JEE Physics Mechanical Properties Of Fluids MCQs Set 40

Multiple Choice Questions (MCQs) for JEE Physics: Mechanical Properties Of Fluids

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Multiple Choice Questions

Question 1: 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
(a) \( W/2 \)
(b) \( 2W \)
(c) \( 4W \)
(d) \( 2\tfrac13W \)
Show Answer & Explanation

Answer: (c) \( 4W \)

Explanation:
1. A soap bubble has two surfaces.

Teacher's Note:
1. Key relation used: Work \( =8\pi R^2T\propto R^2 \), so 4W.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 2: A spherical drop of oil of radius 1 cm is broken into 1000 droplets of equal radii. The surface tension of oil is 50 dyne/cm. The work done is
(a) \( 18\pi \) ergs
(b) \( 1800\pi \) ergs
(c) \( 180\pi \) ergs
(d) \( 8000\pi \) ergs
Show Answer & Explanation

Answer: (b) \( 1800\pi \) ergs

Explanation:
1. Radius of each droplet: \( r=R/10 \).
2. Increase in area \( =4\pi(10-1)=36\pi \) cm\(^2\).

Teacher's Note:
1. Work equals surface tension times increase in area.

Question 3: The work done in blowing a soap bubble of radius \( r \) of a solution of surface tension \( T \) will be
(a) \( 8\pi r^2T \)
(b) \( 2\pi r^2T \)
(c) \( 4\pi r^2T \)
(d) \( \tfrac43\pi r^2T \)
Show Answer & Explanation

Answer: (a) \( 8\pi r^2T \)

Explanation:
1. A bubble has an inner and an outer surface.

Teacher's Note:
1. Key relation used: Two surfaces: \( 2\times4\pi r^2T \).
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 4: If two identical mercury drops are combined to form a single drop, then its temperature will
(a) decrease
(b) increase
(c) remain the same
(d) none of these
Show Answer & Explanation

Answer: (b) increase

Explanation:
1. Merging reduces the total area.

Teacher's Note:
1. Key relation used: Surface area falls, so surface energy is released as heat.
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 5: If the surface tension of a liquid is \( T \), the gain in surface energy for an increase in liquid surface by \( A \) is
(a) \( AT^{-1} \)
(b) \( AT \)
(c) \( A^2T \)
(d) \( A^2T^2 \)
Show Answer & Explanation

Answer: (b) \( AT \)

Explanation:
1. Surface tension is energy per unit area.

Teacher's Note:
1. Key relation used: Energy \( =T\Delta A \).
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 6: The surface tension of a soap solution is \( 2\times10^{-2} \) N/m. To blow a bubble of radius 1 cm, the work done is
(a) \( 4\pi\times10^{-6} \) J
(b) \( 8\pi\times10^{-6} \) J
(c) \( 12\pi\times10^{-6} \) J
(d) \( 16\pi\times10^{-6} \) J
Show Answer & Explanation

Answer: (d) \( 16\pi\times10^{-6} \) J

Explanation:
1. Bubble means two surfaces.

Teacher's Note:
1. Key relation used: \( W=8\pi r^2T=8\pi\times10^{-4}\times2\times10^{-2}=16\pi\times10^{-6} \) J.
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 7: A mercury drop of 1 cm radius is broken into \( 10^6 \) small drops of equal size. The energy used is (surface tension of mercury is \( 35\times10^{-3} \) N/m)
(a) \( 4.4\times10^{-3} \) J
(b) \( 2.2\times10^{-4} \) J
(c) \( 8.8\times10^{-4} \) J
(d) \( 10^4 \) J
Show Answer & Explanation

Answer: (a) \( 4.4\times10^{-3} \) J

Explanation:
1. \( R=10^{-2} \) m.
2. \( \Delta A=4\pi(10^{-2}-10^{-4})=0.124 \) m\(^2\).

Teacher's Note:
1. Convert cm to m first.

Question 8: The surface tension of a soap solution is \( 1.9\times10^{-2} \) N/m. Work done in blowing a bubble of 2.0 cm diameter will be
(a) \( 7.6\times10^{-6}\pi \) J
(b) \( 15.2\times10^{-6}\pi \) J
(c) \( 1.9\times10^{-6}\pi \) J
(d) \( 1\times10^{-4} \) J
Show Answer & Explanation

Answer: (b) \( 15.2\times10^{-6}\pi \) J

Explanation:
1. Diameter 2 cm means radius 1 cm.

Teacher's Note:
1. Key relation used: \( W=8\pi r^2T=8\pi\times10^{-4}\times1.9\times10^{-2}=15.2\times10^{-6}\pi \) J.
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 9: What is the ratio of surface energy of 1 small drop and 1 large drop, if 1000 small drops combine to form 1 large drop?
(a) 100 : 1
(b) 1000 : 1
(c) 10 : 1
(d) 1 : 100
Show Answer & Explanation

Answer: (d) 1 : 100

Explanation:
1. Volume conservation gives \( R=10r \).

Teacher's Note:
1. Key relation used: \( R=10r \); energy \( \propto r^2 \), so 1 : 100.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 10: The amount of work done in forming a soap film of size 10 cm \( \times \) 10 cm is (surface tension \( T=3\times10^{-2} \) N/m)
(a) \( 6\times10^{-4} \) J
(b) \( 3\times10^{-4} \) J
(c) \( 6\times10^{-3} \) J
(d) \( 3\times10^{-3} \) J
Show Answer & Explanation

Answer: (a) \( 6\times10^{-4} \) J

Explanation:
1. A soap film has two faces.

Teacher's Note:
1. Key relation used: Two surfaces: \( W=2\times3\times10^{-2}\times0.01=6\times10^{-4} \) J.
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 11: The work done in blowing a soap bubble of 10 cm radius is (surface tension of the soap solution \( =\tfrac{3}{100} \) N/m)
(a) \( 75.36\times10^{-4} \) J
(b) \( 37.68\times10^{-4} \) J
(c) \( 150.72\times10^{-4} \) J
(d) 75.36 J
Show Answer & Explanation

Answer: (a) \( 75.36\times10^{-4} \) J

Explanation:
1. \( 75.36\times10^{-4} \) J.

Teacher's Note:
1. Key relation used: \( W=8\pi r^2T=8\pi(0.1)^2(0.03)=7.536\times10^{-3} \) J.
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 12: A liquid drop of diameter \( D \) breaks up into 27 small drops of equal size. If the surface tension of the liquid is \( \sigma \), then the change in surface energy is
(a) \( \pi D^2\sigma \)
(b) \( 2\pi D^2\sigma \)
(c) \( 3\pi D^2\sigma \)
(d) \( 4\pi D^2\sigma \)
Show Answer & Explanation

Answer: (b) \( 2\pi D^2\sigma \)

Explanation:
1. \( 27r^2=3R^2 \).

Teacher's Note:
1. Key relation used: \( r=R/3 \); \( \Delta A=4\pi R^2(3-1)=8\pi R^2=2\pi D^2 \).
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 13: 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
(a) 1000 : 1
(b) 1 : 1000
(c) 10 : 1
(d) 1 : 10
Show Answer & Explanation

Answer: (d) 1 : 10

Explanation:
1. Surface area drops by a factor of 10.

Teacher's Note:
1. Key relation used: \( R=10r \); final \( 4\pi R^2=100\cdot4\pi r^2 \) vs initial \( 1000\cdot4\pi r^2 \): 1 : 10.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 14: If \( \sigma \) is the surface tension, the work done in breaking a big drop of radius \( R \) into \( n \) drops of equal radius is
(a) \( Rn^{2/3}\sigma \)
(b) \( (n^{2/3}-1)R\sigma \)
(c) \( (n^{1/3}-1)R\sigma \)
(d) \( 4\pi R^2(n^{1/3}-1)\sigma \)
Show Answer & Explanation

Answer: (d) \( 4\pi R^2(n^{1/3}-1)\sigma \)

Explanation:
1. \( nr^2=R^2n^{1/3} \).

Teacher's Note:
1. Key relation used: \( r=Rn^{-1/3} \); \( \Delta A=4\pi R^2(n^{1/3}-1) \).
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 15: When \( 10^6 \) small drops coalesce to make a new larger drop, then the drop
(a) density increases
(b) density decreases
(c) temperature increases
(d) temperature decreases
Show Answer & Explanation

Answer: (c) temperature increases

Explanation:
1. Density is a material property.

Teacher's Note:
1. Key relation used: Energy is released as the surface area decreases.
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 16: 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
(a) 1 : 10
(b) 1 : 15
(c) 1 : 20
(d) 1 : 25
Show Answer & Explanation

Answer: (c) 1 : 20

Explanation:
1. Radius ratio \( \sqrt[3]{8000}=20 \).

Teacher's Note:
1. Key relation used: \( R=20r \); \( 4\pi R^2:8000\cdot4\pi r^2=400:8000=1:20 \).
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 17: The surface energy of a liquid film on a ring of area 0.15 m\(^2\) is (surface tension of liquid \( =5 \) N m\(^{-1}\))
(a) 0.75 J
(b) 1.5 J
(c) 2.25 J
(d) 3.0 J
Show Answer & Explanation

Answer: (b) 1.5 J

Explanation:
1. A film has two free surfaces.

Teacher's Note:
1. Key relation used: Two surfaces: \( 2\times5\times0.15=1.5 \) J.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 18: 8 mercury drops coalesce to form one mercury drop. The energy changes by a factor of
(a) 1
(b) 2
(c) 4
(d) 6
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Energy decreases by a factor 2.

Teacher's Note:
1. Key relation used: Initial area \( 8\times4\pi r^2 \), final \( 4\pi(2r)^2 \): half.
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 19: A mercury drop of 1 cm radius is sprayed into \( 10^6 \) drops of equal size. The energy expended in joules is (surface tension of mercury \( =460\times10^{-3} \) N/m)
(a) 0.057
(b) 5.7
(c) \( 5.7\times10^{-4} \)
(d) \( 5.7\times10^{-6} \)
Show Answer & Explanation

Answer: (a) 0.057

Explanation:
1. \( r=10^{-4} \) m.

Teacher's Note:
1. Key relation used: \( \Delta A=0.124 \) m\(^2\); \( W=0.124\times0.46=0.057 \) J.
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 20: When two small bubbles join to form a bigger one, energy is
(a) released
(b) absorbed
(c) both (a) and (b)
(d) none of these
Show Answer & Explanation

Answer: (a) released

Explanation:
1. Excess energy is released.

Teacher's Note:
1. Key relation used: Surface area decreases.
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 21: A film of water is formed between two straight parallel wires of length 10 cm 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 \( =7.2\times10^{-2} \) N/m)?
(a) \( 7.22\times10^{-6} \) J
(b) \( 1.44\times10^{-5} \) J
(c) \( 2.88\times10^{-5} \) J
(d) \( 5.76\times10^{-5} \) J
Show Answer & Explanation

Answer: (b) \( 1.44\times10^{-5} \) J

Explanation:
1. Two surfaces of the film.

Teacher's Note:
1. Key relation used: \( \Delta A=2\times0.1\times0.001=2\times10^{-4} \) m\(^2\); \( W=7.2\times10^{-2}\times2\times10^{-4}=1.44\times10^{-5} \) J.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 22: A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy (surface tension of mercury \( =0.465 \) J/m\(^2\))
(a) 23.4 \( \mu \)J
(b) 18.5 \( \mu \)J
(c) 26.8 \( \mu \)J
(d) 16.8 \( \mu \)J
Show Answer & Explanation

Answer: (a) 23.4 \( \mu \)J

Explanation:
1. Single surface for a drop.

Teacher's Note:
1. Key relation used: Each droplet has \( r=1 \) mm; \( \Delta A=4\pi(8r^2-R^2)=5.03\times10^{-5} \) m\(^2\); \( \Delta E=0.465\times5.03\times10^{-5}=23.4\ \mu\)J.
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 23: The radius of a soap bubble is increased from \( R \) to \( 2R \). The work done in this process in terms of surface tension \( S \) is
(a) \( 24\pi R^2S \)
(b) \( 48\pi R^2S \)
(c) \( 12\pi R^2S \)
(d) \( 36\pi R^2S \)
Show Answer & Explanation

Answer: (a) \( 24\pi R^2S \)

Explanation:
1. Two surfaces.

Teacher's Note:
1. Key relation used: \( W=2S\cdot4\pi(4R^2-R^2)=24\pi R^2S \).
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 24: A liquid film is formed in a loop of area 0.05 m\(^2\). The increase in its potential energy is (\( T=0.2 \) N/m)
(a) \( 5\times10^{-2} \) J
(b) \( 2\times10^{-2} \) J
(c) \( 3\times10^{-2} \) J
(d) none of these
Show Answer & Explanation

Answer: (b) \( 2\times10^{-2} \) J

Explanation:
1. Two surfaces of the film.

Teacher's Note:
1. Key relation used: \( 2\times0.2\times0.05=2\times10^{-2} \) J.
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 25: If two soap bubbles of equal radii \( r \) coalesce then the radius of curvature of the interface between the two bubbles will be
(a) \( r \)
(b) 0
(c) infinity
(d) \( 1/2r \)
Show Answer & Explanation

Answer: (c) infinity

Explanation:
1. \( \Delta P=4T/r \) on both sides.

Teacher's Note:
1. Key relation used: The excess pressures are equal so the interface is flat.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 26: The meniscus of mercury in the capillary tube is
(a) convex
(b) concave
(c) plane
(d) uncertain
Show Answer & Explanation

Answer: (a) convex

Explanation:
1. The mercury surface bulges upward.

Teacher's Note:
1. Key relation used: Mercury does not wet glass; angle of contact is obtuse.
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 27: If the work done in increasing the size of a soap film from 10 cm \( \times \) 6 cm to 10 cm \( \times \) 11 cm is \( 2\times10^{-4} \) J, the surface tension of the film is
(a) \( 2\times10^{-2} \) N m\(^{-1}\)
(b) \( 2\times10^{-4} \) N m\(^{-1}\)
(c) \( 2\times10^{-6} \) N m\(^{-1}\)
(d) \( 2\times10^{-8} \) N m\(^{-1}\)
Show Answer & Explanation

Answer: (a) \( 2\times10^{-2} \) N m\(^{-1}\)

Explanation:
1. Two surfaces of the film.

Teacher's Note:
1. Key relation used: \( \Delta A=2\times10\times5\ \text{cm}^2=0.01 \) m\(^2\); \( T=2\times10^{-4}/0.01=2\times10^{-2} \) N/m.
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

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Multiple Choice Questions (MCQs) for JEE Physics Mechanical Properties Of Fluids

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