165 + Interview Questions in PROPERTIES-OF-MATTER IN GENERAL AWARENESS Page 2 InterviewSolution

51.	A cylinder with a movable piston contains air under a pressure `p_(1)` and a soap bubble of radius `r`. The pressure `p_(2)` to which the air should be compressed by slowly pushing the piston into the cylinder for the soap bubble to reduce its size by half will be : (The surface tension is `sigma`, and the temperature `T` is maintained constant)A. `p_(1) + (4T)/(r)`B. `4p_(1) +(12T)/(r)`C. `8 p_(1) + (24T)/(r)`D. `p_(1) + (2T)/(r)`
Answer» Correct Answer - C

Discussion

52.	The radius of a soap bubble is increased from `(1)/(sqrtpi)` cm to `(2)/(sqrtpi)` cm. If the surface tension of water is `30dynes` per cm, then the work done will beA. 180 ergsB. 360 ergsC. 720 ergsD. 960 ergs
Answer» Correct Answer - C `W=8piT(r_2^2-r_1^2)=8piT[((2)/(sqrtpi))^2-((1)/(sqrtpi))^2]` `W=8xxpixx30xx(3)/(pi)=720erg`

Discussion

53.	A wooden block is floating in a liquid. About `50%` of its volume is inside the liquid when the vessel is stationary, Percentage volume immersed when the vessel moves upwards with acceleration `a=g/2` isA. `75 %`B. `25%`C. `50%`D. `33.33%`
Answer» Correct Answer - C

Discussion

54.	The time period of a simple pendulum of infinte length isA. infiniteB. `2pisqrt(R//g)`C. `2pi sqrt(g//R)`D. `(1)/(2pi) sqrt(R//g)`
Answer» Correct Answer - B

Discussion

55.	In the above question the value of stress is:A. `(Wl)/(2xA)`B. `(Wl)/(4xA)`C. `(2Wl)/(lA)`D. `(4xW)/(lA)`
Answer» Correct Answer - A From the figure it is clean that `2Tcostheta=W` `impliesT=(W)/(2costheta)` Now `costheta=(x)/(sqrt(l^2+x^2))=l(x)/([1+(x^2)/(2l^2)])` If `(x^2)/(2l^2)` is neglected being small, then `costheta=(x)/(l)` `T=(W)/(2x)` Stress`=(T)/(A)=(((Wl)/(2x)))/(A)(Wl)/((2x)/(A))=(Wl)/(2Ax)` Correct choice is (a).

Discussion

56.	A soap bubble has radius `r` The work done in increasing its radius to three times its original radius, without any rise of temperature, is (Given Surface tension of soap solution is `T`)A. `12pir^2T`B. `16pir^2T`C. `64pir^2T`D. `48pir^2T`
Answer» Correct Answer - C `W=[2xx4pi(3r)^2-2xx4pir^2]T=64pir^2T`

Discussion

57.	An air bubble of 1 cm radius is rising at a steady rate of `2.00ms^-1` through a liquid of density `1.5gcm^-3`. Neglect density of air. If `g=1000cms^-2`, then the coeffieciet of viscosity of the liquid isA. `0.166xx10^3` PoiseB. `1.66xx10^-3` poiseC. `166xx10^3` poiseD. `16.6xx10^3` poise
Answer» Correct Answer - B Here `B=6pietarv` `implies(4)/(3)pir^3rho_1g=6pietarv` Put the values and get `eta=1.66xx10^3` poise

Discussion

58.	A uniform solid sphere of relative density 5 is released in water filled in a long vertical tube. Its terminal velocity achived is `V`. If another uniform solid sphere of same material but double the radius is released in the same water then the terminal velocity achived will be.A. `V`B. `4V`C. `(V)/(4)`D. `2V`
Answer» Correct Answer - B Initially the terminal velocity `V` of sphere of radius `a` is `W_(eff)=6pietaaV` (1)`(W_(eff)=`weight`-`Buoyant force) As the radius is doubled, mass is increased to 8 times and new terminal velocity will be `8W_(eff)=6pieta2aV^(`)` .(2) From (1) and (2) `V^(`)=4V`

Discussion

59.	A sphere of radius R and density `rho_1` is dropped in a liquid of density `sigma`. Its terminal velocity is `v_1`. If another sphere of radius `R` and density `rho_2` is dropped in the same liquid, its terminal velocity will be:A. `((rho_2-sigma)/(rho_1-sigma))v_1`B. `((rho_1-sigma)/(rho_1-sigma))v_1`C. `((rho_1)/(rho_2))v_1`D. `((rho_2)/(rho_1))v_1`
Answer» Correct Answer - A `6pietaRv_1=(4)/(3)piR^3(rho_1-sigma)` `6pietaRv_2=(4)/(3)piR^3(rho_2-sigma)` `(v_2)/(v_1)=((rho_2-sigma)/(rho_1-sigma))`

Discussion

60.	Two uniform solid balls of same density and if radii `r` and `2r` are dropped in air and fall vertically downwards. The terminal velocity of the ball with radius `r` is `1cms^-1`, then the terminal velocity of the ball of radius `2r` will be (neglect buoyant force on the balls.)A. `0.5cms^-1`B. `4cms^-1`C. `1cms^-1`D. `2cms^-1`
Answer» Correct Answer - B At equilibrium `mg=6pietarv_0` or `rho(4pi)/(3)g=6pietarv` `(v_r)/(v_(2r))=((r )^2)/((2r)^2)` or `v_2r=(v_r)xx4=4(cm)/(s)`

Discussion

61.	Two indetical spherical drops of water are falling (vertically downwards) through air with a steady velocity of `5(cm)/(sec)`. If both the drops coalesce (combine) to form a new spherical drop, the terminal velocity of the new drop will be (neglect buoyant force on the drops.)A. `5xx2(cm)/(sec)`B. `5xxsqrt2(cm)/(sec)`C. `5xx(4)^((1)/(3))(cm)/(sec)`D. `(5)/(sqrt2)(cm)/(sec)`
Answer» Correct Answer - C When two drops of radius `r` each combine to form a big drop, the radius of big drop will be given by `(4)/(3)piR^3=(4pi)/(3)r^2+(4pi)/(3)r^3` or `R^3=2r^2` or `R=2^((1)/(3))r` Now `(V_R)/(V_r)=((R )/(R ))^2=2^((2)/(3))=4^((1)/(3))` `V_R=5xx4^((1)/(3))(cm)/(s)`

Discussion

62.	If a liquid neither rises nor depresses in a capillary then it means thatA. angle of contact is `0^(@)`B. angle of contact may by `90^(@)`C. surface tension of the liquid must be zeroD. None of these
Answer» Correct Answer - B

Discussion

63.	A siphon tube is discharging a liquid of density `900(kg)/(m^(3))` as shown in figure. `(P_(0)=1.01xx10^(5) N//m^(2))` Pressure at point C isA. `5.5xx10^(4) N//m^(2)`B. `6.5xx10^(4) N//m^(2)`C. `8.0xx10^(4) N//m^(2)`D. `10.5xx10^(4) N//m^(2)`
Answer» Correct Answer - B

Discussion

64.	A siphon tube is discharging a liquid of specific gravity 0.9 from a reservoir as shown in the figure. (a) Find the velocity of the liquid through the siphon. (b) Find the pressure at the highest point B. (c ) Find the pressure at point C.A. 6 m/sB. 8 m/sC. 10 m/sD. 12 m/s
Answer» Correct Answer - C

Discussion

65.	A siphon tube is discharging a liquid of density `900(kg)/(m^(3))` as shown in figure. `(P_(0)=1.01xx10^(5) N//m^(2))` Pressure at point B isA. `4.25xx10^(4) N//m^(2)`B. `6.25xx10^(4) N//m^(2)`C. `2.50xx10^(4) N//m^(2)`D. `2.0xx10^(5) N/m^(2)`
Answer» Correct Answer - A

Discussion

66.	On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross section 2`mm^2`, the force required is [Y for steel `=2.2xx10^11(N)/(m^2)`]A. `1.1xx10^5N`B. `1.1xx10^4N`C. `1.1xx10^3N`D. `1.1xx10^2N`
Answer» Correct Answer - D `Y=(Fl)/(Atrianglel)` or `F=(YAtrianglel)/(l)` or `F=(2.2xx10^11xx2xx10^-6xx0.5xx10^-3)/(2)=1.1xx10^2N`

Discussion

67.	A lift is tied with thick iron and its mass is 314 kg. What should be the minimum diameter of wire if the maximum acceleration of lift is `1.2(m)/(sec^2)` and the maximum safe stress of the wire is `1xx10^7(N)/(m^2)`?A. 2 cmB. 1 cmC. 1.5 cmD. none of these
Answer» Correct Answer - A The tension T in the rope of the lift when it goes upward is given by `T=m(g+a)=314xx11N` Let r be the radius of the wire, Then maximum stress will be `(T)/(pir^2)` Hence `(T)/(pir^2)` `=1.1xx10^7` or `r^2=(T)/(pixx1.4xx10^8)=(314xx11)/(3.14xx(1.1xx10^7))=(1)/(10^4)` Now `r=(1)/(10^2)m=1cm` Diameter of the wire `=2r=2cm`.

Discussion

68.	The mean density of sea water is `rho`, and bulk modulus is B. The change in density of sea water in going from the surface of water in going from the surface of water to a depth of `h` isA. `(B rho^(2))/(gh)`B. `B rho gh`C. `(rho^(2)gh)/(B)`D. `(rho gh)/(B)`
Answer» Correct Answer - C

Discussion

69.	A sample of a liquid has an initial volume of 1.5 L The volume is reduced by 0.2 mL, when the pressure increases by 140 kPz. What is the bulk modulus of the liquid?A. `1.05xx10^9Pa`B. `3.05xx10^9Pa`C. `2.10xx10^9Pa`D. `5.10xx10^9Pa`
Answer» Correct Answer - A `B=-(triangleP)/((triangleV)/(V))=-(VtriangleP)/(triangelV)=-(1.5xx140xx10^3)/(-0.2xx10^-3)` `=1.05xx10^9Pa`

Discussion

70.	A sample of a liquid has an initial volume of 1.5 L The volume is reduced by 0.2 mL, when the pressure increases by 140 kPz. What is the bulk modulus of the liquid?A. `1.05xx10^9Pa`B. `1.1xx10^9Pa`C. `1.2xx10^9Pa`D. `1.4xx10^9Pa`
Answer» Correct Answer - A `B=-(triangleP)/((triangleV)/(V))=-(VtriangleP)/(triangelV)=-(1.5xx140xx10^3)/(-0.2xx10^-3)` `=1.05xx10^9Pa`

Discussion

71.	A metallic wire of diameter `d` is lying horizontally o the surface of water. The maximum length of wire so that is may not sink will beA. `sqrt((2T)/(pi rho g))`B. `sqrt((4T)/(rho g)`C. `sqrt((T)/(pi rho g))`D. `sqrt((T rho)/(pi g))`
Answer» Correct Answer - A

Discussion

72.	Two liquids which do not react chemically are placed in a bent tube as shown in figure. The height of the liquids above their surface of separation areA. directly proportional to their densitiesB. inversely proportional to their densitiesC. directly proprotional to square of their densitiesD. equal
Answer» Correct Answer - B

Discussion

73.	A wire frame ABCD has a soap film. The wire BC can slide on the frame without friction and it is in equilibrium in the position shown in the figure. Find m, if T is the surface tension of the liquid. A. `(2Tl)/(g)`B. `(Tl)/(g)`C. `(Tl^(2))/(2g)`D. `(Tl)/(2g)`
Answer» Correct Answer - A

Discussion

74.	A square wire frame of length `l` is dipped in a solution. When the frame is taken out, a liquid film is formed. What is the algebraic sum of all the force acting on the frame due to surface tension of the liquid? (given`sigma=`surface tension of the liquid).A. `4sigmal`B. `8sigmal`C. `10sigmal`D. `12sigmal`
Answer» Correct Answer - B It may be noted that the soap film has two free surfaces. So the effective length is `8l`

Discussion

75.	A small steel ball of mass `m` and radius `r` is falling under gravity through a viscous liquid of coefficient of viscosity `eta`. If `g` is the value of acceleration due to gravity. Then the terminal velocity of the ball is proportional to (ignore buoyancy)A. `(mg(eta))/(r )`B. `mg(eta)r`C. `(mgr)/(eta)`D. `(mg)/(reta)`
Answer» Correct Answer - D Since the steedl ball is given to be small therefore Now, `6pietarv_0=mg` or `v_0pro(mg)/(etar)`

Discussion

76.	A force of `3.14 N` is required to drag a sphere of radius `4 cm` with a speed of `5 ms^(-1)` in a medium in gravity free space. Find the coefficient of the viscosity of the medium.A. `8.3xx10^-5poise`B. `4.3xx10^-5poise`C. `7.4xx10^-5poise`D. `2.3xx10^-5poise`
Answer» Correct Answer - A `F=6pietarv` `eta=(F)/(6pirv)=(3.14)/((6)(3.14)(5xx10^2))=8.3xx10^-5poise`

Discussion

77.	For a planet moving around the sun in an elliptical orbit, which of the following quantities remain constant ?A. The total energy of the sun plus planet systemB. The angular momentum of the planet about the sunC. The force of attraction between the twoD. The linear momentum of the planet
Answer» Correct Answer - A::B

Discussion

78.	A satellite going round the earth in a circular orbit loses some energy due to a collision. Its speed is `upsilon` and distance from the earth is d.A. d will increase, `upsilon` will increaseB. d will increase, `upsilon` will decreaseC. d will decrease, `upsilon` will decreaseD. d will decrease, `upsilon` will increase.
Answer» Correct Answer - D

Discussion

79.	A satellite close to the earth is in orbit above the equator with a period of rotation of `1.5` hours. If it is above a point `P` on the equtor at some time, it will be above `P` again after time ...........A. 1.5 hoursB. 1.6 hours if it is rotating from west to eastC. 24/17 hours if it is rotating from west to eastD. 24/17 hours if it is rotating from east to west
Answer» Correct Answer - B::D

Discussion

80.	A block of gelatine is 60 mm be 60 mm be 20 mm when unstressed. A force of 0.245 N is applied tangentially to the upper surface, causing a 5 mm displacement relative to the lower surface. Following observations are made regarding the block. (i) the shearing strees develop on surface is 68.1 Pa (ii) the shearing strain develop on surface is 0.25 Pa (iii) the shear modulus of material is `272(N)/(m^2)` Regarding above statements we can say:A. only statements (i) and (iii) are correctB. only statements (ii) and (iii) are correctC. only statements (i) and (ii) are correctD. all the statements are correct
Answer» Correct Answer - D (i) Stress`=(F)/(A)=(0.245)/(36xx10^-4)=68.1Pa` (ii) Strain`=theta=(r )/(h)=(5)/(20)=0.25` (iii) Shear modulus`=S` `((F)/(A))/(tantheta)=(68.4)/(0.25)=272.4(N)/(m^2)`

Discussion

81.	A capillary tube of length(i) `l = 60 cm`,(ii) `l = 50 cm`and radius `r = 1//4 mm` is immersed vertically into water. Find the capillary rise in both cases. Angle of contact `= 0^@`. Take coefficient of surface tension as `72 "dyne"//cm, g = 1000 cm s^(-2)`.A. `30cm`B. `35cm`C. `40cm`D. `50cm`
Answer» Correct Answer - D `h=h_0=(2Tcostheta)/(rhogr)` `=(2(72)cos0^@)/((1)(1000)((1)/(40)))=57.6cm` Since `l(=50cm)lth_0` `h=50cm`

Discussion

82.	When load of 5 kg is hung on a wire than extension of 3 m takes place, then work done will beA. `75 "joule"`B. `60 "joule"`C. `50 "joule" `D. `100 "joule" `
Answer» Correct Answer - A `W=(1)/(2)Fl=(1)/(2)xxMgxxl=(1)/(2)xx5xx10xx3=75J`

Discussion

83.	If the potential energy of a spring is `V` on stretching it by `2cm`, then its potential energy when it is stretched by `10cm` will beA. `(V)/(25)`B. `5V`C. `(V)/(5)`D. `25V`
Answer» Correct Answer - D `U=(1)/(2)((YA)/(L))l^2` `because` `Upropl^2` `(U_2)/(U_1)=((l_2)/(l_1))^2=((10)/(2))^2impliesU_2=25U_1` i.e., potential energy of the spring will be `25V`

Discussion

84.	A soap film measure `10cmxx6cm`. It is increased to `10cmxx12cm`. IF surface tension is `30xx10^-3` newtor per metre. Then the work done isA. `3.6xx10^0J`B. `3.6xx10^-2J`C. `3.6xx10^-4J`D. `3.6xx10^3J`
Answer» Correct Answer - C `W=2[120xx10^-4-60xx10^-4]30xx10^-3J` `=2xx60xx10^-4xx30xx10^-3J=3.6xx10^-4J`

Discussion

85.	A ball floats on the surface of water in a container exposed to the atmosphere. Volume `V_(1)` of its volume is inside the water. The container is now covered and the air is pumped out. Now let `V_(2)` be the volume immersed in water. ThenA. `V_(1) = V_(2)`B. `V_(1) gt V_(2)`C. `V_(2) gt V_(1)`D. `V_(2) = 0`
Answer» Correct Answer - C

Discussion

86.	An ice cube of size `a = 10 cm` is floating in a tank (base area `A = 50 cm xx 50 cm`) partially filled with water. The change in gravitational potential energy, when ice melts completely is (density of ice is `900 kg//m^(2)`)A. `- 0.072 J`B. `- 0.24 J`C. `- 0.016 J`D. `-0.045 J`
Answer» Correct Answer - D

Discussion

87.	A satellite revolves round the earth. Air pressure inside the satellite is maintained at 76 cm of mercury. What will be the height of mercury column in a barometer tube 1m long placed inn the satellite?A. 76 cmB. 90 cmC. zeroD. None of these
Answer» Correct Answer - B

Discussion

88.	In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as (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 but reason is false. (d) If assertion and reason are false. Q. A needle placed carefully on the surface of water may float, whereas a ball of the same material will always sink. Reason: The buoyancy of an object depends both on the material and shape of the object.
Answer» Correct Answer - C Needle floats due to surface tension there is no role of buoyant force in its floating. Buoyant force `=Vsigmag` where `V=` volume of body submerged in liquid `sigma=`density of liquid. i.e., the buoyancy of an object depends on the shape of the object.

Discussion

89.	A drop of water of mass `m` and density `rho` is placed between two weill cleaned glass plates, the distance between which is `d`. What is the force of attraction between the plates? `(T=` surface tension)A. `(Tm)/(2 rho d^(2))`B. `(4Tm)/(rho d^(2))`C. `(2Tm)/(rho d^(2))`D. `(Tm)/(rho d^(2))`
Answer» Correct Answer - C

Discussion

90.	What is the height to which a liquid rises between two long parallel plates, a distance d apart ? (Surface tension of liquid is T and density is `rho`)A. `(4T)/(rho g d)`B. `(2T)/(rho g d)`C. `(T)/(rho g d)`D. `(T)/(2 rho g d)`
Answer» Correct Answer - B

Discussion

91.	In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as (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 but reason is false. (d) If assertion and reason are false. Q. Assertion: If length of a rod is doubled the breaking load remains unchanged. Reason: Breaking load is equal to the elastic limit.
Answer» Correct Answer - C Breaking load depends on the sea of cross section and is independent length of rod. Hence, Breaking load`=` breaking stress`xx`cross sectional area.

Discussion

92.	In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as (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 but reason is false. (d) If assertion and reason are false. Q. Assertion: Bulk modulus of elasticity B represents incompressibility of the material Reason: `B=-(triangleP)/((triangleV)/(V))`, where symbols have their usual meaning.
Answer» Correct Answer - B Bulk modulus of elasticity measures how good the body is to regain its original volume on being compressed. Therefore it represents incompressibility of the material Bulk modulus`B=-(triangleP)/((triangleV)/(V))=-((triangleP)V)/(triangleV)` Where `triangleP=` Change in pressure and `triangleV=` change in volume Hence, negative sign implies that when the pressure increases, volume decreases and vice-versa.

Discussion

93.	The approximate depth of an ocean is `2700m`. The compressibility of water is `45.4xx10^(-11)Pa^-1` and density of water is `10^3(kg)/(m^3)`. What fractional compression of water will be obtained at the bottom of the ocean?A. `0.8xx10^-2`B. `1.0xx10^-2`C. `1.2xx10^-2`D. `1.4xx10^-2`
Answer» Correct Answer - C Bulk modulus `\|B\|=(trianglep)/(((triangleV)/(V)))` `B=(1)/(compresibility(K))` `(1)/(K)=(rrhog)/(((triangleV)/(V)))` `((triangleV)/(V))=(hrhog)K` `implies(triangleV)/(V)=2.7xx10^3xx10^3xx10xx45.4xx10^(-11)` `=1.23xx10^-2`

Discussion

94.	The area of cross-section of the wider tube shown in fig., is `800cm^(2)` . If a mass of 12 kg is placed on the massless piston, what is the difference in the level of water in two tubes. A. 10 cmB. 6 cmC. 15 cmD. 2 cm
Answer» Correct Answer - C

Discussion

95.	When temperature of a gas is `20^@C` and pressure is changed from `p_1=1.01xx10^5Pa` to `p_2=1.165xx10^5Pa` then the volume changed by `10%`. The bulk modulus isA. `204.8xx10^5Pa`B. `102.4xx10^5Pa`C. `51.2xx10^5Pa`D. `1.55xx10^5Pa`
Answer» Correct Answer - D `K=(trianglep)/((triangleV)/(V))=((1.165-1.01)xx10^5)/((10)/(100))=(0.155xx10^5)/((1)/(10))` `=1.55xx10^5Pa`

Discussion

96.	The pressure applied from all direction on a cube is P. How much its temperature should be raised to maintain the original volume ? The volume elasticity of the cube is `beta` and the coefficient of volume expansion is `alpha`A. `(P)/(alphabeta)`B. `(Palpha)/(beta)`C. `(Pbeta)/(alpha)`D. `(alphabeta)/(P)`
Answer» Correct Answer - A If coefficient of volume expansion is `alpha` and rise in temperature is `triangletheta` then`triangleV=Valphatrianglethetaimplies(triangleV)/(V)=alphatriangletheta` Volume elasticity `beta=(P)/((triangleV)/(V))=(P)/(alphatriangletheta)impliestriangle=(P)/(alphabeta)`

Discussion

97.	A large tank is filled waith water (density `= 10^(3) kg//m^(3)`). A small hole is made at a depth 10 m below water surface. The range of water issuing out of the hole is R on ground. Approximately what extra pressure must be applied on the water surface so that the range becomes 2R (take `1 atm = 10^(5) Pa` and `g = 10 m//s^(2)`) A. 9 atmB. 4 atmC. 5 atmD. 3 atm
Answer» Correct Answer - D

Discussion

98.	Soap helps in cleaning clothes, becauseA. Chemicals of soap changeB. it increases the surface tension of the solutionC. It absorbs the dirtD. It lower the surface tension of the solution.
Answer» Correct Answer - D Soap helphs to lower the surface tension of solution, thus soap get stick to the dust particles and grease and these are removed by action of water.

Discussion

99.	Assuming the xylem tissues through which water rises from root to the branches in a tree to be of uniform cross-section find the maximum radius of xylem tube in a `10m` high coconut tree so that water can rise to the top. (Surface tension of water`=0.1(N)/(m)`, Angle of contact of water with xylem tube`=60^@`)A. `1cm`B. `1mm`C. `10mum`D. `1mum`
Answer» Correct Answer - D `rhohgpir^2=2pirScostheta` `impliesr=(2Scostheta)/(rhogh)=(2xx1xx0.5)/(10^3xx10xx10)=10^-6m`

Discussion

100.	Water rises to a height of `16.3cm` in a capillary of height `18cm` above the water level. If the tube is cut at a height of `12cm`A. Water will come as a fountain from the capillary tube.B. Water will stay at a height of `12cm` in the capillary tubeC. The height of the water in the capillary will be `10.3cm`D. Water will flow down the sides of the capillary tube
Answer» Correct Answer - B Because if the length available is less than required then water will rise upto available height and adjust its radius of curvature.

Discussion

Explore topic-wise InterviewSolutions in .

The radius of a soap bubble is increased from `(1)/(sqrtpi)` cm to `(2)/(sqrtpi)` cm. If the surface tension of water is `30dynes` per cm, then the work done will beA. 180 ergsB. 360 ergsC. 720 ergsD. 960 ergs

A wooden block is floating in a liquid. About `50%` of its volume is inside the liquid when the vessel is stationary, Percentage volume immersed when the vessel moves upwards with acceleration `a=g/2` isA. `75 %`B. `25%`C. `50%`D. `33.33%`

The time period of a simple pendulum of infinte length isA. infiniteB. `2pisqrt(R//g)`C. `2pi sqrt(g//R)`D. `(1)/(2pi) sqrt(R//g)`

In the above question the value of stress is:A. `(Wl)/(2xA)`B. `(Wl)/(4xA)`C. `(2Wl)/(lA)`D. `(4xW)/(lA)`

A soap bubble has radius `r` The work done in increasing its radius to three times its original radius, without any rise of temperature, is (Given Surface tension of soap solution is `T`)A. `12pir^2T`B. `16pir^2T`C. `64pir^2T`D. `48pir^2T`

An air bubble of 1 cm radius is rising at a steady rate of `2.00ms^-1` through a liquid of density `1.5gcm^-3`. Neglect density of air. If `g=1000cms^-2`, then the coeffieciet of viscosity of the liquid isA. `0.166xx10^3` PoiseB. `1.66xx10^-3` poiseC. `166xx10^3` poiseD. `16.6xx10^3` poise

If a liquid neither rises nor depresses in a capillary then it means thatA. angle of contact is `0^(@)`B. angle of contact may by `90^(@)`C. surface tension of the liquid must be zeroD. None of these

A siphon tube is discharging a liquid of density `900(kg)/(m^(3))` as shown in figure. `(P_(0)=1.01xx10^(5) N//m^(2))` Pressure at point C isA. `5.5xx10^(4) N//m^(2)`B. `6.5xx10^(4) N//m^(2)`C. `8.0xx10^(4) N//m^(2)`D. `10.5xx10^(4) N//m^(2)`

A siphon tube is discharging a liquid of specific gravity 0.9 from a reservoir as shown in the figure. (a) Find the velocity of the liquid through the siphon. (b) Find the pressure at the highest point B. (c ) Find the pressure at point C.A. 6 m/sB. 8 m/sC. 10 m/sD. 12 m/s

A siphon tube is discharging a liquid of density `900(kg)/(m^(3))` as shown in figure. `(P_(0)=1.01xx10^(5) N//m^(2))` Pressure at point B isA. `4.25xx10^(4) N//m^(2)`B. `6.25xx10^(4) N//m^(2)`C. `2.50xx10^(4) N//m^(2)`D. `2.0xx10^(5) N/m^(2)`

On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross section 2`mm^2`, the force required is [Y for steel `=2.2xx10^11(N)/(m^2)`]A. `1.1xx10^5N`B. `1.1xx10^4N`C. `1.1xx10^3N`D. `1.1xx10^2N`

A lift is tied with thick iron and its mass is 314 kg. What should be the minimum diameter of wire if the maximum acceleration of lift is `1.2(m)/(sec^2)` and the maximum safe stress of the wire is `1xx10^7(N)/(m^2)`?A. 2 cmB. 1 cmC. 1.5 cmD. none of these

The mean density of sea water is `rho`, and bulk modulus is B. The change in density of sea water in going from the surface of water in going from the surface of water to a depth of `h` isA. `(B rho^(2))/(gh)`B. `B rho gh`C. `(rho^(2)gh)/(B)`D. `(rho gh)/(B)`

A sample of a liquid has an initial volume of 1.5 L The volume is reduced by 0.2 mL, when the pressure increases by 140 kPz. What is the bulk modulus of the liquid?A. `1.05xx10^9Pa`B. `3.05xx10^9Pa`C. `2.10xx10^9Pa`D. `5.10xx10^9Pa`

A sample of a liquid has an initial volume of 1.5 L The volume is reduced by 0.2 mL, when the pressure increases by 140 kPz. What is the bulk modulus of the liquid?A. `1.05xx10^9Pa`B. `1.1xx10^9Pa`C. `1.2xx10^9Pa`D. `1.4xx10^9Pa`

A metallic wire of diameter `d` is lying horizontally o the surface of water. The maximum length of wire so that is may not sink will beA. `sqrt((2T)/(pi rho g))`B. `sqrt((4T)/(rho g)`C. `sqrt((T)/(pi rho g))`D. `sqrt((T rho)/(pi g))`

Two liquids which do not react chemically are placed in a bent tube as shown in figure. The height of the liquids above their surface of separation areA. directly proportional to their densitiesB. inversely proportional to their densitiesC. directly proprotional to square of their densitiesD. equal

A wire frame ABCD has a soap film. The wire BC can slide on the frame without friction and it is in equilibrium in the position shown in the figure. Find m, if T is the surface tension of the liquid. A. `(2Tl)/(g)`B. `(Tl)/(g)`C. `(Tl^(2))/(2g)`D. `(Tl)/(2g)`

A force of `3.14 N` is required to drag a sphere of radius `4 cm` with a speed of `5 ms^(-1)` in a medium in gravity free space. Find the coefficient of the viscosity of the medium.A. `8.3xx10^-5poise`B. `4.3xx10^-5poise`C. `7.4xx10^-5poise`D. `2.3xx10^-5poise`

For a planet moving around the sun in an elliptical orbit, which of the following quantities remain constant ?A. The total energy of the sun plus planet systemB. The angular momentum of the planet about the sunC. The force of attraction between the twoD. The linear momentum of the planet

A capillary tube of length(i) `l = 60 cm`,(ii) `l = 50 cm`and radius `r = 1//4 mm` is immersed vertically into water. Find the capillary rise in both cases. Angle of contact `= 0^@`. Take coefficient of surface tension as `72 "dyne"//cm, g = 1000 cm s^(-2)`.A. `30cm`B. `35cm`C. `40cm`D. `50cm`

When load of 5 kg is hung on a wire than extension of 3 m takes place, then work done will beA. `75 "joule"`B. `60 "joule"`C. `50 "joule" `D. `100 "joule" `

If the potential energy of a spring is `V` on stretching it by `2cm`, then its potential energy when it is stretched by `10cm` will beA. `(V)/(25)`B. `5V`C. `(V)/(5)`D. `25V`

A soap film measure `10cmxx6cm`. It is increased to `10cmxx12cm`. IF surface tension is `30xx10^-3` newtor per metre. Then the work done isA. `3.6xx10^0J`B. `3.6xx10^-2J`C. `3.6xx10^-4J`D. `3.6xx10^3J`

An ice cube of size `a = 10 cm` is floating in a tank (base area `A = 50 cm xx 50 cm`) partially filled with water. The change in gravitational potential energy, when ice melts completely is (density of ice is `900 kg//m^(2)`)A. `- 0.072 J`B. `- 0.24 J`C. `- 0.016 J`D. `-0.045 J`

A satellite revolves round the earth. Air pressure inside the satellite is maintained at 76 cm of mercury. What will be the height of mercury column in a barometer tube 1m long placed inn the satellite?A. 76 cmB. 90 cmC. zeroD. None of these

A drop of water of mass `m` and density `rho` is placed between two weill cleaned glass plates, the distance between which is `d`. What is the force of attraction between the plates? `(T=` surface tension)A. `(Tm)/(2 rho d^(2))`B. `(4Tm)/(rho d^(2))`C. `(2Tm)/(rho d^(2))`D. `(Tm)/(rho d^(2))`

What is the height to which a liquid rises between two long parallel plates, a distance d apart ? (Surface tension of liquid is T and density is `rho`)A. `(4T)/(rho g d)`B. `(2T)/(rho g d)`C. `(T)/(rho g d)`D. `(T)/(2 rho g d)`

The approximate depth of an ocean is `2700m`. The compressibility of water is `45.4xx10^(-11)Pa^-1` and density of water is `10^3(kg)/(m^3)`. What fractional compression of water will be obtained at the bottom of the ocean?A. `0.8xx10^-2`B. `1.0xx10^-2`C. `1.2xx10^-2`D. `1.4xx10^-2`

The area of cross-section of the wider tube shown in fig., is `800cm^(2)` . If a mass of 12 kg is placed on the massless piston, what is the difference in the level of water in two tubes. A. 10 cmB. 6 cmC. 15 cmD. 2 cm

When temperature of a gas is `20^@C` and pressure is changed from `p_1=1.01xx10^5Pa` to `p_2=1.165xx10^5Pa` then the volume changed by `10%`. The bulk modulus isA. `204.8xx10^5Pa`B. `102.4xx10^5Pa`C. `51.2xx10^5Pa`D. `1.55xx10^5Pa`

Soap helps in cleaning clothes, becauseA. Chemicals of soap changeB. it increases the surface tension of the solutionC. It absorbs the dirtD. It lower the surface tension of the solution.