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Physics MCQs for NEET — Practice Questions with Answers

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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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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)

$192 π \times 10^{- 4} J$

$280 π \times 10^{- 4} J$

$200 π \times 10^{- 4} J$

None of these

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Explanation

(a) $W = 8 {πr}^{2} \times T = 8 π \times (0.2)^{2} \times 0.06 = 192 π \times 10^{- 4} J$

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A liquid film is formed in a loop of area 0.05 $m^{2}$ . Increase in its potential energy will be (T = 0.2 N/m)

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

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

$3 \times 10^{- 2} J$

None of these

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Explanation

(b) Increment in Potential energy = $T \times ∆ A$

$= 0.02 \times 2 \times 0.05 = 2 \times 10^{- 2} J$

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In order to float a ring of area 0.04 $m^{2}$ in a liquid of surface tension 75 N/m, the required surface energy will be

3 J

6.5 J

1.5 J

4 J

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Explanation

(a) $E = T \times ∆ A = 75 \times 0.04 = 3 J$

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If two soap bubbles of equal radii r coalesce then the radius of curvature of interface between two bubbles will be

Infinity

1/2r

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Explanation

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When the temperature is increased the angle of contact of a liquid

Increases

Decreases

Remains the same

First increases and then decreases

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Explanation

(b) Cohesive force decreases so angle of contact decreases.

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