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Physics MCQs for NEET — Practice Questions with Answers

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The potential energy of a molecule on the surface of liquid compared to one inside the liquid is -

Zero

Smaller

The same

Greater

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Explanation

(d)

When the surface area of the liquid is increased molecules from the interior of the liquid rise to the surface. This work is stored in the molecules in the form of ofpotential energy. Thus, the potential energy of the molecules lying on the surface is greater than that of the molecules in the interior of the liquid.

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Two droplets merge with each other and forms a large droplet. In this process

Energy is liberated

Energy is absorbed

Neither liberated nor absorbed

Some mass is converted into energy

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Explanation

(a) When two droplets merge with each other, their surface energy decreases.

$W = T (∆ A)$ =(negative) i.e. energy is released

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Radius of a soap bubble is 'r', surface tension of soap solution is T. Then without increasing the temperature, how much energy will be needed to double its radius

$4 {πr}^{2} T$

$2 {πr}^{2} T$

$12 {πr}^{2} T$

$24 {πr}^{2} T$

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Explanation

(d) $W = 8 πT (R_{2}^{2} - R_{1}^{2}) = 8 π [(2 r)^{2} - (r)^{2}] = 24 {πr}^{2} T$

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Work done in splitting a drop of water of 1 mm radius into $10^{6}$ droplets is (Surface tension of water $= 72 \times 10^{- 3} J / m^{2}$ )

$9.58 \times 10^{- 5} J$

$8.95 \times 10^{- 5} J$

$5.89 \times 10^{- 5} J$

$5.98 \times 10^{- 5} J$

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Explanation

(b) Work done in splitting a water drop of radius R into n drops of equal size

$= 4 {πR}^{2} T (n^{1 / 3} - 1)$

$= 4 π \times (10^{- 3})^{2} \times 72 \times 10^{- 3} \times (10^{6 / 3} - 1) = 4 π \times 10^{- 6} \times 72 \times 10^{- 3} \times 99 = 8.95 \times 10^{- 5} J$

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The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (T= surface tension of the solution)

$4 π (D^{2} - d^{2}) T$

$8 π (D^{2} - d^{2}) T$

$π (D^{2} - d^{2}) T$

$2 π (D^{2} - d^{2}) T$

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Explanation

(d)

$W = T \times 8 π (r_{2}^{2} - r_{1}^{2}) = T \times 8 π (\frac{D^{2}}{4} - \frac{d^{2}}{4}) = 2 π (D^{2} - d^{2}) T$

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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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