Surface Energy
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A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle at a constant speed. The speed of the stone is 4 m/sec. The tension in the string will be 52 N, when the stone is
At the bottom of the circle
the string makes an angle of 45 to the vertical
At the top of the circle
the string is horizontal
The radii of two soap bubbles are r 1 and r 2. In isothermal conditions, two meet together in vaccum. Then the radius of the resultant bubble is given by
r =(r1 + r2)/2
r =r1(r1r2+r2)
r2 =r12+r22
r =r1+r2
The work done in blowing a soap bubble of radius , given that the surface tension of soap solution is is
- r3
- r2
- r
- 1r
(Given: Surface tension of liquid, T=75 dyne/cm)
- 25π erg
- 29.4π erg
- 32π erg
- 28π erg
The volume of a spherical body is decreased by 10-3% when it is subjected to pressure of 40 atmospheres. Find the bulk modulus of body.
(1 atm = 1.01 x 103 N/m2).
4.04 x 106 N/m2
4.04 x 102 N/m2
4.04 x 103 N/m2
4.04 x 1011 N/m2
- 0.250 N m−1
- 0.125 N m−1
- 0.275 N m−1
- 0.325 N m−1
- 21/3:2
- 21/2:2
- 22/3:1
- 2:1
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)
19 erg
Zero
46 erg
74 erg
- 5π×10−12 J
- 4.2π×10−12 J
- 3.6π×10−12 J
- 1.8π×10−12 J
n drops of liquid each with surface energy E, join to form a single drop. In this process
No energy will be absorbed
Energy absorbed is E(n - n2/3)
Energy released will be E(n - n2/3)
Energy released will be E(22/3 - 1)
- Energy =3VT(1r+1R) is released
- Energy =3VT(1r+1R) is absorbed
- Energy =3VT(1r−1R) is released
- Energy =3VT(1r−1R) is absorbed
- Can't say
- Hydrogen gas
- Oxygen gas
- Both leak at equal rates
- liberated
- absorbed
- neither liberated nor absorbed
- 3.6π×10−4 J
- 0.6π×10−4 J
- 1.6π×10−4 J
- 2.6π×10−4 J
- 21/3:2
- 21/2:2
- 22/3:1
- 2:1
- 18 N/m
- 14 N/m
- 164 N/m
- 132 N/m