InterviewSolution
Saved Bookmarks
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.
| 1. |
A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a freely falling elevator, the length of water column in the capillary tube will beA. `20cm`B. `4cm`C. `10cm`D. `8cm` |
|
Answer» Correct Answer - A In the condition of weightlessness, water rise to the whole of the available length. |
|
| 2. |
The end of a capillary tube with a radius `r` is immersed in water. Is mechanical energy conserved when the water rises in the tube? The tube is suficiently long. If not calculate the energy change. |
|
Answer» In the equilibrium position `(theta=0^@` for water and glass) `2piTcos0^@=pir^(2)hrhog` or `h=(2T)/(rhogr)` Work done by surface tension `k=(2pirT)xxh=(4piT^(2))/(rhog)` The potential energy of water in the tube `U=(pir^(2)hro)gh//2,` it is multiplied by `h//2` because the centrte of gravity of the water and the capillary tube is at a height `h//2` `:. U=(2piT^(2))/(rhog)` Thus it is seen that the mechanical energy is not conserved. therefore, mechanical enegy loss `=(4piT^(2))/(rhog)-(2piT^(2))/(rhog)=(4piT^(2))/(rhog)` This energy is converted into heat. |
|
| 3. |
The lower end of a capillary tube is at a depth of `12 cm` and water rises `3 cm` in it. The mouth pressure required to blow an air bubble at the lower end will be `x cm` of water column, where `x` isA. `12`B. `15`C. `3`D. `9` |
|
Answer» Correct Answer - B Pressure due to `15 cm` long liquid column needs to be balanced. |
|
| 4. |
A hollow sphere has a small hole in it. On lowering the sphere in a tank of water, it is observed that water enters into the hollow sphere at a depth of `40 cm` below the surface. Surface tension of water is `7 xx 10^(-2) N//m`. The diameter of the hole isA. `1/28mm`B. `1/21mm`C. `1/14mm`D. `1/7mm` |
|
Answer» Correct Answer - C `40/100xx1000xx9.8=(2xx7xx10^(2))/R` or `R=(14xx10^(-2)xx10)/(40xx1000xx9.8)=(14xx1000)/(40xx1000xx9.8)m` `=1/28mm` Diameter `= 2R=1/14mm` |
|
| 5. |
Spherical particles of pollen are shaken up in water and allowed to settle. The depth of water is `2xx10^(-2)m`. What is the diameter of the largest particles remaining in suspension one hour later? Density of pollen `=1.8xx10^(3)kgm^(-3)` viscosity of water `=1xx10^(-2)` poise and density of water `=1xx10^(-5)kgm^(-3)` |
|
Answer» Terminal velocity `v=(2r^(2))/9 ((rho-sigma)g)/eta`…….i But we know `v=s/t` `:. s/t=2/9(r^(2)(rho-sigma)g)/etaimpliesr^(2)(9s)/(2t) eta/((rho-sigma)g)` Given `s=2xx10^(-2)m, t=1h=3600s` `:. Eta=1xx10^(-2)` poise `=1xx10^(-3)kgm^(-1)s^(-1)` Substituting given values we get `r^(2)=9/2xx(2xx10^(-2))/3600xx(1xx10^(-3))/((1.8xx10^(3)-1xx10^(3))xx10)=9/36xx1/810^(-10)` `=1/32xx10^(-10)` `:. r=sqrt(100/32)xx10^(-6)m=1.77xx10^(-6)m` Diameter `D=2r=2xx1.77mum=3.54mum` |
|
| 6. |
Statement I: Dust particles generally settle down in a closed room.Statement II: The terminal velocity is inversely proportional to the square of their radii.A. Statement I is true, statement II is true and Statement II is a correct explanation for Statement I.B. Statement I is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
|
Answer» Correct Answer - C Since the particles of dust are like spheres of very small radii and when it acquires the terminal velocity they begin through air. As from the definition, the terminal velocity of dust particles is directly proportional to the square of its radii. Thus, the terminal velocity of dust particle is very small and so they settle down in a closed room after some time. |
|
| 7. |
A barometer contains two uniform capillaries of radii `1.4xx10^(-3)m` and `7.2xx10^(-4)m`. If the height of liquid in narrow tube is `0.2m` more than that in wide tube, calculate the true pressure difference. Density of liqid `=10^(3)kg//m^(3)`, surface tension `=72xx10^(-3)N//m` and `g=9.8ms^(-12)`. |
|
Answer» Let the presures in wide and narrow limbs of `P_(1)` and `P_(2)`, respectively. If `R_(1)` and `R_(2)` be the radii of meniscus in wide and narrow limb pressure just below hhe meniscus of wide tube `=P_(1)- (2T)/(R_(1))` and pressure just below the meniscus of narrow limb `=P_(2)=(2T)/(R_(2))`. Therefore, difference of these pressures `(P_(1)-(2T)/(R_(1)))-(P_(2)-(2T)/(R_(2)))=hrhog` Therefore, true pressure difference, `P_(1)-P_(2)=hrhog-2T(1/(R_(2))-1/(R_(1)))` For the water and glass surface, takingg in the angle of contact `theta` to be zero, we have `R_(1)=(r_(1))/(costheta)~~r_(1)` and `R_(2)=(r_(2))/(costheta)~~r_(2)` where `r_(1)` and `r_(2)` are radii of wide and narrow limbs, respectively. `:.P_(1)-P_(2)=hrhog-2T(1/(r_(2))-1/(r_(1)))=0.2xx10^(3)xx9.8-2xx72` `xx10^(-3)xx(1/(7.2xx10^(-4))-1/(1.44xx10^(-3)))` `=1.96xx10^(3)-0.10xx10^(3)=1.86xx10^(3)N//m^(2)=1860N//m^(2)` |
|
| 8. |
Between a plate of area `100 cm^(2)` and another plate of area `100 m^(2)` there is a `1 mm`, thick layer of water, if the coefficient of viscosity of water is `0.01` poise, then the force required to move the smaller plate with a velocity `10 cms^(-1)` with reference to large plate isA. `100 dyn`B. `10^(4)dyn`C. `10^(6) dyn`D. `10^(9)dyn` |
|
Answer» Correct Answer - A `F=(0.01xx100xx10)/0.1dyn=100dyn` |
|
| 9. |
A plate of area `100 cm^(2)` is placed on the upper surface of castor oil, `2mm` thick. Taking the coefficient of viscosity to be `15.5` poise, calculate the horizontal force neccesary to move the plate with a velocity `3cm^(-1)`. |
|
Answer» The (horizontal tangential) viscous force is given by `F=-etaA(dv)/(dy)` Given `eta=15.5` poise, `A=100cm^(2)` `(dv)/(dx)=(v_(2)-v_(1))/(x_(2)-x_(1))=((0-3)cm//s)/((2-0)mm) =-3//0.2=-15 s^(-1)` `F=15.5xx100xx(-15)=2.235xx10^(4)dyn=0.2325N` |
|
| 10. |
In the figure shown, forces of equal magnitude are applied to the two ends of a uniform rod. Consider `A` as the cross-sectional area of the rod. For this situation, mark out the incorrect statements. A. The rod is in compressive stress.B. The numerical value of stress developed in the rod is equal to `F//A`.C. The stress is defined as internal force developed at any cross section per unit area.D. none of these |
|
Answer» Correct Answer - D All of the statements are factual statements. Stress is defined as internal force developed in the system per unit cross sectional area and is not defined as force applied per unit corss section section area although in equilibrium, stress is numerically equal to applied for per unit area. |
|
| 11. |
Two wires `A` and `B` have equal lengths and are made of the same material , but diameter of wire `A` is twice that of wire `B`. Then, for a given load,A. The extension of `B` will be four times that of `A`B. the extension of `A` and `B` will be equalC. the strain in `B` is four times that in `A`D. the strains in `A` and `B` will be equal |
|
Answer» Correct Answer - A::C Area of cross section `A=(pid^(2))/4` where `d` is the diameter of the wire. Therefore `l=(4FL)/(pid^(2)Y)` since `F, L` and `Y` are the same for wires `A` and `B` `:. Lprop1/(d^(2))` i.e., the extension is inversely proportional to the square of the diameter. Hence choice a is correct.The strain is `l/L=(4F)/(pid^(2)Y)` Thus strain `prop1/(d^(2))` i.e., the extension is inversely proportional material to the square of the diameter. Hence choice a is correct. The strain is `l/L=(4F)/(pid^(2)Y)` Thus, strain`prop1/(d^(2))` Hence the correct choice are a and c. |
|
| 12. |
A hydraulic press contains `0.25 m^(3)(250L)` of oil. Find the decrease in volume of the oil wen it is subjected to a pressure increase `/_p=1.6xx10^(7)Pa`. The bulk modulus of the oil is `B=5.0xx10^(9)Pa`. |
|
Answer» `B=-(/_p)/(/_V//V_0)=-((0.25m^(3))(1.6xx10^(7)Pa))/(5.0xx10^(9)Pa)` `=-8.0xx10^(-4)m^(3)=-0.80L` Even through the pressure increase is very large, the fractional change in volume is very small. `:./_V//V_0=(-8.0xx10^(-4)m^(3))/(0.25m^(3))=-0.0032` or `-0.32%` |
|
| 13. |
A mercury drop of radius `R` is sprayed into `n` droplets of equal size. Calculate the energy expended if surface tension of mercury is `T`. |
|
Answer» Correct Answer - `4piT(Nr_(1)^(2)-R^(2))` `/_U=N(4pir_(1)^(2))T-4R^(2)T` ………..i `+4piT(Nr_(1)^(2)-R^(2))` By mass conservation. `N(2/3pir_(1)^(3)rho)=4/3piR^(3)rhog` `r_(1)=R/((N)^(1/3)`…..ii Substitute `r_(1)` from eqn ii in eqn i |
|
| 14. |
Statement I: Small liquid drops assume sphereical shape. Statement II: Due to surface tension liquid drops tend to have minimum surface area.A. Statement I is true, statement II is true and Statement II is a correct explanation for Statement I.B. Statement I is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
|
Answer» Correct Answer - A The free surface of a liquid drop tries to acquire minimum surface area due to surface tension. Since for a given volume, the surface area of sphere is minimum, the liquid drops acquire sphere is minimum, the liquid drops acquire spherical shape. |
|
| 15. |
A large number of droplets, each of radius a, coalesce to form a bigger drop of radius `b`. Assume that the energy released in the process is converted into the kinetic energy of the drop. The velocity of the drop is `sigma =` surface tension, `rho =` density)A. `[(sigma)/(rho)(1/a-1/b)]^(1/2)`B. `[(2sigma)/(rho)(1/a-1/b)]^(1/2)`C. `[(3sigma)/(rho)(1/a-1/b)]^(1/2)`D. `[(6sigma)/(rho)(1/a-1/b)]^(1/2)` |
|
Answer» Correct Answer - D Energy realeased `=[nxx4pia^(2)=-4pib^(2)]sigma` Now `nxx4/3pir^(3)=4/3pib^(3)` or `n=(b^(3))/(a^(3))` Therefore energy released is `=[(b^(3))/(a^(3))=4pia^(2)-4pib^(2)]sigma=4pib^(2)[b/a-1]sigma` Now `1/2(4/3pib^(3))rhov^(2)=4pib^(2)[b/a-1]sigma` or `v=[(6sigma)/(rho)(1/a-1/b)]^(1/2)` |
|
| 16. |
Statement I: A small drop of mercury is spherical F bigger drops are oval in shape.Statement II: Surface tension of liquid decreases with increase in temperature.A. Statement I is true, statement II is true and Statement II is a correct explanation for Statement I.B. Statement I is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
|
Answer» Correct Answer - B In a small drop, the force due to the surface tension is very large as compared with its weight and hence it is spherical in shape. A big drop becomes oval in shape due to its large weight. The surface tension of liquid decreases with increase of temperature. |
|
| 17. |
A drop of liquid of density `rho` is floating half-immersed in a liquid of density `d`. If `rho` is the surface tension the diameter of the drop of the liquid isA. `sqrt((sigma)/(g(2rho-d)))`B. `sqrt((2sigma)/(g(2rho-d)))`C. `sqrt((6sigma)/(g(2rho-d)))`D. `sqrt((12sigma)/(g(2rho-d)))` |
|
Answer» Correct Answer - D `2pirsigma+1/2xx4/3pir^(3)dg=4/3rhog` or `2pirsigma=(pir^(3)g)/3[4rho-2d]` or `r^(2)=(3xx2pisigma)/(pig(2rho-2d)` or `r^(2)=(3sigma)/(g(2rho-d)` or `r=sqrt((3sigma)/(g(2rho-d)))` Diameter `=2r=sqrt((12sigma)/(g(2rho-d))` |
|
| 18. |
A paper disc of radius `R` from which a hole of radius `r` is cut out is floating in a liquid of the surface tension `S`. The force on the disc due to the surface tension isA. `Sxx2piR`B. `Sxx2pir`C. `Sxx2pi(R-r)`D. `Sxx2pi(R+r)` |
|
Answer» Correct Answer - D Surface tension force `F=Sxx"length" F=Sxx[2piR+2pir]` as liquid surface is on the both sides. |
|
| 19. |
A Copper wire and steel of the same diameter and length are connected end to end and a force is applied, which stretches their combined length by 1 cm. The two wires will haveA. same stress and same strainB. same stress and different strainsC. different stresses and same strainD. different stresses and different strains |
|
Answer» Correct Answer - B Tension is any of the wire is same at all points as cross sectional area are the same and tension also. So stress `(=F//A)` in both the wires would be the same but from Strain `= "Stress"/"Strain"` as `Y` is different for both, so train is different in both. |
|
| 20. |
Statement I: More is the cohesive force, more is the surface tension. Statement II: More cohesive force leads to more shrinking of liquid surface.A. Statement I is true, statement II is true and Statement II is a correct explanation for Statement I.B. Statement I is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
|
Answer» Correct Answer - A Surface tension can be understood as a property of a surface due to which it tries to acquire the minimum possible area. If cohesive force is more, the liquid is having the capability to shrink its surface more, i.e., surface area of the liquid is less and hence more is the surface tension. |
|
| 21. |
The bulk modulus of water is `2.0xx10^(9) N//m^(2)`. The pressure required to increase the density of water by `0.1%` isA. `2xx10^(9)N//m^(2)`B. `2x10^(8)N//m^(2)`C. `2xx10^(N)//m^(2)`D. `2xx10^(4)N//m^(2)` |
|
Answer» Correct Answer - C The density would increase by `0.1%` if the volume decreases by `0.1 %` `K=(/_P)/(/_V//V)` `/_P=K(/_V)/V=2xx10^(9)xx0.1/100=2xx10^(6)N//m^(2)` |
|
| 22. |
A glass rod of radius `r_(1)` is inserted symmetrically into a vertical capillary tube of radius `r_(2)` such that their lower ends are at the same level. The arrangement is now dipped in water. The height to which water will rise into the tube will be (`sigma =` surface tension of water, `rho = ` density of water)A. `(2sigma)/((r_(2)-r_(1))rhog)`B. `sigma/((r_(2)-r_(1))rhog)`C. `(2sigma)/((r_(2)+r_(1))rhog)`D. `(2sigma)/((r_(2)^(2)+r_(1)^(2))rhog)` |
|
Answer» Correct Answer - A Total upward force due to surface tension `=sigma(2r_(1)+2pi_(2))`. This supports the weight of the liquid column of height `h`. Weight of liquid column` =[pir_(2)^(2)-pir_(1)^(2)]rhog` Equating we get `hpi(r_(2)+r_(1))(r_(2)-r_(1))rho=2pisigma(r_(1)+r_(2))` |
|
| 23. |
Statement I: Surface tension has the same units as fore gradient. Statement II: Surface tension is the force gradient along the surface of liquid.A. Statement I is true, statement II is true and Statement II is a correct explanation for Statement I.B. Statement I is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
|
Answer» Correct Answer - A The surface tension of liquid can be measured as the force per unit length (force gradient) on an imaginary line drawn on the liquid surface, which acts perpendicularly to the line on its either side at every point and tangentially to the liquid surface. Therefore, surface tension and force gradient have the same dimensions. |
|
| 24. |
A horizontal aluminium rod of diameter `4.8 cm` projects `5.3 cm` from a wall. A `1200 kg` object is suspended from the end of the rod. The shear modulus of aluminium is `3.0xx10^(10)N//m^(2)`. Nelecting the mass of te rod find a shearing stress on the rod and b the vertical deflection of the end of the rod. |
|
Answer» a. Shearing stress `=F/A=(Mg)/(pir^(2))=(1200xx9.8)/(3.14xx(0.024)^(2))` `=6.5xx10^(6)Nm^(-2)` b. Let `/_x` be the vertical defelection of the rod. Then shear modulus, `G=(F/A)/(/_x/l)` Thus `/_x=((F/A)l)/G=(6.5xx10^(6)xx0.53)/(3.0xx10^(10))=1.1xx10^(-5)M` |
|
| 25. |
An cube is shifted to a depth of `100 m` is a lake. The change in volume is `0.1%`. The bulk modulus of the material isA. `10Pa`B. `10^(4)Pa`C. `10^(7)Pa`D. `10^(9)Pa` |
|
Answer» Correct Answer - D `10m` column of water exerts nearly `1` atmosphere pressure, i.e., `10xx10^(-5)Pa` or `10^(6)Pa`. Now `K=((/_P)V)/(/_V)=(10^(6)xx100)/0.1Pa=10^(9)Pa` |
|
| 26. |
A brass rod of length `1 m` is fixed to a vertical wall at one end, with the other end keeping free to expand. When the temperature of the rod is increased by `120^@C`, the length increases by `3 cm`. What is the strain? |
|
Answer» After the rod expands to the new length, there are no elastic forces developed internally in it. So strain `=0` |
|
| 27. |
The shear modulus for a metal is `50000 Mpa`. Suppose that a shear fore of `200 N` is applied ot the upper surface of a cube of this metal that is `3.0 cm` on each edge. How far will the top surface be displaced? |
|
Answer» Shearing strain`=/_L//L=(F/A)S` `=2000/[L^(2)(5x10^(10))]=(4xx10^(-9))/L^(2)` Solving for `/_L` with `L=0.030m`, we have `/_L=1.33xx10^(-7)m=0.133mum` |
|
| 28. |
Two blocks of masses `1 kg` and `2 kg` are connected by a metal wire goijng over a smooth pulley as shown in figure. The breaking stress of the metal is `(40//3pi)xx10^(6)N//m^(2)`. If `g=10ms^(-12)`, then what should be the minimum radlus of the wire used if it is not to break? A. `0.5mm`B. `1mm`C. `1.5mm`D. `2mm` |
|
Answer» Correct Answer - B `T=(2m_(1)m_(2))/(m_(1)+m_(2))g=(2xx1xx2)/(1+2)xx10N=40/3N` If `r` is the minimum radius, then Breaking stress `=(40/3)/(pir^(2))` or `40/(3pi)xx10^(6)=40/(3pir^(2))` or `r^(2)=1/(10^(6))` or `r=1/(10^(3))m` or `r=1/(10^(3))xx10^(3)m=1mm` |
|
| 29. |
A cube with a mass `= 20 g` wettable water floats on the surface of water. Each face of the cube is `alpha = 3 cm` long. Surface tension of water is `70 dyn//cm`. The distance of the lower face of the cube from the surface of water is (`g= 980 cm s^(-12)`)A. `2.3cm`B. `4.6cm`C. `9.7cm`D. `12.7cm` |
|
Answer» Correct Answer - A If `y` is the required distance, then `a^(2)yrhog=mg+4asigma` or `y=(mgxx4asigma)/(a^(2)rhog)=(20xx980+4xx3xx70)/(3xx3xx1xx980)cm=2.3cm` |
|
| 30. |
A small but heavy block of mass `10 kg` is attached to a wire `0.3 m` long. Its breaking stress is `4.8 xx 10^(7) N//m^(2)`. The area of the cross section of the wire is `10^(-6) m^(2)`. The maximum angular velocity with which the block can be rotated in the horizontal circle isA. `4rad//s`B. `8rad//s`C. `10rad//s`D. `32rad//s` |
|
Answer» Correct Answer - A `mromega^(2)=` Breaking stress `xx` cross sectional area `=10xx0.3omega^(2)=4.8xx10^(7)xx10^(-6)=48` or `omega^(2)=48/3=16` or `omega=4rad//s` |
|
| 31. |
A body of mass `M` is attached to the lower end of a metal wire, whose upper end is fixed . The elongation of the wire is `l`.A. Loss in gravitational potential energy of `M` is `Mgl`B. Elastic potential energy stored in the wire is `(Mgl)/2`C. Elastic potential energy stored in the wire is `Mgl`D. Elastic potential energy stored in the wire is `(Mgl)/3` |
|
Answer» Correct Answer - A::B Loss in gravitational potential energy of `M` is `Mgl` as `M` falls down by `l`. Elastic potential energy stored in wire is `U=1/2xx"Stress"xx"Strain"xx"Volume"` `=1/2 1/2 xx(Mg)/Axxl/LxxAL=(Mgl)/2` Now work done by gravityy force is not equal to elastic potential energy stored in wire. This is due to the fact that some work has been done against air friction etc., which increases the internal energy of wire. |
|
| 32. |
A glass tube of circular cross section is closed at one end. This end is weighted and the tube floats vertically in water, heavy end down. How far below the water surface is the end of the tube? Given: outer radius of the tube is `0.14 cm`, mass of weighted tube is `0.2 g`, surface tension of water `73 dyn//cm` and `g = 980 cms^(-12)`. |
|
Answer» Let `I` be the length of the tbe inside water. The forces acting on the tube are: a. Upthrust of water acting upward `=pirho^(2)lxx1xx980` `=22/7x(0.14)^(2)lxx980=60.368l`dyne b. Weight of the system acting downwards `=mg=0.2xx980=196`dyne c. Force of surface tension acting downward `=2pirhoT=64.24`dyne Since the tube is in eqilibrium the upward force is balanced by the downward force. That is `60.368l=196+64.24=260.24` `l=4.31cm` |
|
| 33. |
A massive stone pillar `20 m` high and of uniform cross section rests on a rigid base and supports a vertical load of `5.0 xx 10^(5)N` at its upper end. If the compressive stress in the pillar is not exceed `16 xx10^(6)N//m`, what is the minimum cross-sectional area of the pillar? (Density of the stone `=2.5xx10^(3) kg//m^(3)` take `g= 10 N//kg)`A. `0.15m^(2)`B. `0.25m^(2)`C. `0.35m^(2)`D. `0.45m^(2)` |
|
Answer» Correct Answer - D `(20axx2.5xx10^(3)xx10+5xx10^(5))/a=1.6xx10^(6)` or `500a+500=1600a` or `100a=500` or `a=5/11m^(2)=0.45m^(2)` |
|
| 34. |
A liquid is containe in a vertical tube of semicircular cross section figure.The contact angle is zero. The force of surface tension on the curved part and on the flat part are in ratio A. `2:pi`B. `1:pi`C. `3:pi`D. `2.7:pi` |
|
Answer» Correct Answer - A Effective length of flat that `=2r` Effective length of curved part is `pir`. `F=sigmal` So required ratio is `2:pi` |
|
| 35. |
A wire is suspended vertically from a rigid support. When loaded with a steel weight in air, the wire extends by `16cm`. When the weight is completely immersed in Water, the extension is reduced to `14 cm`. The relative density of the material of the weight isA. `2g//cm^(3)`B. `6g//cm^(3)`C. `8g//cm^(3)`D. `16 g//cm^(3)` |
|
Answer» Correct Answer - C `Y=(Fl)/(a/_l), Y, l` and `a` constants `:. F=/_l` In the first case `Vrhog=16` In the second case `(Vrho-Vxx1xxg)=14` Dividing `=(Vg(rho-1))/(Vgrho)=14/16=7/8` or `(rho-1)/rho=7/8` or `8rho-7rho` or `rho=8g//cm^(3)` |
|
| 36. |
A liquid of specific gravity `1.5` is observed to rise `3.0 cm` in a capillary tube of diameter `0.50 mm` and the liquid wets the surface of the tube. Calculate the excess pressure inside a spherical bubble of `1.0 cm` diameter blown from the same liquid. |
|
Answer» The surface tension of liquid is `T=(rhrog)/2` `((0.025cm)(3.0cm)(1.5gm//cm^(3))(980cm//sec)^(2))` `=55"dyne"//cm` Hence excess pressure inside a spherical bubble `/_p=(4T)/R=(4xx55"dyne"//cm)/((0.5cm))=440dyne//cm^(2)` |
|
| 37. |
If two soap bubbles of different radii are connected by a tubeA. air follows from the larger bubble into smaller bubble till both bubbles acquire-same sizeB. air follows from the smaller bubble into larger bubble and the larger bubble grows in size with decrease in size of the smaller bubbleC. air does not flow but the sizes of the bubbles changesD. sizes of the bubbles remain unchanged |
|
Answer» Correct Answer - B When two bubbles come into contact. They try to minimize the surface area for which air flows from bubble having higher pressure (smaller radius) to bubble having lower pressure. |
|
| 38. |
Two soap bubbles, one of radius `50 mm` and the other of radius `80 mm`, are brought in contact so that they have a common interface. The radius of the curvature of the common interface isA. `0.003 m`B. `0.133m`C. `1.2m`D. `8.9m` |
|
Answer» Correct Answer - B `[P_(0)+(4sigma)/(R_(2))]=[P_(0)+(4sigma)/(R_(1))](4sigma)/R` or `1/R=1/R-2-1/(R_(2))` `R=(R_(1)R_(2))/(R_(1)-R_(2))=(50xx80)/30mm=400/3mm` `=400/(2xx1000)m=4/30m=0.133m` |
|
| 39. |
Statement I: Spraying of water causes cooling.Statement II: For an isolated system, surface energy increase on the expense of internal energy.A. Statement I is true, statement II is true and Statement II is a correct explanation for Statement I.B. Statement I is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
|
Answer» Correct Answer - A On spraying, surface area increases and hence surface energy, this increase in surface energy is on the expense of decrease in internal energy and hence temperature decreases. |
|
| 40. |
Statement I: While blowing a soap bubble. to increase the size of soap bubble, we have to increase the air pressure within the soap bubble.Statement II: To increase the size of soap bubble more air has to be pushed into the bubble.A. Statement I is true, statement II is true and Statement II is a correct explanation for Statement I.B. Statement I is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
|
Answer» Correct Answer - D Excess pressure inside a soap bubble is given by `p_(i)-p_(0)=4S//r` as `p_(i)=p_(0)prop1/r`, so as `r` increase `p_(1)` decreases (`p_(0)` constant) therefore Statement I is wrong. Statement II is correct as more air has to be forced in .to cause the bubble to grow but contrary to the process of blowing up a rubber balloon, the excess pressure required to force air into the bubble decrease with the bubble size. |
|
| 41. |
Calculate the work down against surface tension in blowing a soap bubble from a radius of `10 cm` to `20 cm`, if the surface tension of soap solution in `25xx10^(-3)N//m`. |
|
Answer» Original total surface area `=2xx4pir_(1)^(2)=2xx4pixx(0.1)^(2)m^(2)` (as bubble has two surfaces). Final total surface area`=2xx4pir_(2)^(2)=2xx4pixx(0.2)^(2)m^(2)` Therefore, extension in area `=2xx4pi[(0.2)^(2)-(0.1)^(2)]=0.24pim^(2)` Now, work done `W = `surface tension `xx` extension in area `=25xx10^(-3)xx0.14pi=6pixx10^(-3)J` |
|
| 42. |
Why is moisture retained longer in the soil if it is harrowed? |
| Answer» When the soil is not harrowed, there are large capillaries in it. Water in the soil risesup the capillary holes to the surface from where it evaporates continously. Thus the soil continously loses water. When harrowed, all these capillaries are destroyed, and so capillary suction stops altogether. Thus, water is retained longer in the soil. | |
| 43. |
Four rods `A, B, C` and `1`) of the same length and material but of different radii `r, rsqrt(2),rsqrt(3)` and `2r`, respectively, are held between two rigid walls. The temperature of all rods is increased through the same range. If the rods do not bend, thenA. the stress in the rods `A, B, C` and `D` is in the ratio `1:2:3:4`B. the forces on them exerted by the wall are in the ratio `1:2:3:4`C. the energy stored in the rods due to elasticity is in the ratio `1:2:3:4 `D. it is independent of area like surface tension while friction depends |
|
Answer» Correct Answer - B::C Thermal force `=Yaalphad theta=Ypir^(2)alphad theta` `r_(1)=r, r_(2)=rsqrt(2), r_(3)=rsqrt(3), r_(4)=2r` `F_(1):F_(2):F_(3):F_(4)=1:2:3:4` Thermal stress `=Yalphad theta` As `Y` and a are same for all the rods, hence stress developed in each rod will be the same. As strain `=alphad theta`, so strain will also be the same. `E=` energy stored `=1/2Y("strain")^(2)xxAxxL` `:. E_(1):E_(2):E_(3):E_(4)=1:2:4` so, option b and c are correct. |
|
| 44. |
A vertical glass capillary tube, open at both ends, contains some water. Which of the following shapes may not be taken by the water -in the tube?A. B. C. D. |
|
Answer» Correct Answer - A::B::C The two free liquid surfaces must provide a net upward force due to surface tension to balances the weight of liquid column. |
|
| 45. |
The length of a needle floating on water is `2.5 cm`. The minimum force in addition to its weight needed to lift the needle above the surface of water will be (surface tension of water is `0.072 N//m`)A. `3.6xx10^(-3)N`B. `10^(-2)N`C. `9xx10^(-4)N`D. `6xx10^(-4)N` |
|
Answer» Correct Answer - A Surface tension force `F=Sxx2l=0.072xx2xx2.5/100=3.6xx10^(-3)N` |
|
| 46. |
A sphere of brass released in a long liquid column attains a terminal speed `v_(0)`. If the terminal speed is attained by a sphere of marble of the same radius and released in the same liquid is `nv_(0)`, then the value of `n` will be (Given: The specific gravities of brass, marble and liquid are `8.5, 2.5` and `0.8`, respectively)A. `5/17`B. `17/77`C. `1/31`D. `17/5` |
|
Answer» Correct Answer - B For the same radius, terminal velocity is proportional to the density difference. |
|
| 47. |
Statement I: A raindrop after failing through some height attains a constant velocity. Statement II: At constant velocity, the viscous drag is just equal to its weight.A. Statement I is true, statement II is true and Statement II is a correct explanation for Statement I.B. Statement I is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
|
Answer» Correct Answer - A When a raindrop falls in air (viscous medium), after falling through the same height, the viscous drag balances the weight of the drop. Through the rest of its height, velocity is constant or it attains a terminal velocity. |
|
| 48. |
A small sphere falls from rest in a viscous liquid. Due to friction, heat is prodced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity. |
|
Answer» Viscous force on a falling sphere in a liquid `F=pietav_(T)` where `v_(T)=(2.9)r^(2)(rho-sigma)g.eta` is terminal velocity `rho=` density of sphere `sigma=`density of liquid Rate of production of heat `=(="power")=Fv_(T)=6pietarv_(1)^(2)` `=6pietar[2/9(r^(2)(rho-sigma)g)/eta]^(2)=8/27(pig^(2)(rho-sigma)^(2))/etar^(5)` clearly `(dQ)/(dt)propr^(5)` |
|
| 49. |
A glass rod of diameter `d_(1)=1.5 mm` in inserted symmetricaly into a glas capillary with inside diameter `d_(2)=2.0` mm. Then the whole arrangement is vertically oriented and bruoght in contact with the surface of water. To what height will the liquid rise in the capillary? Surface tension of water `=73xx10^(-3)N//m` |
|
Answer» If `R` is radius of meniscs, then `(2T)/R=hrhog` Here `R=(r_(2)-r_(1))/(costheta)` `theta` being angle of contact `r_(1)=` radius of glass rod `r_(2)=`radius of capillary, `(2Tcostheta)/(r_(2)-r_(1))=hrhog` or `h=(2Tcostheta)/((r_(1)-r_(2))rhog)` Here `r_(1)=(d_(1))/2, r_(2)=(d_(2))/2` `:. h=(4Tcostheta)/((d_(2)-d_(1))rhog)` Substituting given values and `theta~=0^@` for water glass interface we have `h=(4xx73xx10^(-3)cos0^@)/((2.0-1.5)xx10^(-3)xx10^(3)xx9.8)=60xx10^(-3)m=6cm` |
|
| 50. |
A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four lime the density of the material of the ball. The frictional force of the liquid on the rising ball is greater than the weight of the ball by a factor ofA. `2`B. `3`C. `4`D. `6` |
|
Answer» Correct Answer - B `Vrhog+F=V(4rho)g` or `F=3Vrhog=3mg` or `F/(mg)=3` |
|