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Quiz 6
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Q.1
The work done in blowing a bubble of radius R is W, then the work done in making a bubble of radius 2R is
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a)W/2
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b) 2W
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c)4W
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d)21/3W
Explanation
Work ∝ (radius)2 Thus W ∝ R2 W2∝(2R)2∴ W2 ∝ 4R2on taking ratio W2=4W Answer: (c)
Q.2
A soap bubble of diameter 8cm is formed in air. The surface tension of liquid is 30 dyne/cm. The excess pressure inside the soap bubble is [ PET 1990]
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a) 150 dyne/ cm2
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b) 30 dyne/ cm2
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c)3×10-3 dyne/ cm2
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d)12 dyne/ cm2
Explanation
Excess pressure=4T/R R=d/2Excess Pressure=4(30) /4=30dyne/ cm2Answer: (b)
Q.3
Two tubes of same material but of different radii are dipped in a liquid. The height to which a liquid rises in one tube is 2.2 cm and in the other tube is 6.6 cm . The ratio of their radii is [ PET 1990]
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a) 9 : 1
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b) 1 ; 9
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c)3 : 1
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d)1 : 3
Explanation
Height of liquid By taking the ratio of height we get Answer:(c)
Q.4
An oil drop of radius 1cm is divided into 1000 small equal drops of same radius. if the surface tension of oil drop is 50 dynes/cm then the work done is [ PET 1990]
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a) 18 π erg
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b) 180π erg
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c) 1800π erg
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d) 18000π erg
Explanation
If R is radius of bigger drop and r be the radius of small drop, if n is number of small drops then radius small drop r=R n-1/3 Thus Change in surface area=(n)4πr2 - 4πR2 Change in surface area=4π[n×r2- R2] Change in surface area=4π[n× n-2/3R2 - R2] Change in surface area=4πR[n× n-2/3 - 1]Change in surface area=4πR [ n1/3 - 1] Change in surface area=4π×1 [ 10001/3 - 1] Change in surface area==36πwork=Change in area × TW=36π×50=1800πAnswer: (c)
Q.5
The surface tension of soap is T. The work done in blowing a soap bubble of diameter D to that of diameter 2D is : [ MPPMT 1990]
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a)2πD2T
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b) 4πD2T
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c)6πD2T
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d)8πD2T
Explanation
Initial bubble surface area S1=4π(D/2)2 S1=π(D)2 Final bubble surface area S2=4π(2D/2)2 S2=4π(D)2 Increase in surface area ΔS=4π(D)2 - π(D)2 ΔS=3π(D)2 Energy=2ΔS×T Energy=2×3π(D)2 T Energy=6π(D)2TAnswer: (c)
Q.6
The surface tension of a liquid is T, then increase in its energy on increasing the area of the surface by A is [ PET 1991]
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a) AT-1
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b) AT
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c)A2T
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d)A2T2
Explanation
Answer: (b)
Q.7
If a glass rod is dipped in mercury and withdrawn out, the mercury does not wet the rod because [ MPPMT 1995]
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a) angle of contact is acute
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b) cohesion force is more
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c)adhesion force is more
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d)density of mercury is more
Explanation
Answer:(b)
Q.8
At which of the following temperatures, the value of surface tension of water is minimum [ MPPMT 1998]
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a) 4° C
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b) 25° C
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c) 50° C
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d) 75° C
Explanation
Answer: (d)
Q.9
The work done to increase unit area of liquid surface is called surface tension when [ Raj.PMT 1996]
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a)Temperature is constant
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b) Pressure is constant
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c)Volume is constant
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d)Adiabatic condition
Explanation
Answer: (a)
Q.10
A loop of thread is placed over a horizontal film of soap. if the thread is penetrated in the mid it acquires a circular shape of radius R. If the surface tension of soap is T, the tension in the thread is [ Raj.PET 1996]
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a) πR2 / T
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b) πR T
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c)2πR / T
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d)2πR T
Explanation
String has acquired circular shape of radius R but inside the circle there is no soap solution only outer side of ring is in contact with solution. circumference of ring=2πR Surface Tension T=Force/ length Force=T(length)=2πRTAnswer: (d)
Q.11
A soap bubble in vacuum has a radius of 3cm and another soap bubble in vacuum has a radius 4cm. If the two bubbles collaps under isothermal condition then the radius of the new bubble is [ MPPMT 1998]
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a) 2.3 cm
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b) 4.5 cm
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c)5.0 cm
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d)7.0 cm
Explanation
Process is isothermal, according to Boyle's lawP1V1 + P2V2=PV Since bubbles are in vacuum thus pressure difference=pressure inside the bubble=4T/RAnd Volume V=4πR2 Answer:(c)
Q.12
A capillary tube is dipped in water up to length l, the level of water reaches up to height h. Now the end which is inside the water is closed and capillary tube is put outside the water and that closed end is opened if l>h, the height of the remaining water column in the capillary will be [ Raj.PET 1996]
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a) 0
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b) l+h
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c) 2h
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d) h
Explanation
When capillary dipped in water pressure difference developed=2T/r Thus force up ward=(2T/R)(A) Down ward gravitational force=mgh Thus (2T/R)(A)=mgh --eq(1) After taking out the tube as shown in figure total upward force due to surface tension=(4T/r)(A) Let h' be the height of water in the tube thus down ward force=mgh' (4T/r)(A)=mgh' --eq(2) from equation (1) and (2) we get h'=2h Answer: (c)
Q.13
Surface tension of a liquid is found to be influenced by [ ISM Dhanabad 1996]
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a)Its increase with the increase of temperature
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b) nature of the liquid in contact
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c)presence of soap that increases it
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d)its variation with the concentration of liquid
Explanation
Answer: (d)
Q.14
The pressure of air inside a soap bubble of diameter 0.7 cm is 8mm of water above the pressure outside. the surface tension of soap solution is [ MPPMT 1997]
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a) 980 dynes/cm
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b) 68.6 dynes/cm
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c)72 dynes/cm
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d)137.3 dynes/cm
Explanation
Difference in pressure=8mm of water Difference in pressure=hρg Here ρ=density of water Difference in pressure=0.8 ×1×980=784 dynes/cm2 For a soap bubble difference in pressure=4T/r hρg=4T/r 784=4(T) / 0.35T=68.6 dynes/cm Answer: (b)
Q.15
The mass of water which rises in capillary tube of radius R is M, then the mass of water which rises in capillary tube of radius 2R will be [ Raj.PMT 1997]
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a) M
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b) 2M
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c)M/2
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d)4M
Explanation
We know that upward force on the liquid=2πRT cosθ Here R is radius of tube , T is surface tension Downward pull due to gravity=Mg Thus Mg=2πRT cosθ as θ, T, g are constantM ∝ R Answer:(b)
Q.16
If a capillary is dipped in water vertically, the height of water column in the capillary is 5cm. When the capillary is bent according to the figure, the height of water column will become
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a) 5 cm
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b) less than 5cm
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c) more than 5cm
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d) 4 cosα
Explanation
height of water column in the capillary is independent of depth of capillary in water, hence no change Answer: (a)
Q.17
A plate of area 10cm2 is separated from another plate by a 1mm thick layer of glycerine. If the coefficient of viscosity is 20 poise then the force required to move the upper plate with velocity of 1cm/sec over the lower one is [ MPPMT 1991]
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a) 80 dyne
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b) 200 dyne
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c)800 dyne
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d)2000 dyne
Explanation
Viscus force formulaHere A=10 cm2, dv=1 cm/sec, dx=0.1 cm and η=20 Answer: (d)
Q.18
Water is flowing through a horizontal pipe in stream line flow. At the narrowest part of the pipe [ MPPMT 1991]
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a) Velocity is maximum and pressure is minimum
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b) Pressure is maximum and velocity is minimum
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c)Both the pressure and velocity are minimum
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d)Both the velocity and pressure are minimum
Explanation
According to equation of continuity AV=av and velocity is inversely proportional to pressure Answer:(a)
Q.19
The clouds float in the atmosphere because of
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a) their low temperature
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b) their low viscosity
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c) their low density
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d) creation of low pressure
Explanation
Clouds can hold an enormous amount of water. When this water falls as rain it clearly has a significant mass so why don't clouds fall? In fact, the small water droplets that make up clouds do fall slowly. However, the drag force of the air dominates over the gravitational force for small particles. The drag force increases as the size of an object decreases. The force needed to move a sphere through a viscous medium is given by Stokes's law, F = 6πηRv. Here, R is the radius of the sphere, v is the velocity, and η is the viscosity. The viscosity of air is about 0.018×10-3 Pa·s and the viscosity of water is about 1.8×10-33 Pa·s. Answer: (b)
Q.20
Water is flowing through a tube of non-uniform cross-section. If the radius of the tube at the entrance and exit is 3 : 2 then the ratio of velocity of liquid entering and leaving the tube is [ PET 1991]
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a)8 : 27
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b) 4 : 9
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c) 1 : 1
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d) 9 : 4
Explanation
According to equation of continuity AV=av A/a=v/V R2 / r2=v/V 9/4=v/V V/v=4/9Answer: (b)
Q.21
The terminal speed of rain drop of radius 0.3mm in air is 1 m/sec. If the coefficient of viscosity of air is 18×10-5 poise, then the viscous force on the drop is [ PET 1990]
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a) 101.73 ×10-4 dyne
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b) 101.73 ×10-5 dyne
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c)16.95 ×10-5 dyne
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d)16.95 ×10-4 dyne
Explanation
Viscous force on the sphere F=6πηrv Answer: (a)
Q.22
The ratio of radii of the two sphere is 1:2 when they are dropped in a viscous liquid. The ratio of their terminal speed is [ PET 1990]
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a)1 : 1
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b)2 : 1
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c)1 : 2
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d)1 : 4
Explanation
Terminal speed formula is here d=density of the body σ=density of the medium η=coefficient of viscosity of the medium r=radius of sphere From the formula v∝ r2 Answer:(d)
Q.23
A fluid is flowing in a cylindrical pipe of internal diameter 4 cm with a velocity of 5 m/s. If this tube is joined with another tube of internal diameter 2cm, then the velocity of flow of liquid in the smaller tube will be [ PET 1990]
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a) 10 m/s
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b) 20 m/s
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c) 40 m/s
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d) 5 m/s
Explanation
According to equation for continuity AV=av Thus R2V=r2v 22×5=(1)2vv=20 m/s Answer: (b)
Q.24
Two drops of small radius are falling in air with velocity of 5 cm/s. If both the drops collapse then, the terminal velocity will be : [ MPPMT 1990]
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a)10 cm/s
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b) 2.5 cm/s
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c)5×(4)1/3 cm/s
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d)5×√2 cm/s
Explanation
Radius of big drop R=(2)1/3r Formula for terminal velocity as all other terms except radius are constant v ∝ r3Answer: (c)
Q.25
The level of water in a tank is 5m high. A hole of area 1 cm2 is made in the bottom of the tank. The rate of leakage of water from the hole is ( g=10 m/s2 [MPPMT 1990]
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a) 10-3 m3 / sec
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b) 10 -4 m3 / sec
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c)10 m3 / sec
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d)10 -2 m3 / sec
Explanation
Q=avQ=a √(2gh)Q=10-4×(2×10×h) Q=10-3 m3/secAnswer: (a)
Q.26
For streamline flow of fluid Bernoulli's theorem states that following remains constant [ MPPMT 1992]
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a)
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b)
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c)
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d)
Explanation
Answer:(c)
Q.27
A fluid of density 'd' and viscosity η is flowing through a pipe of radius r with a velocity 'v'. Reynold number R is [ MPPMT 1989]
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a)
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b)
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c)
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d)
Explanation
Answer: (b)
Q.28
Water is filled upto a height H in a tank. A hole is made at depth 'h' below the water level. The velocity of efflux of the fluid is [ PMT 1989]
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a)2gh
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b) gh
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c)√(2gh)
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d)√(gh)
Explanation
Answer: (c)
Q.29
A small sphere of radius R and density d is falling freely in a fluid of uniform density. It will fall with a terminal velocity which is proportional to [ MPPMT 1987]
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a) R2
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b) R
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c)R-1
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d)R-2
Explanation
Formula for terminal velocity Answer: (a)
Q.30
A solid ball of volume V is dropped in a viscous liquid. It experiences a viscous force F. If the solid ball of volume 2V of same material is dropped in the same fluid, then the viscous force acting on it will be [ PET 1989]
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a)nF / 2
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b) F/2
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c)2F
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d)2nF
Explanation
Viscous force F=6πηrvbut volume v ∝ r2∴ F ∝ V ∴ On doubling the volume of the drop, the viscous force is doubled Answer:(c)
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