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Physics MCQs for NEET — Practice Questions with Answers

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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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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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