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A liquid drop of diameter D breaks upto into 27 small drops of equal size. If the surface tension of the liquid is $σ$ , then change in surface energy is

${πD}^{2} σ$

$2 {πD}^{2} σ$

$3 {πD}^{2} σ$

$4 {πD}^{2} σ$

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Explanation

(b)

Work done = $4 {πR}^{2} T (n^{1 / 3} - 1) = 4 π {(\frac{D}{2})}^{2} ({σn}^{1 / 3} - 1) = {πD}^{2} σ (27^{1 / 3} - 1) = 2 {πD}^{2} σ$

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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

1000 : 1

1 : 1000

10 : 1

1 : 10

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Explanation

(d) As volume remain constant therefore $R = n^{1 / 3} r \frac{surface energy of one big drop}{surface energy of n drop} = \frac{4 {πR}^{2} T}{n \times 4 {πr}^{2} T} \frac{R^{2}}{{nr}^{2}} = \frac{n^{2 / 3} r^{2}}{{nr}^{2}} = \frac{1}{n^{1 / 3}} = \frac{1}{(1000)^{1 / 3}} = \frac{1}{10}$

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If $σ$ be the surface tension, the work done in breaking a big drop of radius R in n drops of equal radius is

${Rn}^{2 / 3} σ$

$(n^{2 / 3} - 1) Rσ$

$(n^{1 / 3} - 1) Rσ$

$4 {πR}^{2} (n^{1 / 3} - 1) σ$

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Explanation

$W = T ∆ A = σ (4 n π r^{2} - 4 π R^{2}) = 4 π σ (n r^{2} - R^{2}) A g a i n R^{3} = n r^{3} W = 4 π σ (n . n^{- 2 / 3} R^{2} - R^{2}) H e n c e, W = 4 π R^{2} (n^{1 / 2} - 1) σ$

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A big drop of radius R is formed by 1000 small droplets of water, then the radius of small drop is

R/2

R/5

R/6

R/10

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Explanation

(d) $\frac{4}{3} {πR}^{3} = 1000 \times \frac{4}{3} {πr}^{3}$ (As volume remains constant)

$R^{3} = 1000 r^{3} \Rightarrow R = 10 r \Rightarrow r = \frac{R}{10}$

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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

1 : 10

1 : 15

1 : 20

1 : 25

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Explanation

$\frac{surface energy of one big drop}{surface energy of 8000 small drop} = \frac{4 {πR}^{2} T}{8000 4 {πr}^{2} T}$

$= \frac{R^{2}}{8000 r^{2}} = \frac{(20 r)^{2}}{8000 r^{2}} = \frac{1}{20}$

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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

$2 \times 10^{- 2} {Nm}^{- 1}$

$2 \times 10^{- 4} {Nm}^{- 1}$

$2 \times 10^{- 6} {Nm}^{- 1}$

$2 \times 10^{- 8} {Nm}^{- 1}$

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Explanation

(a) $T = \frac{W}{∆ A} = \frac{2 \times 10^{- 4}}{2 \times (50 \times 10^{- 4})} = 2 \times 10^{- 2} N / m$

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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 $460 \times 10^{- 3} N / m$ )

0.057

5.7

$5.7 \times 10^{- 4}$

$5.7 \times 10^{- 6}$

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Explanation

(a) $W = T ∆ A = 4 {πR}^{2} T (n^{1 / 3} - 1)$

$= 4 \times 3.14 \times (10^{- 2})^{2} \times 460 \times 10^{- 3} \times [(10^{6})^{1 / 3} - 1] = 0.057$

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When two small bubbles join to form a bigger one, energy is

Released

Absorbed

Both (a) and (b)

None of these

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Explanation

(a)

When water droplets merge to form a bigger drop the total surface area decreases. Since the molecules in the surface have greater potential energy, the potential energy of surface molecules in the bigger drop decreases from before the merger. In other terms, the surface energy per unit area is equal to the surface tension. Since the surface tension remains the same, the surface energy will be less in the merged bigger drop. Hence the energy is liberated in the process.

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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 = $7.2 \times 10^{- 2} N / m$ )

$7.22 \times 10^{- 6} Joule$

$1.44 \times 10^{- 5} Joule$

$2.88 \times 10^{- 5} Joule$

$5.76 \times 10^{- 5} Joule$

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Explanation

(b) Increment in area of soap film = $A_{2} - A_{1}$

$= 2 \times [(10 \times 0.6) - (10 \times 0.5 \times)] \times 10^{- 4} = 2 \times 10^{- 4} m^{2}$

Work done = $T \times ∆ A$

$= 7.2 \times 10^{- 2} \times 2 \times 10^{- 4} = 1.44 \times 10^{- 5} J$

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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 $0.465 J / m^{2}$ )

$23.4 μJ$

$18.5 μJ$

$26.8 μJ$

$16.8 μJ$

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Explanation

(a) Increase in surface energy or work done in splitting a big drop $4 {πR}^{2} T (n^{1 / 3} - 1)$

$\Rightarrow W = 4 π \times (2 \times 10^{- 3})^{2} \times 0.465 (8^{1 / 3} - 1) = 23.4 μJ$

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