done
clear
D)
8000\[\pi \]ergs done
clear
View Solution play_arrow
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question_answer14)
The work done in blowing a soap bubble of radius r of the solution of surface tension T will be [DPMT 1999; MP PMT 2003]
A)
\[8\pi {{r}^{2}}T\] done
clear
B)
\[2\pi {{r}^{2}}T\] done
clear
C)
\[4\pi {{r}^{2}}T\] done
clear
D)
\[\frac{4}{3}\pi {{r}^{2}}T\] done
clear
View Solution play_arrow
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question_answer15)
If two identical mercury drops are combined to form a single drop, then its temperature will [RPET 2000]
A)
Decrease done
clear
B)
Increase done
clear
C)
Remains the same done
clear
D)
None of the above done
clear
View Solution play_arrow
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question_answer16)
If the surface tension of a liquid is T, the gain in surface energy for an increase in liquid surface by A is [MP PET 1991; RPMT 2002]
A)
\[A{{T}^{-1}}\] done
clear
B)
\[AT\] done
clear
C)
\[{{A}^{2}}T\] done
clear
D)
\[{{A}^{2}}{{T}^{2}}\] done
clear
View Solution play_arrow
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question_answer17)
The surface tension of a soap solution is\[2\times {{10}^{-2}}N/m.\] To blow a bubble of radius 1 cm, the work done is [MP PMT 1989]
A)
\[4\pi \times {{10}^{-6}}J\] done
clear
B)
\[8\pi \times {{10}^{-6}}J\] done
clear
C)
\[12\pi \times {{10}^{-6}}J\] done
clear
D)
\[16\pi \times {{10}^{-6}}J\] done
clear
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question_answer18)
A mercury drop of 1 cm radius is broken into \[{{10}^{6}}\] small drops. The energy used will be (surface tension of mercury is \[35\times {{10}^{-3}}N/cm)\] [Roorkee 1984]
A)
\[4.4\times {{10}^{-3}}J\] done
clear
B)
\[2.2\times {{10}^{-4}}J\] done
clear
C)
\[8.8\times {{10}^{-4}}J\] done
clear
D)
\[{{10}^{4}}J\] done
clear
View Solution play_arrow
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question_answer19)
The surface tension of a liquid at its boiling point [MP PMT 1980]
A)
Becomes zero done
clear
B)
Becomes infinity done
clear
C)
is equal to the value at room temperature done
clear
D)
is half to the value at the room temperature done
clear
View Solution play_arrow
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question_answer20)
Surface tension of a soap solution is \[1.9\times {{10}^{-2}}N/m.\]. Work done in blowing a bubble of 2.0 cm diameter will be [MP PMT 1991]
A)
\[7.6\times {{10}^{-6}}\pi \]joule done
clear
B)
\[15.2\times {{10}^{-6}}\pi \]joule done
clear
C)
\[1.9\times {{10}^{-6}}\pi \]joule done
clear
D)
\[1\times {{10}^{-4}}\pi \]joule done
clear
View Solution play_arrow
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question_answer21)
The surface tension of liquid is 0.5 N/m. If a film is held on a ring of area 0.02 m2, its surface energy is [CPMT 1977]
A)
\[5\text{ }\times \text{ }{{10}^{-2}}joule\] done
clear
B)
\[2.0\text{ }\times \text{ }{{10}^{-2}}joule\] done
clear
C)
\[4\text{ }\times \text{ }{{10}^{-4}}joule\] done
clear
D)
\[0.8\text{ }\times \text{ }{{10}^{-1}}joule\] done
clear
View Solution play_arrow
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question_answer22)
What is ratio of surface energy of 1 small drop and 1 large drop, if 1000 small drops combined to form 1 large drop [CPMT 1990]
A)
100 : 1 done
clear
B)
1000 : 1 done
clear
C)
10 : 1 done
clear
D)
1 : 100 done
clear
View Solution play_arrow
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question_answer23)
The amount of work done in forming a soap film of size \[10\,cm\times 10\,cm\]is (Surface tension \[T=3\times {{10}^{-2}}\,N/m)\] [MP PET 1994; MP PET 2000]
A)
\[6\times {{10}^{-4}}\,J\] done
clear
B)
\[3\times {{10}^{-4}}\,J\] done
clear
C)
\[6\times {{10}^{-3}}\,J\] done
clear
D)
\[3\times {{10}^{-4}}\,J\] done
clear
View Solution play_arrow
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question_answer24)
The work done in blowing a soap bubble of 10 cm radius is (Surface tension of the soap solution is \[\frac{3}{100}N/m\]) [MP PMT 1995; MH CET 2002]
A)
\[75.36\times {{10}^{-4}}joule\] done
clear
B)
\[37.68\times {{10}^{-4}}joule\] done
clear
C)
\[150.72\times {{10}^{-4}}joule\] done
clear
D)
\[75.36joule\] done
clear
View Solution play_arrow
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question_answer25)
A liquid drop of diameter D breaks upto into 27 small drops of equal size. If the surface tension of the liquid is s, then change in surface energy is [DCE 2005]
A)
\[\pi {{D}^{2}}\sigma \] done
clear
B)
\[2\pi {{D}^{2}}\sigma \] done
clear
C)
\[3\pi {{D}^{2}}\sigma \] done
clear
D)
\[4\pi {{D}^{2}}\sigma \] done
clear
View Solution play_arrow
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question_answer26)
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 is [MP PET 1997; KCET 1999]
A)
1000 : 1 done
clear
B)
1 : 1000 done
clear
C)
10 : 1 done
clear
D)
1 : 10 done
clear
View Solution play_arrow
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question_answer27)
The work done in increasing the size of a soap film from 10 cm× 6 cm to 10 cm × 11 cm is 3 ×10-4 joule. The surface tension of the film is [MP PET 1999; JIPMER 2001, 02; MP PMT 2000; AIIMS 2000]
A)
\[1.5\times {{10}^{-2}}N/m\] done
clear
B)
\[3.0\times {{10}^{-2}}N/m\] done
clear
C)
\[6.0\times {{10}^{-2}}N/m\] done
clear
D)
\[11.0\times {{10}^{-2}}N/m\] done
clear
View Solution play_arrow
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question_answer28)
If s be the surface tension, the work done in breaking a big drop of radius R in n drops of equal radius is [Bihar CEET 1995]
A)
\[R{{n}^{2/3}}\sigma \] done
clear
B)
\[({{n}^{2/3}}-1)R\sigma \] done
clear
C)
\[({{n}^{1/3}}-1)R\sigma \] done
clear
D)
\[4\pi {{R}^{2}}({{n}^{1/3}}-1)\sigma \] done
clear
E)
\[\frac{1}{{{n}^{1/3}}-1}\sigma R\] done
clear
View Solution play_arrow
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question_answer29)
A big drop of radius R is formed by 1000 small droplets of water, then the radius of small drop is [AFMC 1998; Pb. PMT 2000]
A)
R/2 done
clear
B)
R/5 done
clear
C)
R/6 done
clear
D)
R/10 done
clear
View Solution play_arrow
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question_answer30)
When \[{{10}^{6}}\]small drops coalesce to make a new larger drop then the drop [RPMT 1999]
A)
Density increases done
clear
B)
Density decreases done
clear
C)
Temperature increases done
clear
D)
Temperature decreases done
clear
View Solution play_arrow
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question_answer31)
Which of the following statements are true in case when two water drops coalesce and make a bigger drop [Roorkee 1999]
A)
Energy is released done
clear
B)
Energy is absorbed done
clear
C)
The surface area of the bigger drop is greater than the sum of the surface areas of both the drops done
clear
D)
The surface area of the bigger drop is smaller than the sum of the surface areas of both the drops done
clear
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question_answer32)
8000 identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is [EAMCET (Engg.) 2000]
A)
1 : 10 done
clear
B)
1 : 15 done
clear
C)
1 : 20 done
clear
D)
1 : 25 done
clear
View Solution play_arrow
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question_answer33)
The surface energy of liquid film on a ring of area \[0.15\ {{m}^{2}}\] is (Surface tension of liquid \[=5N{{m}^{-1}})\] [EAMCET (Engg.) 2000]
A)
0.75 J done
clear
B)
1.5 J done
clear
C)
2.25 J done
clear
D)
3.0 J done
clear
View Solution play_arrow
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question_answer34)
8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of [DCE 2000]
A)
1 done
clear
B)
2 done
clear
C)
4 done
clear
D)
6 done
clear
View Solution play_arrow
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question_answer35)
If work done in increasing the size of a soap film from \[10\ cm\times 6\ cm\] to \[10\ cm\times 11\ cm\] is \[2\times {{10}^{-4}}J,\] then the surface tension is [AIIMS 2000]
A)
\[2\times {{10}^{-2}}N{{m}^{-1}}\] done
clear
B)
\[2\times {{10}^{-4}}N{{m}^{-1}}\] done
clear
C)
\[2\times {{10}^{-6}}N{{m}^{-1}}\] done
clear
D)
\[2\times {{10}^{-8}}N{{m}^{-1}}\] done
clear
View Solution play_arrow
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question_answer36)
A mercury drop of radius 1cm is sprayed into \[{{10}^{6}}\]drops of equal size. The energy expended in joules is (surface tension of Mercury is \[460\times {{10}^{-3}}N/m)\] [EAMCET 2001]
A)
0.057 done
clear
B)
5.7 done
clear
C)
\[5.7\times {{10}^{-4}}\] done
clear
D)
\[5.7\times {{10}^{-6}}\] done
clear
View Solution play_arrow
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question_answer37)
When two small bubbles join to form a bigger one, energy is [BHU 2001]
A)
Released done
clear
B)
Absorbed done
clear
C)
Both (a) and (b) done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer38)
A film of water is formed between two straight parallel wires of length 10cm each separated by 0.5 cm. If their separation is increased by 1 mm while still maintaining their parallelism, how much work will have to be done (Surface tension of water =\[7.2\times {{10}^{-2}}N/m)\] [MP PET 2001]
A)
\[7.22\times {{10}^{-6}}\,Joule\] done
clear
B)
\[1.44\times {{10}^{-5}}\,Joule\] done
clear
C)
\[2.88\times {{10}^{-5}}\,Joule\] done
clear
D)
\[5.76\times {{10}^{-5}}\,Joule\] done
clear
View Solution play_arrow
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question_answer39)
A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is \[0.465\ J/{{m}^{2}}\]) [UPSEAT 2002]
A)
\[23.4\mu J\] done
clear
B)
\[18.5\mu J\] done
clear
C)
\[26.8\mu J\] done
clear
D)
\[16.8\mu J\] done
clear
View Solution play_arrow
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question_answer40)
Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is [AIIMS 2003; DCE 2003]
A)
\[1:{{2}^{1/3}}\] done
clear
B)
\[{{2}^{1/3}}:1\] done
clear
C)
2 : 1 done
clear
D)
1 : 2 done
clear
View Solution play_arrow
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question_answer41)
Radius of a soap bubble is increased from R to 2R work done in this process in terms of surface tension is [BHU 2003, RPET 2001; CPMT 2004]
A)
\[24\pi {{R}^{2}}S\] done
clear
B)
\[48\pi {{R}^{2}}S\] done
clear
C)
\[12\pi {{R}^{2}}S\] done
clear
D)
\[36\pi {{R}^{2}}S\] done
clear
View Solution play_arrow
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question_answer42)
The work done in blowing a soap bubble of radius 0.2 m is (the surface tension of soap solution being 0.06 N/m) [Pb. PET 2002]
A)
\[192\pi \times {{10}^{-4}}J\] done
clear
B)
\[280\pi \times {{10}^{-4}}J\] done
clear
C)
\[200\pi \times {{10}^{-3}}J\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer43)
A liquid film is formed in a loop of area 0.05 m2. Increase in its potential energy will be (T = 0.2 N/m) [RPMT 2002]
A)
\[5\times {{10}^{-2}}J\] done
clear
B)
\[2\times {{10}^{-2}}J\] done
clear
C)
\[3\times {{10}^{-2}}J\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer44)
In order to float a ring of area 0.04 m2 in a liquid of surface tension 75 N/m, the required surface energy will be [RPMT 2003]
A)
3 J done
clear
B)
6.5 J done
clear
C)
1.5 J done
clear
D)
4 J done
clear
View Solution play_arrow
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question_answer45)
If two soap bubbles of equal radii r coalesce then the radius of curvature of interface between two bubbles will be [J&K CET 2005]
A)
r done
clear
B)
0 done
clear
C)
Infinity done
clear
D)
1/2r done
clear
View Solution play_arrow