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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

18 ergs

180 ergs

1800 ergs

8000 ergs

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Explanation

(c)

$W = 4 {πR}^{2} T (n^{1 / 3} - 1) = 4 π \times 1 \times 50 (10^{3 / 3} - 1) = 1800 π erg$

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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

$2 {πR}^{2} T$

$3 {πR}^{2} T$

$4 {πR}^{2} T$

$2 {πRT}^{2}$

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Explanation

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The radius of a soap bubble is increased from $\frac{1}{\sqrt{π}}$ cm to $\frac{2}{\sqrt{π}}$ cm. If the surface tension of water is 30 dynes per cm, then the work done will be

180 ergs

360 ergs

720 ergs

960 ergs

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Explanation

(c)

$W = 8 πT (r_{2}^{2} - r_{1}^{2}) = 8 πT [{(\frac{2}{\sqrt{π}})}^{2} - {(\frac{1}{\sqrt{π}})}^{2}] ∴ W = 8 \times π \times 30 \times \frac{3}{π} = 720 erg$

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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

W/2

$2 \frac{1}{3} W$

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Explanation

If radius becomes double then work done will become four times.

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If the surface tension of a liquid is T, the gain in surface energy for an increase in liquid surface by A is

${AT}^{- 1}$

$AT$

$A^{2} T$

$A^{2} T^{2}$

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Explanation

(b) Surface energy = surface tension × increment in area
= $T \times A$

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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

$4 π \times 10^{- 6} J$

$8 π \times 10^{- 6} J$

$12 π \times 10^{- 6} J$

$16 π \times 10^{- 6} J$

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Explanation

(d) $W = 8 {πR}^{2} T = 8 \times π \times (10^{- 2})^{2} \times 2 \times 10^{- 2} = 16 π \times 10^{- 6} J$

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The surface tension of a liquid at its boiling point

Becomes zero

Becomes infinity

is equal to the value at room temperature

is half to the value at the room temperature

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Explanation

(a)

As the temperature increases, surface tension decreases. At the boiling point, every molecule of a liquid is in motion from bottom to surface, and due to a rise in temperature, molecules lose the adhesion, and hence, surface tension becomes zero.

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The surface tension of liquid is 0.5 N/m. If a film is held on a ring of area 0.02 $m^{2}$ , its surface energy is

$5 \times 10^{- 2} joule$

$2 \times 10^{- 2} joule$

$4 \times 10^{- 4} joule$

$0.8 \times 10^{- 1} joule$

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Explanation

(b) Surface energy = $T \times ∆ A = 0.5 \times 2 \times (0.02) = 2 \times 10^{- 2} J$

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What is ratio of surface energy of 1 small drop and 1 large drop, if 1000 small drops combined to form 1 large drop ?

100 : 1

1000 : 1

10 : 1

1 : 100

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Explanation

(d) Volume of liquid remain same i.e. volume of 1000 small drops will be equal to volume of one big drop

$n \frac{4}{3} {πr}^{3} = \frac{4}{3} {πR}^{3} \Rightarrow 1000 r^{3} = R^{3} \Rightarrow R = 10 r ∴ \frac{r}{R} = \frac{1}{10} \frac{surface energy of one small drop}{surface energy of one big drop} = \frac{4 {πr}^{2} T}{4 {πR}^{2} T} = \frac{1}{100}$

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The amount of work done in forming a soap film of size is 10 cm X 10 cm(Surface tension $T = 3 \times 10^{- 2} N / m$ )

$6 \times 10^{- 4} J$

$3 \times 10^{- 4} J$

$6 \times 10^{- 3} J$

$3 \times 10^{- 3} J$

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Explanation

(a) $E = T \times ∆ A = 3 \times 10^{- 2} \times 2 (100 \times 10^{- 4}) = 6 \times 10^{- 4} J$

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