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Physics NEET MCQ
Quiz 5
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Q.1
in case of liquids which do not wet the walls of the containing vessel, the force of adhesion is
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a) Less than √2 times the force of cohesion
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b) More than √2 times the force of cohesion
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c)Less than 1/√2 times the force of cohesion
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d)More than 1/√ times the force of cohesion
Explanation
Answer:(c)
Q.2
A capillary tube is dipped in a water container, so that loss in weight of the capillary tube is
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a) Equal to the upward buoyant force
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b) Less than the upward buoyant force
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c) More than the upward buoyant force
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d) Half of the buoyant force
Explanation
Answer: (b)
Q.3
The lower end of a capillary tube is at a depth of 12cm and the water rises 3cm in it. the mouth pressure required to blow an air bubble at the lower end will be X cm of water column, where X is
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a)3
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b) 9
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c)12
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d)15
Explanation
Answer: (d)
Q.4
Water proofing agent changes the angle of contact
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a) from an obtuse to acute value
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b) from an acute to obtuse value
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c)from obtuse to π/2
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d)from acute to π/2
Explanation
Answer: (b)
Q.5
What will be the height of the liquid column in a capillary tube on the surface of moon
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a) six times that on the surface of earth
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b) (1/6) of what was on earth's surface
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c)It will remain unchanged
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d)None of the preceding is true
Explanation
Answer:(a)
Q.6
A long capillary tube with both ends open is filled with water and then set in a vertical position in air. What is the length of the liquid column remaining in the tube? ( surface tension of water is 70 dyne/cm and the radius of the capillary bore is 1mm)
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a) 1.5 cm
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b) 3 cm
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c) zero
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d) 4.5 cm
Explanation
As shown in figure weight of liquid column is balance by pressure difference created due to surface tension 2T/r at upper end lower end Thus 4T/r=hdg , here d is density of liquid Answer: (b)
Q.7
A given mass of a metal is moulded into solids of different shapes. Its surface area is least when it is
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a)A circular disc
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b) A parabolic of revolution
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c)A right cone
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d)A sphere
Explanation
Answer: (d)
Q.8
Two water droplets merge with each other to form a large droplet. In this process [ Raj.PMT 1997]
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a) Energy is liberated
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b) Energy is absorbed
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c)Energy is neither liberated nor absorbed
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d)Some mass is converted into energy
Explanation
When big drops is formed from small drops, surface area decreases, As the surface area decreases, energy corresponding to surface area, the surface tension energy, also decreases. So excess energy is liberated. When big drop is divided in small drop, sum of the surface area of all drops increases, thus total energy of surfaces increases. It may take that energy from the work done on the drop to divide it in small drop Answer: (a)
Q.9
Under a pressure head, the rate of orderly volume flow of liquid through capillary tube is Q, if the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become [ MPPMT 1988]
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a) Q/4
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b) Q/16
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c)Q/32
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d)Q/8
Explanation
Rate of volume flow Q ∝ r4 /l∴ on reducing the radius to r/2, and lengthg is now doubled the volume flowing through the tube becomes Q/32 Answer:(c)
Q.10
A ring is cut from a platinum tube having 8.5 cm internal and 8.7 cm external diameter. it is supported horizontally from a pair of balance so that it comes in contact with the water in glass vessel. If an extra 3.97 gm weigh is required to pull it away from water, the surface tension of water is [ MNR 1992]
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a) 73.4 dyne/cm
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b) 70.80 dyne/cm
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c)65.35 dyne/cm
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d)60.00 dyne/cm
Explanation
Ring have two perimeters thus total=2πr1+2πr2 r1=8.5/2r1=8.7/2Surface tension force=T× total perimeter S.T force=[2πr1+2πr2]TS.T force=2π[r1 +r2]T 3.97×103=2×3.14[8.5 /2 + 8.7/2]×T3.97×103=(22/7)[8.5 + 8.7]×T3.97×103=54.057 × TT=73.42 dyne Answer:(a)
Q.11
The excess of pressure in a soap bubble of radius R and surface tension T is given by [ BHU 1995]
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a) P=2T/R
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b) P=4T/R
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c) P=T/R
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d) P=6T/R
Explanation
Answer: (b)
Q.12
Consider a liquid contained in a vessel. the liquid-solid adhesive force is very weak as compared to the cohesive force in the liquid. the shape of the liquid-surface near the solid shall be [ MNR 1994]
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a) horizontal
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b) almost vertical
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c)concave
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d)convex
Explanation
Answer: (d)
Q.13
S.T. of water is 5N/m. If the film is formed on a ring of area 0.02m2, then its surface energy is [ PET 1989]
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a) 5×10⁻² Joule
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b) 2.5×10⁻² Joule
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c)2×10⁻¹ Joule
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d)3×10⁻¹
Explanation
T=Energy / (2×Area) Energy=T×(2×Area)E=5×(2×0.02)=0.2 JAnswer: (c)
Q.14
If work done in blowing a bubble of volume V is W then the work done in blowing a soap bubble of volume 2V will be [ PET 1898]
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a) W
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b) 2W
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c)√ W
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d)(2)2/3 W
Explanation
Work=2×ΔS × T W1=2S1TW2=2S2TW1 / W2=S1 / S2W1 / W2=1 / (22/3)W2=W1 (22/3) Answer:(d)
Q.15
The force required to drag a circular flat plate of radius 5cm on the surface of water is (S.T. of water is 75 dynes/cm) [ MPPMT 1991]
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a) 30 dynes
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b) 60 dynes
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c) 750 dynes
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d) 750π dynes
Explanation
W= T (2πr) W=75×2×π×5 W=750π Answer: (d)
Q.16
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.17
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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d)12 dyne/ cm2
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c)3×10⁻³ dyne/ cm2
Explanation
Excess pressure=4T/R R=d/2Excess Pressure=4(30) /4=30dyne/ cm2Answer: (b)
Q.18
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.19
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.20
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.21
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.22
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.23
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.24
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.25
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.26
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.27
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.28
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.29
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.30
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)
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