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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.
| 101. |
What is the excess pressure inside a bubble of soap solution of radius 5.00mm, given that the surface tension of soap solution at the temperature `(20^(@)C)` is `2.50xx10^(-2)Nm^(-1)`? If an air bubble of the same dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1atm. is `1.01 xx 10^(5)Pa`). |
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Answer» Here surface tension of soap solution at room temperature `T=2.50xx10^(-2)Nm^(-1)`, radius of soap bubble `r=5.00mm=5.00xx10^(-3)m` `therefore` Excess pressure inside soap bubble, `P=P_(i)-P_(0)=(4T)/(r)` `=(4xx2.50xx10^(-2))/(5.00xx10^(-3))=20.0Pa` When an air bubble of radius `r=5.00xx10^(-3)` m is formed at a depth `h=40.0cm=0.4m` inside a container containing a soap solution of relative density 1.20 or density `rho=1.20xx10^(3)kgm^(-3)`, then excess pressure `P=P_(i)-P_(0)=(2T)/(r)` `thereforeP_(i)=P_(0)+(2T)/(r)=(P_(a)+h rhog)+(2T)/(r)` `=[1.01xx10^(5)xx0.4xx1.2xx10^(3)xx9.8+(2xx50xx10^(-2))/(5.00xx10^(-3))]Pa` `=(1.01xx10^(5)+4.7xx10^(3)+10.0)Pa` `cong1.06xx10^(5)Pa`. |
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| 102. |
The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. What is the ratio between the volume of the first and the second bubble?A. `9:1`B. `1:3`C. `1:9`D. `3:1` |
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Answer» Correct Answer - C Pressure , `P=(4S)/(r)` or `P prop (1)/(r)` `therefore (P_(1))/(P_(2))=(r_(2))/(r_(1))=(3)/(1)" "…(i)` or `r_(2)=3r_(1)` Also `(A_(1))/(A_(2))=(4pir_(1)^(2))/(4pir_(2)^(2))=((r_(1))/(r_(2)))^(2)=((r_(1))/(3r_(1)))^(2)" "`(Using (i)) `=(1)/(9)` |
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| 103. |
The tangential force or viscous force on any layer of the liquid is directly proportional to the velcoity gradient `dupsilon//dx.` Then the direction of velcoity gradient isA. perpendicular to the direction of flow of liquid.B. parallel to the direction of flow of liquid.C. opposite to the direction of flow of the liquid.D. independent of the direction of flow of liquid. |
| Answer» Correct Answer - A | |
| 104. |
Statement I: As radius of soap bubble increases, the insude pressure increases. Statement II: Excess pressure in soap bubble is inversely propotional to radius.A. Both statement 1 and statemet 2 are true and statement 2 is the correct explanation of statement 1.B. Both statement 1 and statement 2 are true but statement 2 is not the correct explanantion of statement 7C. Statement 1 is true but statement 2 is false.D. Statement 1 is false but statement 2 is false. |
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Answer» Correct Answer - D `P_(i)-P_(0)=(4T)/(r)` |
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| 105. |
Statement-1 : The value of moduli of elasticity is directly proportional to stress. Statement-2 : The value of moduli of elasticity is inversely proportional to strain. Statement-3 : The value of moduli of elasticiy is independent of magnitude of stress and strain.A. T T FB. F F TC. T F FD. F F F |
| Answer» Correct Answer - B | |
| 106. |
A lead sphere is steadily is steadily sinking in glycerine whose viscosity is equal to `eta=13.9` P.What is the maximum diameter of the sphere at which the flow around that sphere still remains laminar? It is known that the transition to the turbulent flow correspond to reynolds number `R_(e)=0.5`. (here the charactrstic length is taken to be the sphere diameter). (in mm) |
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Answer» Correct Answer - 5 `v_(t)=(2r^(2)rhog)/(9eta)` |
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| 107. |
Find the excess pressure inside a liquid drop of radius 2 cm, if the surface tension of water is 0.073 N `m^(-1)` |
| Answer» Correct Answer - 7.3 N `m^(-2)` | |
| 108. |
The cylinderical tube of a spray pump has a cross-section of `8.0cm^(2)` one end of which has 40 fine holes each of diameter 1.0mm. If the liquid flow inside the tube is 1.5 m per minute, what is the speed of ejection of the liquid through the holes? |
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Answer» Total cross-sectional area of 40 hiles, `a_(2)` `=40xx(22)/(7)xx((1xx10^(-3))^(2))/(4)m^(2)` `=(22)/(7)xx10^(-5)m^(2)` cross-sectional are of tube `a_(1)=8xx10^(-4)m^(2)` speed inside the tube, `v_(1)=1.5m" "mi n^(-1)=(1.5)/(60)ms^(-1)` speed of ejection `v_(2)=?` speed of ejection `v_(2)=?` Using `a_(2)v_(2)=a_(1)v_(1)` we get `v_(2)=(a_(1)v_(1))/(a_(2))=(8xx10^(-4)xx(1.5)/(60)xx7)/(22xx10^(-5))ms^(-1)=0.64ms^(-1)`. |
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| 109. |
The cylinderical tube of a spray pump has a cross-section of `6 cm^(2)` one of which has 50 holes each of diameter 1 mm. If the liquid flow inside the tube is 1.2 m per minutes, then the speed of ejection of the liquid through the holes isA. `2.1 m s^(-1)`B. `0.31 ms^(-1)`C. `0.96 m s^(-1)`D. `3.4 m s^(-1)` |
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Answer» Correct Answer - B Here, area of cross section of tube `a_(1)=6 cm^(2)=6xx10^(-4) m^(2)` Number of holes =50 Diameter of each hole, D=1mm, `10^(-3)`m Radius of hole, `r=(D)/(2)=(1)/(2)xx10^(-3)=5xx10^(-4)`m Area of cross-section of each hole `=pir^(2)=pi(5xx10^(-4))^(2) m^(2)` Speed of liquid inside the tube, `v_(1)=1.2 m "min"^(-1)=(1.2)/(60) m s^(-1)=0.02 m s^(-1)` Let the velocity of ejection of the liquid through the hole `v_(2)` As `a_(1)v_(1)=a_(2)v_(2)` or `v_(2)=(a_(1)v_(1))/(a_(2))=(6xx10^(-4)xx0.02)/(50xxpi(5xx10^(-4))^(2))=0.31 m s^(-1)` |
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| 110. |
A liquid of density `rho` is flowing with a speed v through a pipe of cross sectional area A. The pipe is bent in the shape of a right angles as shown. What force should be exerted on the pipe at the corner to keep it fixed? A. `(sqrt(2)SV)/(rho)`B. `sqrt(2)SV^(2)rho`C. `sqrt(3)(SV(2)rho)/(2)`D. `sqrt(3)SV^(2)rho` |
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Answer» Correct Answer - B Take x-axis along the flow and y-axis perpendicular to it `V_("initial")=Vveci,V_("final")=Vvecj,|Deltavec(V)|=sqrt(V^(2)+V^(2))=sqrt(2)V` `F=m(Deltav)/(t)=m(Deltav)/(t)=rhoxxS(l)/(t)xxsqrt(2)V=sqrt(2)SV^(2)rho` |
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| 111. |
A homogeneous solid cylinder of length L(LltH/2), cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure `P_0`. Then density D of solid is given by |
| Answer» Correct Answer - 5 | |
| 112. |
A solid cylinder of height h and mass m floats in a liquid of density `rho` such that three fourth of its height is immersed. Now, the cylinder is released inside a liquid of density `(rho)/(4)`, contained in a downward accelerated vessel. Find the magnitude of acceleration of vessel A, for which cylinder sinks with relative acceleration of `(A)/(3)` w.r.t. vessel. |
| Answer» Correct Answer - `(2g)/(3)` | |
| 113. |
During blood transfusion the needle is inserted in a vein where the gauge pressure is `2000Pa` . At what height must the blood container be placed so that blood may just enter the vein? Density of whole blood = `1.06xx10^(3)Kg//m^(3)`. |
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Answer» `h=P//rhog=200//(1.06xx10^(3)xx9.8)=0.1925m` The blood may just enter the vein if the height at which the gblood container be kept must be slightly greater than `0.1925m` i.e., 0.2m. |
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| 114. |
Two vessels have the same base area but differnent shapes. The first vessel takes twice the vloume of water that the second vessel requires to fill up to a paricular common height . Is the force exerted by water on the base of the vessel the same in the two case? If so, why do the vessels filled with water to that same height give different reading on a weighting scale ? |
| Answer» Pressure (and therefore force) on the two equal base areas are identical. But force is exertd by water on the sides of the vessels also, which has a non-zero vertical component when sides of the vessel are not perfectly normal to the base. This net vertical component of force by water on the sides of the vessel is g reater for the first vessel than the second. hence, the vessels weigh different even when the force on the base is the same in the two cases. | |
| 115. |
A bubble having surface tension T and radius R is formed on a ring of radius `b(b lt lt R)`. Air is blown inside the tube with velocity v as shown. The air molecule collides perpendicularly with the wall of the bubble and stops. Calculate the radius at which the bubble separates from the ring. A. `(4T)/(rhov^(3))`B. `(4T)/(rhov)`C. `(2T)/(rhov^(2))`D. `(4T)/(rhov^(2))` |
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Answer» Correct Answer - D The bubble will separate from the tube when thrust force exerted by the air is equal to the force due to excess pressure. `i.e., rhoAv^(2)=((4T)/(r))A or r=(4T)/(rhov^(2))` |
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| 116. |
A glass capillary tube of inner diameter 0.28 mm is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that water level in the tube is same as the vessel in `(N)/(m^(2))` is (surface tension of water `=0.07(N)/(m)` atmospheric pressure `=10^(5)(N)/(m^(2))`A. `10^(3)`B. `99xx10^(3)`C. `100xx10^(3)`D. `101xx10^(3)` |
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Answer» Correct Answer - 4 `P_("total")=P_(0)+(2T)/(R)` |
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| 117. |
There is a hole at the bottom of a large open vessel. If water is filled upto a height h, it flows out in time t. if water is filled to a height 4h, it will flow out in timeA. `4t`B. `(t)/(4)`C. `(t)/(2)`D. `2t` |
| Answer» `t=sqrt((2)/(g))(A)/(a)(sqrt(H))implies(t_(1))/(t_(2))=(sqrt(H_(1)))/(sqrt(H_(2)))` | |
| 118. |
A long cylinderical glass vessel has a small hole of radius r at its bottom. The depth to which the vessel can be lowered vertically in a deep water (surface tension S) without any water entering inside isA. `(2T)/(r)`B. `(2T)/(rgd)`C. `(T)/(rgd)`D. `(rgd)/(2)` |
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Answer» Correct Answer - 2 `h=(2T)/(rdg)` |
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| 119. |
There is a small hole at the bottom of tank filled with water. If total pressure at the bottom is `3 atm(1 atm=10^(5)Nm^(-2))`, then find the velocity of water flowing from hole.A. `sqrt(400) m//s`B. `sqrt(200) m//s`C. `sqrt(600) m//s`D. `sqrt(500) m//s` |
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Answer» Correct Answer - A Pressure due to water in the tank `=3atm-1atm=2atm=20m` of water column height of the water in the tank is `h=20m` velocity of efflux`=sqrt(2gh)=sqrt(2xx10xx20)=sqrt(400)m//sec` |
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| 120. |
A vessel has a small hole at its bottom. If water can be poured into it upto a height of 7 cm without leakage `(g=10ms^(-2)`) the radius of the hole is (surface tension of water is `0.7Nm^(-1)`)A. 2 mmB. 0.2 mmC. 0.1 mmD. 0.4 mm |
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Answer» Correct Answer - B `T=(rhdg)/(2)` |
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| 121. |
A large tank filled with water to a height `h` is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from h to `(h)/(2)` and from `(h)/(2)` to zero isA. `sqrt(2)`B. `(1)/(sqrt(2))`C. `sqrt(2)-1`D. `(1)/(sqrt(2)-1)` |
| Answer» `t=sqrt((2)/(g))(A)/(a)(sqrt(H_(1))-sqrt(H_(2)))implies(t_(1))/(t_(2))=(sqrt(h)-sqrt((h)/(2)))/(sqrt((h)/(2))-sqrt(0))` | |
| 122. |
Water is maintained at a constant level of 4.9 m in a big tank. The tank has a small hole to the wall near the bottom. The bottom of the tank is 2.5 above the ground level. The horizontal distace at which water touches are ground isA. 19.6 mB. 7 mC. 35 mD. 78.4 m |
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Answer» Correct Answer - 2 Range `=V_("efflux")xxsqrt((2H)/(g))` `sqrt(2gh)xxsqrt((2H)/(g))=2sqrt(gH)` |
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| 123. |
A cylindrical tank has a hole of `2cm^(2)` at its bottom if the water is allowed to flow into tank from a tube above it at the rate of `100cm^(3)//s` then find the maximum height upto which water can rise in the tank (take `=g=10ms^(-2)`)A. `2.5xx10^(-2)m`B. `1.25xx10^(-2)m`C. `5.5xx10^(-2)m`D. `3.5xx10^(-2)m` |
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Answer» Correct Answer - B `Q_("in")=Q_("out")`,`Q=asqrt(2gh)` |
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| 124. |
The level of water in a tank is 5 m high. A hole of area of cross section 1 `cm^(2)` is made at the bottom of the tank. The rate of leakage of water for the hole in `m^(3)s^(-1)` is `(g=10ms^(-2))`A. `10^(-3)`B. `10^(-4)`C. `10`D. `10^(-2)` |
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Answer» Correct Answer - 1 `Q=Asqrt(2gh)` |
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| 125. |
A spherical ball of radius R is floating at the interface of two liquids with densities `rho` and `2rho`. The volumes of the ball immersed in two liquids are equal. Answer the following questions: Find the force exerted by the liquid with density `2rho` on the ballA. `piR^(2)rhog(H+(2R)/(3))`B. `(2)/(3)piR^(2)rhog`C. `(4)/(3)piR^(2)rhog`D. `2piR^(2)rhog(H+(2R)/(3))` |
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Answer» Correct Answer - D `piR^(2)2Hrhog+(2)/(3)piR^(3)2rhog=F` `2piR^(2)Hrhog+(4piR^(3)rhog)/(3)=F,2piR^(2)rhog(H+(2R)/(3))=F` |
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| 126. |
A thin triangular glass tube containing immiscible liquids A, B, C of densities `rho, 2rho, 3rho` is at rest in vertical plane. Find x. |
| Answer» Correct Answer - `(1)/(3)` | |
| 127. |
A cylinder stands vertical in two immiscible liquids of densities `rho` and `2rho` as shown. Find the difference in pressure at point A and B:A. `2pigh`B. `3rhogh`C. `4rhogh`D. none |
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Answer» Correct Answer - B Difference in pressure at A and B`=4rhogh-rhogh=3rhogh` |
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| 128. |
A water barrel stands on a table of height `h`. If a small holes is punched in the side of the barrel at its base, it is found that the resultant stream of water strikes the ground at a horizontal distance `R` from the table. What is the depth of water in the barrel?A. `(4h)/(R^(2))`B. `4hR^(2)`C. `(R^(2))/(4h)`D. `(h)/(4R^(2))` |
| Answer» Correct Answer - C | |
| 129. |
Equal volumes of two immiscible liquids of densities `rho and 2 rho` are filled in a vessel as shown in figure. Two small holes are punched at depth h/2 and 3h/2 from the surface of lighter liquid. If `v_(1) and v_(2)` are the velocities of efflux at these two holes then `((v_(2))/(v_(1)))^(2)` is |
| Answer» Correct Answer - 2 | |
| 130. |
The atmoshperic pressure at sea level is determined by what height of mercury comlumn in a barometer?A. 76 cmB. 7.6 cmC. 86 cmD. 8.6 cm |
| Answer» Correct Answer - A | |
| 131. |
A tube of length 1 and radius R carries a steady flow of fluid whose density is `rho` and viscosity `eta`. The velocity v of flow is given by `v=v_(0)(1-r^(2)//R^(2))` Where r is the distance of flowing fluid from the axis.A. the volume of fluid flowing across the section. Of the tube, in unit time is `2piv_(0)((R^(2))/(4))`B. the kinetic energy of the fluid within the volume of the tube is `K.E=pirholv_(0)^(2)((R^(2))/(6))`C. the frictional force exerted on the tube by the fluid is `F=4pietakv_(0)`D. the pressure difference at the ends of tube is `P=(4etalv_(0))/(R^(2))` |
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Answer» Correct Answer - A::B::C::D The volume of fluid flowing through this section per second `dv=(2pirdr)v_(0)((1-r^(2))/(R^(2)))` Total volume `V=int_(0)^(R)(2pirdr)v_(0)((1-r^(2))/(R^(2)))` `=2piv_(0)((R^(2))/(4))` (ii). The kinetic energy of the fluid within the volme element of thickness dr K.E. of fluid within the tube is `=(1)/(2)(2pil)rhov_(0)^(2)int_(0)^(R)((1-r^(2))/(R^(2)))rdr` we get `K.E pirholv_(0)^(2)((R^(2))/(6))` (iii). The viscous drag exerts a force on the tube `F=-etaA((dv)/(dr))_(e=R)` here `((dv)/(dr))_(r=R)=v_(0)((-2r)/(R))_(r-R)=-2v_(0)//R` `:. F=4pietalv_(0)` (iv). `DeltaP=P_(2)-P_(1)=P` where `P_(1)=0` and `P_(2)=p` `P=("force"(F))/("area"(piR^(2)))=(4etalv_(0))/(R^(2))` |
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| 132. |
A fluid container is containing a liquid of density `rho` is accelerating upward with acceleration a along the inclined plane of inclination `alpha` as shown in the figure. Then the angle of inclination `theta` of free surface isA. `tan^(-1)[(a)/(gcosalpha)]`B. `tan^(-1)[(a+gsinalpha)/(gcosalpha)]`C. `tan^(-1)[(a-gsinalpha)/(g(1+cosalpha))]`D. `tan^(-1)[(a-gsinalpha)/(g(1-cosalpha))]` |
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Answer» Correct Answer - B Apply pseudo force on a particle of mass `m`. Net force along the surface is zero. `macostheta=mgcos[90-(theta-alpha)]` `(a)/(g)costheta=gsin(theta-alpha)` `(a)/(g)=(sinthetacosalpha)/(costheta)-(costhetasinalpha)/(costheta)` `tantheta=((a)/(g)+sinalpha)/(cosalpha)=(a+gsinalpha)/(gcosalpha)` |
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| 133. |
In the arrangement as shown, `m_(B)=3m`, density of liquid is `rho` and density of block `B` is `2rho`. The system is released from rest so that block `B` moves up when in liquid and moves down when out of liquid with the same acceleration. Find the mass of block `A`. A. `(7)/(4)m`B. `2m`C. `(9)/(2)m`D. `(9)/(4)m` |
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Answer» Correct Answer - D Acceleration when block B is in the liquid. `a_(1)=(m_(A)g-(m_(B)g-"upthrust"))/((m_(A)+3m))` `=(m_(A)g-(3mg-(3m)/(2rho)pig))/((m_(A)+3m))(uarr)` Acceleration when block B is outside of the liquid. `a_(2)=(3mg-m_(A)g)/((m_(A)+3m))(darr)` give, `a_(1)=a_(2)` `m_(A)g=(3)/(2)mg=3mg-m_(A)g` `2m_(A)g=(9)/(2)mg,m_(A)=((9)/(4))m` |
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| 134. |
At a point P in a water pipe line the velocity is `1ms^(-1)` and the pressure is `3xx10^(5)pa`. At another point Q the area of cross section is half that of at P and the pressure is `5xx10^(5` pa. The difference of heights between P and Q in metre is `(g=10ms^(-2))`A. 10.5B. 20.15C. 4.5D. zero |
| Answer» `p_(1)+(1)/(2)rhov_(1)^(2)+rhogh_(1)=p_(2)+(1)/(2)rhov_(2)^(2)+rhogh_(2)` | |
| 135. |
The cross-sectional areas of a tube `T_(1)` and the hole in the vessel at B are a and `a//2` respectively. There is a hole in the tube at C (at the level of A) through which liquid in the vessel rises by a height h in the tube. The other liquid heights are shown in the diagram. The plugs at A and B are removed simultaneously. How much horizontal force is required to keep vessel in equilibrium if p is the pressure in the tube and `p_(0)` is the atmospheric pressure? Hole C is closed when plugs are removed.A. `a(p_(0)-p)`B. `(a)/(2)(p_(0)-p)`C. `2a(p_(0)-p)`D. `4a(p_(0)-p)` |
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Answer» Correct Answer - C `p_(0)=p+rhogh,h=((p_(0)-p))/(rhog)` `F_(A)=rhoaV^(2)=rhoa(2gxx(3h)/(2))` `=3rho a g xx ((p_(0) -p))/(rho g) = 3a (p_(0) -p)` `F_(B) = rho (a)/(2) V_(B)^(2) = a (p_(0) - p)` `(as V_(B)^(2) = 2g.((p_(0)-p))/(rho g))` `F_(A) - F_(b) = 2a (p_(0) - p)` |
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| 136. |
A wooden cylinder of length L is partly submerged in a liquid of specific gravity `rho_(1)` with `n^(th) (nlt1)` part of it inside the liquid. Another immiscible liquid of density `rho_(2)` is poured to completely submerge the cylinder. Density of cylinder `rho` is the geometric mean of the densities of the two liquid. Express the density of upper liquid in terms of density of cylinderA. `(rho)/(n)`B. `rhon`C. `(n)/((n+1))`D. none |
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Answer» Correct Answer - B `n=(rho)/(rho_(1))`, or `rho_(1)=(rho)/(n)` but `rho^(2)=rho_(1)rho_(2)=(rho)/(2)rho_(2)` or `rho_(2)=rho.n` |
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| 137. |
Water flows at a speed 5 cm `s^(-1)` through a pipe of radius 2 cm. The visosity of water is 0.001 PI. The Reynolds number and the nature of flow are respectivelyA. 2000, unsteadyB. 1500, turbulentC. 1000, turbulentD. 2500, laminar |
| Answer» Correct Answer - A | |
| 138. |
The vessel shown in the figure has two sections. The lower part is a rectangular vessel with area of cross-section A and height h. The upper part is a conical vessel of height h with base area A and top area `a` and the wals of the vessel are inclined at an angle `30^(@)` witht he vertical.A liquid of density `rho` fills both the sections upto a height 2h. Neglect atmospheric pressure.A. The force F exerted by the liquid on the base of the vessel is `2hrhog .((A +a ))/(2)`B. The pressure P at the base of the vessel is `2hrhog.(A)/(a)`C. The weight of the liquid W is greater than the force exerted by the liquid on the baseD. the walls of the vessel exert a downward force on the liquid. |
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Answer» Correct Answer - D `F_(base)=Pxx"area"=rhog(2h)(A)` |
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| 139. |
A cappillary tube of radiue `r` is lowered into water whose surface tension is `alpha` and density `d`. The liquid rises to a height. Assume that the contact angle is zero. Choose the correct statement (s):A. Magnitude of work done by force of surface tension is `(4pialpha^(2))/(dg)`B. Magnitude of work done by force of surface tension is `(2pialpha^(2))/(dg)`C. Potential energy acquired by the water is `(2pialpha^(2))/(dg)`D. The amount of heat developed is `(2pialpha^(2))/(dg)` |
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Answer» Correct Answer - A::C::D `h=(2alpha)/(dgr)` hence work done by force of surrface tension is `W_(S)=alphaxx2pirxxh=(4pialpha^(2))/(dg)` But centre of mass of liquid in the capillary tube is at a hight `(h)/(2)`. Hence potential energy gained `=(Mgh)/(2)=pir^(2)xxhxxdxxgxx(h)/(2)=(2pialpha^(2))/(gd)` Hence work done by gravity `=(-(2pialpha^(2))/(dg))` Amount of heat developed `=(2pialpha^(2))/(dg)` |
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| 140. |
Statement I: Finer the capillary, greater is the height to which the liquid rises in the tube Statement II: This is in accordance with the ascent formula.A. Both statement 1 and statemet 2 are true and statement 2 is the correct explanation of statement 1.B. Both statement 1 and statement 2 are true but statement 2 is not the correct explanantion of statement 5C. Statement 1 is true but statement 2 is false.D. Statement 1 is false but statement 2 is false. |
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Answer» Correct Answer - A According to the ascent formula `h=(2Tcostheta)/(rpg)implieshalpha(1)/(r)` From the relation radius is less and the height of which the liquid rises will be greater. |
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| 141. |
In the bottom of a vessel with mercury there is a round hole of diameter `d=70mum`. At what maximum thickness of the mercury Layer will the liquid still not flow out through this hole ? `[rho_("mercury")=13600(kg)/(m^(3))]`A. 11 cmB. 21 cmC. 42 cmD. 32 cm |
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Answer» Correct Answer - B `h=(4T)/(rhogd)`. As the height increases the radius of curvature of the liquid layer decreases. |
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| 142. |
Water flows through a frictionless tube with a varying cross-section as shown The variation of Pressure P as the fluid move through the pipe is correctly shown inA. B. C. D. |
| Answer» Correct Answer - A | |
| 143. |
Two vertical parallel glass plates are partially submerged in water. The distance between the plates is d and the length is `l`. Assume that the water between the plates does not reach the upper edges of the plates and the wetting is complete. The water will rise to height (`rho=` density of water and `alpha =` surface tension of water)A. `(2T)/(drhog)`B. `(T)/(2drhog)`C. `(T)/(drhog)`D. None of these |
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Answer» Upward force due to surface tension is balanced by the weight of the liquid which rises in the gap, so `2T.b=bdhrhog` Where `b=` width of the plates `:. h=(2T)/(drhog)` |
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| 144. |
Two vertical parallel glass plates are partially submerged in water. The distance between the plates is d and the length is `l`. Assume that the water between the plates does not reach the upper edges of the plates and the wetting is complete. The water will rise to height (`rho=` density of water and `alpha =` surface tension of water)A. `(2sigma)/(rhogh)`B. `(sigma)/(2rhogd)`C. `(4sigma)/(rhogd)`D. `(5sigma)/(rhogd)` |
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Answer» Correct Answer - A Total upward force dur to surface tension `=2sigmal` Weight of lifted liquid `=(hld)rhog` Equating we get `h=(2sigma)/(rhogd)` |
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| 145. |
If the surface tension of water is 0.06 Nm, then the capillary rise in a tube of diameter 1mm is `(theta=0^(@)`)A. 1.22 cmB. 2.44 cmC. 3.12 cmD. 3.86 cm |
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Answer» Correct Answer - B `h=(2S costheta)/(r rhog)=(2xx0.06xxcos0^(@))/(0.5xx10^(-3)xx10^(3)xx9.8)` `=2.44xx10^(-2)m=2.44 cm` |
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| 146. |
Select the correct optionA. As we move down through a liquid at rest the pressure increasesB. As we move along horizontal through a liquid at rest the pressure remains unchangedC. As we move along horizontal through a liquid subjected to a horizontal acceleration, the pressure increases in the direction of accelerationD. The gauge pressure at a point inside a liquid depends on atmospheric pressure |
| Answer» Correct Answer - A::B | |
| 147. |
In the adjoing figure `P_(1) and P_(2)` are the pressures in broad and narrow tubes with closed ends. Select the correct option A. `P_(1) gt P_(2)`B. `P_(2) - P_(1) prop h^(@)`C. `P_(1) - P_(2) prop h`D. `P_(1) - P_(2) prop (A_(1) - A_(2))^((1)/(2))` |
| Answer» Correct Answer - A::C | |
| 148. |
Two liquid drops have their diameters as 1 mm and 2 mm. The ratio of excess pressures in them isA. `1:2`B. `2:1`C. `4:1`D. `1:4` |
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Answer» Correct Answer - B `DeltaP=(2T)/(r)` |
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| 149. |
A triangular element of the liquid is shown in the fig. `P_(x),P_(y)` and `P_(z)` represent the pressures on the element of the liquid then:A. `P_(x)=P_(y)neP_(z)`B. `P_(x)=P_(y)=P_(z)`C. `P_(x)neP_(y)neP_(z)`D. `P_(x)^(2)+P_(y)^(2)+P_(z)^(2)=` constant |
| Answer» Correct Answer - B | |
| 150. |
A volume V of a viscous liquid flows per unit time due to a pressure head `DeltaP` along a pipe of diameter d and length l. instead of this pipe a set of four pipes each of diameter `(d)/(2)` and length 2l is connected to the same pressure head `DeltaP`. Now the volume of liquid flowing per unit time is:A. `(V)/(16)`B. `(V)/(8)`C. `(V)/(4)`D. `V` |
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Answer» Correct Answer - B `V=(piPr^(4))/(8 l eta),Vprop(r^(4))/(l)` `(V_(1))/(V_(2))=((r_(1))/(r_(2)))^(4)xx(l_(2))/(l_(1))=32,V_(2)=(V)/(32)` For 4 pipes, `V_(2)^(1)=4V_(2)=4xx(V)/(32)=(V)/(8)` |
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