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Mechanical Properties of Fluids MCQs for NEET — Physics Questions with Answers

Practice free Mechanical Properties of Fluids (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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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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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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