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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Which of the following is a characteristic of turbulent now?A. velocity more than critical velocityB. irregular flowC. molecule crossing from one layer to the otherD. 1,2,3 |
| Answer» Correct Answer - D | |
| 52. |
Assertion: The flow is turbulent for Reynolds number greater than `2000` Reason: Turbulence dissipates kinetic energy in the form of heat.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true and reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B The flow is streamline or laminar for Reynolds number less than 1000 and flow is turbulent for Reynolds number greater than 2000. |
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| 53. |
In turbulent flow the velocity of the liquid molecules in contact with the walls of the tube.A. is zeroB. is maximumC. is equal to critical velocityD. may have any value |
| Answer» Correct Answer - A | |
| 54. |
What should be the average velocity of water in a tube of radius 2.5 mm so that the flow is just turbulent? Viscosity of water is 0.001 Pa s. Hint : For flow to be just turbulent `R_(e)` = 2000 |
| Answer» Correct Answer - 0.6 m `s^(-1)` | |
| 55. |
The angle of contact of a liquid in a capillary tube is `60^(@)`. The shape of the meniscus isA. ConcaveB. ConvexC. FlatD. None of these |
| Answer» Correct Answer - A | |
| 56. |
Select the incorrect option (T = surface tension, R = radius of drop, bubble, meniscus)A. Excess pressure inside a liquid drop in air is equal to `(2T)/(R)`B. Excess presure inside a liquid bubbles in air is equal to `(2T)/(R)`C. Excess pressure for spherical liquid meniscus is `(4T)/(R)`D. Excess pressure for cylindrical bubble in air is `(4T)/(R)` |
| Answer» Correct Answer - A::C::D | |
| 57. |
The liquid meniscus in a capillary tube will be convex, if the angle of contact isA. greater that `90^(@)`B. less than `90^(@)`C. equal to `90^(@)`D. equal to zero |
| Answer» Correct Answer - A | |
| 58. |
A vertical U-tube contains a liquid of density` rho` and surface tension T. if the radius of the meniscus of liquid in the limbs of the U-tube are `R_(1)` and `R_(2)` find the difference in the liquid column in the limbs.A. `Deltah=(T(R_(1)-R_(2)))/(rhogR_(1)R_(2))`B. `Deltah=(2T(R_(1)-R_(2)))/(rhogR_(1)R_(2))`C. `Deltah=(2T(R_(1)+R_(2)))/(rhogR_(1)R_(2))`D. `Deltah=(4T(R_(1)-R_(2)))/(rhogR_(1)R_(2))` |
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Answer» Correct Answer - B Let the heights of liquid column in the limbs are `h_(1)` and `h_(2)` using the formula `Deltap=(2T)/(R)` for the meniscus in the limbs, we have the pressures at the points A and B gives as `P_(A)=P_(0)-(2T)/(R_(1))` and `P_(B)=P_(0)-(2T)/(R_(2))` Using the formula `Deltap=rhogh`, we have the pressures at A and C `P_(A)=(P_(C))-P_(B)=rhog(h_(2)-h_(1))`, substituting `P_(A)` from eq. (i) from eq (ii) in eq (iii), we have `[p_(0)-(2T)/(R_(1))]-[p_(0)-(2T)/(R_(2))]=rhogDeltah` atmospheric pressure. Suppose a and b are the radio of the this gives `Deltah=(2T(R_(1)-R_(2)))/(rhogR_(1)R_(2))` |
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| 59. |
An inverted u-tube has its two limbs in water and kerosene contained in two beakers. If water rises to a height of 10 cm to what height does kerosene (density `=0.8gm//c c)` rise in te other limb?A. 10 cmB. 12.5 cmC. 15 cmD. 20 cm |
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Answer» Correct Answer - B `P_(1)=P_(2)impliesh_(1)rho_(1)g=h_(2)rho_(2)gimpliesh_(1)rho_(1)=h_(2)rho_(2)` |
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| 60. |
Fig, shows a U-tube of uniform cross-sectional area `A` accelerated with acceleration a as shown. If `d` is the separation between the limbs. Then the difference in the levels of the liquid in the `U-tube` is A. `(ad)/(g)`B. `(g)/(ad)`C. `adg`D. `ad+g` |
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Answer» Correct Answer - A `tantheta=(a)/(g)=(h)/(d)` |
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| 61. |
A block of mass 1 kg and density `0.8g//cm^(3)` is held stationary with the help of a string as shown in figure. The tank is accelerating vertically upwards with an acceleration a `=1.0m//s^(2)`. Find (a) the tension in the string, (b) if the string is now cut find the acceleration of block. (Take `g=10m//s^(2)` and density of water `=10^(3)kg//m^(3))`.A. `T=2.2N`B. `T=2.75N`C. `T=3N`D. `T=2.4N` |
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Answer» Correct Answer - B `F="upthrust force"=Vrho_(w)(g+a)` `=(("mass of block")/("Density of block"))rho_(w)(g+a)` `=(1)/(800)(1000)(11)=13.75N` `F-T-W=ma,13.75-T-10=1(1)` `T=2.75N` |
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| 62. |
A spherical solild of volume V is made of a material of density `rho_(1)`. It is falling through a liquid of density `rho_(2)(rho_(2) lt rho_(1))`. Assume that the liquid applies a viscous froce on the ball that is proportional ti the its speed v, i.e., `F_(viscous)=-kv^(2)(kgt0)`. The terminal speed of the ball isA. `sqrt((Vg(rho_(1)-rho_(2)))/(k))`B. `(Vgrho_(1))/(k)`C. `sqrt((Vgrho_(1))/(k))`D. `(Vg(rho_(1)-rho_(2)))/(k)` |
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Answer» Correct Answer - A The force acting on the ball are gravity force, buoyancy force and viscous force. When ball acquires terminal speed, it is in dynamic equilibrium,let termina speed of ball is `V_(T)`. so, `vrho_(2)g+kv_(T)^(2)=Vrho_(1)g,V_(T)=sqrt((V(rho_(1)-rho_(2))g)/(k))` |
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| 63. |
A spherical solild of volume V is made of a material of density `rho_(1)`. It is falling through a liquid of density `rho_(2)(rho_(2) lt rho_(1))`. Assume that the liquid applies a viscous froce on the ball that is proportional ti the its speed v, i.e., `F_(viscous)=-kv^(2)(kgt0)`. The terminal speed of the ball isA. `sqrt((Vgrho_(1))/(k))`B. `(Vg(rho_(1)-rho_(2)))/(k)`C. `sqrt((Vg(rho_(1)-rho_(2)))/(k))`D. `(Vgrho_(1))/(k)` |
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Answer» Correct Answer - C `mg=F_(b)+F_(v)` |
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| 64. |
Eight raindrops each of radius R fall through air with teminal velocity 6 cm `s^(-1)`. What is the terminal velocity of the bigger drop formed by coalescing these drops together ?A. 18 cm `s^(-1)`B. 24 cm `s^(-1)`C. 15 cm `s^(-1)`D. 20 cm `s^(-1)` |
| Answer» Correct Answer - B | |
| 65. |
The surface tension of soap solution is 0.3 `(N)/(m)`. The work done in blowing a soap bubble of surface area `40cm^(2)`, (in J) isA. `1.2xx10^(-4)`B. `2.4xx10^(-4)`C. `12xx10^(-4)`D. `24xx10^(-4)` |
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Answer» Correct Answer - 2 In case of soap bubble `W=Txx2xxDeltaA=0.03xx2xx40xx10^(-4)=2.4xx10^(-4)J` |
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| 66. |
The work done is blowing a soap bubble of volume `V` is `W`. The work done in blowing a soap bubble of volume `2V` isA. `W`B. `2^((2)/(3))W`C. `3^((2)/(3))W`D. `2W` |
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Answer» Correct Answer - B `W=8pir^(2)T,V=(4)/(3)pir^(3),Valphar_(3),(W_(2))/(W_(1))=[(V_(2))/(V_(1))]^((2)/(3))` |
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| 67. |
The surface tension of a soap solution is 0.05 `Nm^(-1)` How much work is done to produce a soap bubble of radius 0.03 m?A. `1.8xx10^(-2)J`B. `2.1xx10^(-3)J`C. `1.5xx10^(-2)J`D. `1.1xx10^(-3)J` |
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Answer» Correct Answer - D `U=AxxT` and `w=DeltaU` |
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| 68. |
The amount of work done in blowing up a soap bubble of radius 2 cm is [Given surface tension of soap solution = 4 `xx 10^(-2)` N `m^(-1)`]A. `1.28 xx 10^(-4)` JB. `32 xx 10^(-4)` JC. `1.08 xx 10^(-4)` JD. `4.02 xx 10^(-4)` J |
| Answer» Correct Answer - D | |
| 69. |
The surface tension of a soap solution is `30 xx 10^(-3) N m^(-1)`. How much work is done to increase the radius of a soap bubble from 2 cm to 3 cm? |
| Answer» Correct Answer - `3.77 xx 10^(-4)` J | |
| 70. |
A soap bubble (surface tension T) is charged to a uniform charged density `sigma`. At equilibrium the radius of the bubble is given by `(Nepsilon_(0)T)/(sigma^(2))`. The value of N is [Assume that atmosphere is not present] |
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Answer» Correct Answer - 8 At equilibrium just before bursting the force due to surface tension is balanced by electrostatic repulstion so `(4T)/(r)xx"area"=(sigma^(2))/(2epsilon_(0))xx"area"impliesr=(8epsilon_(0)T)/(sigma^(2))` |
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| 71. |
In the absence of any external force, liquid drop are sperical in shape. It is due toA. Surface tensionB. Energy conservationC. PressureD. Viscosity |
| Answer» Correct Answer - A | |
| 72. |
What should be the pressure inside a small air bubble of `0.1 mm` radius situated just below the surface of water? Surface tension of water `=72xx10^(-3)N//m` and atmospheric pressure `=1.013xx10^(5)N//m^(2)`A. `1.44xx10^(2)Pa`B. `1.44xx10^(3)Pa`C. `1.44xx10^(4)Pa`D. `1.44xx10^(5)Pa` |
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Answer» Correct Answer - 2 `P=P_(0)+(2T)/(r)` |
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| 73. |
Find the total pressure inside a sperical air bubble of radius 0.2 mm. The bubble is at a depth 5 cm below the surface of a liquid of density `2 xx 10^(3)` kg `m^(-3)` and surface tension 0.082 N `m^(-1)`. (Given : atmospheric pressure = 1.01 `xx 10^(5) Nm^(-2)`) Hint : `P_(i) = P_(o) + (2S)/(R)` `P_(o)` = atmospheric pressure + gauge pressure of liquid column. |
| Answer» `1.028 xx 10^(5) N m^(-2)` | |
| 74. |
The depth of water at which air bubble of radius 0.4 mm remains in equilibrium is `(T_("water")=72xx10^(-3)N//m)`A. `3.67cm`B. `3.67m`C. `6.37cm`D. `5.32cm` |
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Answer» Correct Answer - A `hdg=(2T)/(r)` |
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| 75. |
Water rises in a capillary tube to a height `H`, when the capillary tube is vertical. If the same capillary is now inclined to the vertical the length of water column in it willA. increaseB. decreaseC. will not changeD. may increase or decrease depending on the angle of inclination. |
| Answer» Correct Answer - A | |
| 76. |
A hemispherical bowl just floats without sinking in a liquid of density `1.2xx10^(3)kg//m^(3)`. If outer diameter and the density of the bowl are `1m` and `2 xx 10^(4) kg//m^(3)` respectively, then the inner diameter of bowl will beA. 1.91 mB. 0.5 mC. 0.98 mD. 1.75 m |
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Answer» Let `d` the inner diameter weight fo displaced liquid`=` weight of the body `m^(1)g=mg` Vol. of displaced. Liquid`xx` density of liquid `=` vol. of the body `xx` density of the body `(4pi)/(3)((1)/(2))^(3)xx1.2xx10^(3)=(4pi)/(3)[((1)/(2))^(3)-((d)/(2))^(3)]xx2xx10^(4)` |
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| 77. |
Statement-1 : In a capillary tube, excess pressure is balanced by hydrostatic pressure and not by weight. And Statement-2 : The vertical height of a liquid column in capillaries of different shapes and sizes will be same if the radius of meniscus remains same.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - B | |
| 78. |
A tortoise is just sinking in water of density `rho` The tortoise is assumed to be3 a hemisphere of radius `R`. Q. Find the total hydrostatic forceA. `rhogpiR^(3)`B. `sqrt((13)/(3))rhogpiR^(3)`C. `(2)/(3)rhogpiR^(3)`D. `sqrt((16)/(3))rhogpiR^(3)` |
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Answer» Correct Answer - B Hence the net hydrostatic force on the tortoise is `F=sqrt(F_(h)^(2)+F_(v)^(2))=sqrt((rhogpiR^(3))^(2)+((2)/(3)rhogpiR^(3))^(2))` `=sqrt((13)/(3))rhogpiR^3` |
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| 79. |
When equal volumes of two metals are mixed together the specific gravity of alloy is 4. When equal masses of the same two metals are mixed together the specific gravity of the alloy not becomes 3. find specific gravity of each metal? (specific gravity `=("density of substance")/("density of water"))` |
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Answer» In case of mixture, `rho_(mix)=(m_(1)+m_(2))/(V_(1)+V_(2))` When equal volumes are mixed, `4=(Vrho_(1)+Vrho_(2))/(V+V)=(rho_(1)+rho_(2))/(2)` ..(i) When equal masses are mixed, `3=((m+m))/((m)/(rho_(1))+(m)/(rho_(2)))=(2rho_(1)rho_(2))/(rho_(1)+rho_(2))` ..(ii) Therefore from (i) and (ii) specific gravity of the metals are 2 and 6. |
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| 80. |
Figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height `h_(0)` and pressure `2p_(0)` where `rh_(0)` is the atmospheric pressure. There is a hole in the wall of the tank at a depth `h_(1)` below the top from which water comes out. A long vertical tube is connected as shown. Find the speed with which water comes out of the holeA. `(1)/(rho)[p_(0)-rhog(h_(1)-2h_(0))]^(1//2)`B. `[(2)/(rho)[p_(0)+rhog(h_(1)-h_(0))]]^(1//2)`C. `[(3)/(rho)[p_(0)+rhog(h_(1)+h_(0))]]^(1//2)`D. `[(4)/(rho)[p_(0)-rhog(h_(1)-h_(0))]]^(1//2)` |
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Answer» Correct Answer - B `2P_(0)=(h_(2)+h_(0))rhog+p_(0)` (since liquids at the same level have the same pressure) `P_(0)=h_(2)rhog+h_(0)rhog,h_(2)rhog=P_(0)-h_(0)rhog` `h_(2)=(P_(0))/(rhog)+(h_(0)rhog)/(rhog)=(P_(0))/(rhog)-h_(0)` KE to the water `=` pressure energy of the water at that layer `(1)/(2)mV^(2)=mxx(P)/(rho)` `V^(2)=(2P)/(rho)=(2)/(rho)[P_(0)+rhog(h_(1)+h_(0))]` `V=[(2)/(rho){P_(0)+rhog(h_(1)-h_(0))}]^(1//2)` We know `2P_(0)+rhog(h_(1)-h_(0))=P_(0)+rhogX` `impliesX=(P_(0))/(rhog)+(h_(1)-h_(0))=h_(2)+h_(1)` i.e., X is `h_(1)` meter below the top or X is `h_(2)` above the top. |
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| 81. |
Figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height `h_(0)` and pressure `2p_(0)` where `rh_(0)` is the atmospheric pressure. There is a hole in the wall of the tank at a depth `h_(1)` below the top from which water comes out. A long vertical tube is connected as shown. Find the height `h_(2)` of the water in the long tube above top initially.A. `(3p_(0))/(rhog)-(h_(0))/(3)`B. `(2p_(0))/(rhog)-(h_(0))/(2)`C. `(p_(0))/(rhog)-h_(0)`D. `(p_(0))/(2rhog)-2h_(0)` |
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Answer» Correct Answer - C `2P_(0)=(h_(2)+h_(0))rhog+p_(0)` (since liquids at the same level have the same pressure) `P_(0)=h_(2)rhog+h_(0)rhog,h_(2)rhog=P_(0)-h_(0)rhog` `h_(2)=(P_(0))/(rhog)+(h_(0)rhog)/(rhog)=(P_(0))/(rhog)-h_(0)` KE to the water `=` pressure energy of the water at that layer `(1)/(2)mV^(2)=mxx(P)/(rho)` `V^(2)=(2P)/(rho)=(2)/(rho)[P_(0)+rhog(h_(1)+h_(0))]` `V=[(2)/(rho){P_(0)+rhog(h_(1)-h_(0))}]^(1//2)` We know `2P_(0)+rhog(h_(1)-h_(0))=P_(0)+rhogX` `impliesX=(P_(0))/(rhog)+(h_(1)-h_(0))=h_(2)+h_(1)` i.e., X is `h_(1)` meter below the top or X is `h_(2)` above the top. |
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| 82. |
A soap bubble of radius 6 cm and another bubble of 8 cm coalesce under isothermal xonditions in vacuum. The radius of the new bubble isA. 3 cmB. 4 cmC. 10 cmD. 7 cm |
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Answer» Correct Answer - 3 `r=sqrt(r_(1)^(2)+r_(2)^(2))` |
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| 83. |
statement 1 is false, statement 2 is true. Statement-1: A soft plastic bag weights the same when empty or when filled with air and measured in vacuum. Statement-2: The same results will be observed when measured in air.A. Statement-I is true, statement-2 true and statements-2 is a correct explanation for statements 1B. Statement 1 is true, statement 2 is true, statement-2 is not a correct explanation for statement 1C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - D If the bag is weighed in vacuum its weight will be `W_(0)` when empty and `(W_(0)+V_(rhog))` when filled with air. But weighed in air, its weight will be `W_(0)` when empty and `(W_(0)+Vrhog-Vrhog)=W_(0)` when filled with air due to upthrust of air `(Vrhog)` |
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| 84. |
Two blocks A and B float in water. If block A floats with `(1)/(4)` th of its volume immersed and block B floats with `(3)/(5)` th of its volume immersed, the ratio of their densities isA. `5:12`B. `12:5`C. `3:20`D. `20:3` |
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Answer» Correct Answer - 1 `(V_("in"))/(V)=(rho("body"))/(rho_("liquid"))`, same equation for two bodies. |
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| 85. |
A small piece of wire of length 4 cm is floating on the surface of water. If a force of 560 dynes in excess of itss apparent weight is requried to pull it up from the surface find the surface tension of water. |
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Answer» Length of wire `l=4cm` contact length of solid with liquid surface `L=2l=8cm`. Surface tension `T=(F)/(L)impliesT=(560)/(8)=70"dyne"//cm` `T_("water")=70"dyne" cm^(-1)=0.07Nm^(-1)` |
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| 86. |
The force which tends to destroy the relative motion between liquid layers is known asA. force due to surface tensionB. viscous forceC. gravitational forceD. force of cohesion |
| Answer» Correct Answer - B | |
| 87. |
A spherical steel ball released at the top of along column of glycerin of length `l` falls through a distance `l//2` with accelerated motion and the remaining distance `l//2` with uniform velocity let `t_(1)` and `t_(2)` denote the times taken to cover the first and second half and `w_(1)` and `w_(2)` are the work done against gravity in the two halves, then compare times and work done. |
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Answer» Average velocity in first half of the distance `ltv` while in the second half the average velocity is v therefore `t_(1)gt t_(2)`. The work done against gravity in both halves is mg `l//2 :. t_(1)gt t_(2)` `:. w_(1)=w_(2)` |
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| 88. |
A ball is dropped into coaltar. Its velocity time curve will beA. B. C. D. |
| Answer» Correct Answer - B | |
| 89. |
A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in figure, indicate the one that represents the velocity `(v)` of the pebble as a function of time `(t)`A. B. C. D. |
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Answer» Correct Answer - C When the pebble is falling throught the viscous oil the viscous force is `F=6pietarv` where r is radius of the pebble, v is instantaneous speed, `eta` is coefficient of viscosity. As the force is variable , hence accleration is also variable so v-t graph will not be straight line. First velocity increases and then becomes constant known as terminal velocity. |
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| 90. |
A spherical bal is dropped in a long column of a viscous liquid. The speed of the ball as a function of time may be best represented by the graph A. curve AB. curve BC. curve CD. curve S |
| Answer» Correct Answer - C | |
| 91. |
Two identical lead shots are dropped at the same time in two glass jars containing water and glycerin. The glass jars containing water and gycerin. The lead shot dropped in glycerin descends slowly becauseA. viscous force is more in water than in glycerinB. viscous force is more in glycerin than in waterC. surface tension is more is waterD. surface tension is more in glycerin |
| Answer» Correct Answer - B | |
| 92. |
When a metallic sphere is dropped in a long column of a liquid, the motion of the sphere is opposed by the viscous force of the liquid. If the apparent weight of the sphere equals to the retardation forces on it, the sphere moves down with a velocity called.A. critical velocityB. terminal velcoityC. velocity gradientD. constant velocity |
| Answer» Correct Answer - B | |
| 93. |
When a boat in a river enters the sea water, then itA. sinks a littleB. rises a littleC. remains sameD. will drown |
| Answer» Correct Answer - B | |
| 94. |
After the storm, the sea water waves subside due toA. surface tension of sea-waterB. disappearance of heavy currectsC. The vicosity of sea waterD. gravitational pull of the storm |
| Answer» Correct Answer - C | |
| 95. |
Two soap bubble of radii `r_(1)` and `r_(2)` combime to form a single bubble of radius r under isothermal conditions . If the external pressure is P, prove that surface tension of soap solution is given by `S=(P(r^(3)-r_(1)^(3)-r_(2)^(3)))/(4(r_(1)^(2)+r_(2)^(2)-r^(2)))`.A. `(P_(0)(R^(3)+R_(1)^(3)+R_(2)^(3)))/(4(R^(2)+R_(1)^(2)+R_(2)^(2)))`B. `(P_(0)(R_(1)^(3)+R_(2)^(3)-R^(3)))/(4(R^(2)-R_(1)^(2)-R_(2)^(2)))`C. `P_(0)(R_(1)^(3)+R_(2)^(3)-R^(3))`D. `4P_(0)(R_(1)^(3)+R_(2)^(3)-R^(3))` |
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Answer» According to isothermal process, `PV=P_(1)V_(1)+P_(2)V_(2)` `(P_(0)+(4T)/(R))(4)/(3)piR^(3)=(P_(0)+(4T)/(R_(1)))(4)/(3)piR_(1)^(3)+(P_(0)+(4T)/(R_(2)))(4)/(3)piR_(2)^(3)` `P_(0)(R^(3)-R_(1)^(3)-R_(2)^(3))=4T(R_(1)^(2)+R_(2)^(2)-R^(2))` |
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| 96. |
Two soap bubble of radii `R_(1)` and `R_(2)` are in atmosphere of pressure `P_(0)` at constant temperature. Ratio of masses of air inside them is |
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Answer» `PV=(m)/(M)Rti.e.,mpropPV(m_(1))/(m_(2))=(P_(1)V_(1))/(P_(2)V_(2))` `p_(1)=p_(0)+(4T)/(R_(1)),p_(2)=p_(0)+(4T)/(R_(2))` `=((P_(0)+(4T)/(R_(1)))(4)/(3)piR_(1)^(3))/((P_(0)+(4T)/(R_(2)))(4)/(3)piR_(2)^(3))=((P_(0)+(4T)/(R_(1)))R_(1)^(3))/((P_(0)+(4T)/(R_(2)))R_(2)^(3))` |
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| 97. |
If the surface tension of soap solution is `35"dynes"//cm`, calculate the work done to form an air bubble of diameter 14 mm with that solution. |
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Answer» Surface tension`T=35`dynes `cm^(-1)=0.035Nm^(-1)` `r=(14mm)/(2)=7mm=7xx10^(-3)m` `W=DeltaA xxT=8pir^(2)T=8xx(22)/(7)xx49xx10^(-6)xx0.035` `W=4.312xx10^(-5)J` |
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| 98. |
Surface tension arises from the cohesive force between the surface molecules. Interplay between cohesion and adhesion force make the surface inclined at acute or obtuse angle with the contacting solid surfaces. This causes a capillary rise (or fail) given as `h=(2Tcostheta)/(rhogr)`, where `theta=` angle of contact `T=` surface tension `rho=` density of the liquid, `g=` acceleration due to gravity and `r=` radius of the capillary tube. In capillary rise:A. heat is evolvedB. `U_(gr)` decreaseC. `U_("total")` increaseD. heat is absorbed |
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Answer» Correct Answer - A Work done by surface tension is greater than that of gravity hence heat evolved. |
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| 99. |
A vessel filled with air under pressure `p_(0)` contains a soap bubble of diameter `d`. The air pressure have been reduced `n`-fold, and the bubble diameter increased `r`-fold is isothermally. Find the surface tension of the soap water solution.A. `T=(1)/(2)p_(0)d xx (1-(r^(3))/(n))/(r^(2)-1)`B. `T=(1)/(8)p_(0)d xx (1-r^(3)/(n))/(r^(2)-1)`C. `T=(1)/(4)p_(0)d xx(1-r^(3)/(n))/(r^(2)-1)`D. `T=(1)/(6)p_(0)d xx (1-r^(3)/(n))/(r^(2)-1)` |
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Answer» Correct Answer - B `P_(1)V_(1)=P_(2)V_(2)` `impliesp_(0)(1-(r^(3))/(n))=(8T)/(d)(r^(2)-1)impliesT=(1)/(8)p_(0)d xx (1-(r^(3))/(n))/(r^(2)-1)` |
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| 100. |
A drop of liquid of density `rho` is floating half-immersed in a liquid of density `d`. If `sigma` is the surface tension the diameter of the drop of the liquid isA. `sqrt((3sigma)/(t(2rho-d)))`B. `sqrt((6sigma)/(g(2rho-d)))`C. `sqrt((4sigma)/(g(2rho-d)))`D. `sqrt((12sigma)/(g(2rho-d)))` |
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Answer» Correct Answer - D In equilibrium force due to surface tension `+` force of buoyancy `=` Weight of the spherical liquid drop `2pirT+(2)/(3)+(2)/(3)pir^(3)d_(2)g=(4)/(3)pir^(3)d_(1)g` `T=sigma,d_(1)=rho,d_(2)=d` `2pirsigma+(2)/(3)pir^(3)dg=(4)/(3)pir^(3)rhog` `impliesr^(2)=(3sigma)/(g(2rho-d))implies[(D)/(2)]^(2)=(3sigma)/(g(2rho-d))` |
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