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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. |
Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by `75xx10^-4` newton force due to the weight of the liquid. If the surface tension of water is `6xx`10^-2` newton/metre the inner circumference of the capillary must be:A. `1.25xx10^(-2)m`B. `0.50xx10^(-2)m`C. `6.5xx10^(-2)m`D. `12.5xx10^(-2)m` |
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Answer» Correct Answer - D Weight of liquid = upward force due to surface tension `75xx10^(-4) = 2pirT` Circumference `2pir = (75xx10^(-4))/(T) = (75xx10^(-4))/(6xx10^(-2)) = 0.125 = 12.5xx10^(-2)m` |
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| 52. |
The wooden plank of length 1 m and uniform cross section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. Find the angle `theta` that the plank makes with the vertical in the equilibrium position. (Exclude the case `theta=0`) , |
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Answer» The force acting on the plane are shown in the ure. The height of water level is l=0.5m. The length of the plank is 1.0=2l. The weight of the lane acts through the centre B of the plank. We have OB=l. The buoyant fore F acts through the point A which is the middle point of the dipped part OC of the plank. We have `OA=(OC)/2=l/(2costheta) ` Let the mass per unit length of the plane be rho`. it weight `mg=2lrhog`. The mass of the part OC of the plank `=(l/(costheta))rho`. The mass of water displaced `=1/0.5 l/(costheta)rho=(2/p)/(costheta)` The buoyant force F is therefore `F=(2lrhog)/(costheta)` Now for equilibrium the torque of mg about O should balance the torque of F about O. So,` mg(OB)sintheta=F(OA)sintheta` `or , (2lrho)=((2lrho)/(costheta))(l/(2costheta))` `or, cos^2theta= 1/2` `or costheta=1/sqrt2, or theta=45^@` |
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| 53. |
A block of wood is floating in water in a closed vessel as shown in the figure. The vessel is connected to an air pump. When more air is pushed into the vessel, the block of wood floats with (neglect compressibility of water) A. larger path in the waterB. lesser part in the waterC. same part in the waterD. it will sink |
| Answer» Correct Answer - C | |
| 54. |
A meta cube is plced in an empty vessel. When water is filed in the vessel so that the cubeis completely immersed in the water, the fore on the bottom of the vessel in contact with the cubeA. will increaseB. will decreaseC. will remain the sameD. will become zero |
| Answer» Correct Answer - C | |
| 55. |
A wooden object floats in water kept in as beaker. The object is near a side of the beaker figure. Let `P_1,P_2,P_3` be the pressure at the three points A,B and C of the bottom as shown in the figure. A. `P_1=P_2=P_3`B. `P_1ltP_2ltP_3`C. `P_1gtP_2gtP_3`D. `P_2=P_3=P_1` |
| Answer» Correct Answer - A | |
| 56. |
A wooden object floats in water kept in as beaker. The object is near a side of the beaker figure. Let `P_1,P_2,P_3` be the pressure at the three points A,B and C of the bottom as shown in the figure. A. `P_1 = P_2 = P_3`B. `P_1 lt P_2 lt P_3`C. `P_1 gt P_2 gt P_3`D. `P_2 + P_3 ne P_1` |
| Answer» Correct Answer - A | |
| 57. |
To measure the atmospheric pressure, four different tubes of length 1m, 2m , 3m and 4m are used. If the height of the mercury column in the tubes is `h_(1),h_(2),h_(3),h_(4)` respectively in the four cases, then `h_(1):h_(2):h_(3):h_(4)`A. 1:2:3:4B. 4:3:2:1C. 1:2:2:1D. 1:1:1:1 |
| Answer» Correct Answer - B | |
| 58. |
An open tank `10 m` long and `2 m` deep is filled up to `1.5 m` height of oil of specific gravity `0.82`. The tank is uniformly accelerated along its length from rest to a speed of `20 m//s` horizontally. The shortest time in which the speed may be attained without spilling any oil isA. 20B. 18sC. 10 sD. 5s |
| Answer» Correct Answer - B | |
| 59. |
A rectangular vessel is filled with water and oil in equal proportion (by volume), the oil being twice lighter than water. Show that the force on ach side wall of the vessel will be reduced by one fifth it the vessel is filled only with oil. (Assume atmospheric pressrue is negligible) |
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Answer» Initially `[(0 + rhoga)/(2)] (a)^(2) + [(rhoga + 3rhoga)/(2)] (a^(2)) = 2.5 rhoga^(2)` Finally `[(0 + rhoga)/(2)] (2a^(2)) = 2rhoga^(3)` Difference `= 0.5 rhoga^(3)` [Which is one fifth of initial] |
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| 60. |
A cube of wood suppporting a `200 gm` mass just floats in water. When the mass is removed the cube rises by `2 cm` at equilirbrium. Find side of the cube. |
| Answer» Correct Answer - `10 cm` | |
| 61. |
A barometer made of very narrow tuve (see fig) is placed at normal temperature and pressure. The coefficient of volume expansion of mercury is `0.00018per C^@` and that of the tube is negligible. The temperature of mercury in the barometer is now raised by `1^@C`, but the temperature of the atmosphere does not raised by `1^@C`, but the temperature of the atmosphere does not change. Then the mercury height in the tube remains unchanged. |
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Answer» False, Pressure `P_(1)=P_(2)=1 atm =hrhog`. On changing the temperature `g` will not change and atmosphere pressure wil not change `:.hxxrho=constt.` When temperature is increased the density of `Hg` decreases and hence `h` increases. |
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| 62. |
In the figure shown, the heavy cylinder (radius R) reasting on a smooth surface separates two liquids of densities `2rho` and `3rho`. The height h for the equilibrium of cylinder must be A. `3R//2`B. `Rsqrt((3)/(2))`C. `Rsqrt(2)`D. None of these |
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Answer» Correct Answer - B Torque of hydrostatic force about centre of sphere is already zero as hydrostatic force passes through the centre. `:. F_(LHS) = F_(RHS)` `:. (1)/(2) (2rho) (g) h^(2) = (1)/(2) (3 rho) (g) R^(2)` `:. h sqrt((3)/(2))R` |
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| 63. |
A body floats with one-third of its volume outside water and `3//4` of its volume outside another liquid. The density of another liquid is :A. `9/4 g//c c`B. `4/9 g//c c`C. `8/3 g//c c`D. `3/8 g//c c` |
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Answer» Correct Answer - C Suppose volume and density of the body be `V` and `rho` respectively, the, according to law of flatiron in water. weight = upthrust `V rho g = 2/3 V rho_(w)g`..(i) in liquid `V rhog = 1/4 V rho_(L)g` or `(rho_L)/(rho_w) = (2//3)/(1//4) = 8/3` or, `rho_(L) = 8/3 rho_(w) = 8/3 xx 1 g//c c`. |
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| 64. |
A body floats in water with 40% of its volume outside water. When the same body floats in an oil. 60% of its volume remians outside oil. The relative density of oil isA. `0.9`B. `1.0`C. `1.2`D. `1.5` |
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Answer» Correct Answer - D Using the law of buoyancy, we have `V sigmag=0.6 V sigma_(1)g`, for the part of body outside oil and `Vsigma_(2)g`, for the part of body outside water, hence we get `1=(0.6 sigma_(1))/(0.4 sigma_(2))` So, we have, `(sigma_(1))/(sigma_(2))=(6)/(4)=(3)/(2)=15` |
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| 65. |
A U-tube is partially filled with water. Oil which does not mix with water is next poured into one side, until water rises by 25 cm on the other side. If the density of oil `0.8g//cm^(3)`. The oil level will stand higher than the water byA. `6.25cm`B. `12.50cm`C. `31.75cm`D. `62.50cm` |
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Answer» Correct Answer - B On pouring oil on the left side water sides by `25 cm` from its previous level in the right limb of `U`-tube, creating a difference of levels of water by `50 cm`. Let `h cm` be the height of oil above level `A` in the left limb of `U`-tube. Equating pressures at `A` and `B` we get `P_(A)=P_(B)` `implieshxxrho_("oil")xxg=50rho_("water")xxg` `impliesh=(50xxrho_("water"))/rho_("oil")=(50xx1)/0.8=62.5cm` Difference of level of oil `C` and water `D` in the two limbs `=62.5-50=12.5cm` |
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| 66. |
A body floats with one-third of its volume outside water and `3//4` of its volume outside another liquid. The density of another liquid is :A. `9/4`B. `4/9`C. `8/9`D. `8/3` |
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Answer» Correct Answer - C Let density of body be `sigma` `(sigma)/(rho_w) = 1 -1/3 = 2/3` ..(1) `(sigma)/(rho_l) = 3/4` …(2) From (1) and (2) , `(rho_l)/(rho_w) = 8/9` `implies rho_(l) = 8/9 rho_(w) = 8/9 xx 1 = 8/9 g cm^(-3)`. |
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| 67. |
A tube with both ends open floats vertically in water. Oil with a density 800 kg/`m^3` is poured into the tube. The tube is filled with oil upto the top end while in equilibrium. The portion out of the water is of length 10 cm. The length of oil in the tube is `10 alpha cm`. Find `alpha` (assume effect of surface tension is neglible): A. 50 cmB. 60 cmC. 90 cmD. 100 cm |
| Answer» Correct Answer - A | |
| 68. |
A cylidrical vesel containing a liquid is closed by a smooth piston of mass m as shown in the figure. The area of cross section of the piston is A. If the atmospheric pressure is `P_0`, find the pressure of the liquid just below the piston. ,A. `P_0`B. `P_0 +(mg)/(A)`C. `(mg)/(A)`D. Data is not sufficient. |
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Answer» Correct Answer - B (b) since the piston is in aquarium,, `:. PA = P_0 A + mg or P = P_0 + (mg)/(A)` |
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| 69. |
A hollow spherical body of inner and outer radii 6 cm, and 8 cm respectively floats half submerged in water. Find the density of the material of the sphere. |
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Answer» Correct Answer - C `r_2=8cm, r_1=6cm` `Here m_2=upward thrust `rarr V_(pg)=(V/2)(rho_2)xxg` `[where rho_w=density of water]` `rarr(4/3pir_3^2-4/3pir_1^3)rho=(1/2)rP_2^3xx1` `rarr (r_2^3-r_1^3)xxrho=(1/2)r_2^3xx1` `rarr (8)^3-(6)^3xxrho=(1/2)(8)^3xx1` `rarrrho=512/(2x(512-216))` `=512/(2xx296)` `=0.865gm/cc=865kg/m^3` |
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| 70. |
The volume of the hollow portion of a sphere is `3/4` of the external volume of the sphere. If it floats in a liquid of relatived density `3/2`, half of its external volume immersed, the relative density of the material of the solid is :A. 2B. 3C. `2.4`D. `1.8` |
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Answer» Correct Answer - B Let `R` be external radius of sphere and `r` be radius of `c`. Given `4/3 pi r^(3) = 3/4(4/3 pi R^(3)) , r^(3) = 3/4 R^(3)` Further, `4/3 pi (R^(3)-r^(3))rho g = 1/2 xx 4/3 pi R^(3) xx 3/2 rho_(w)xxg` Using (i) `(rho)/(rho_w) = 3`. |
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| 71. |
In which one of the following cases will the liquid flow in a pipe be most streamlinedA. Liquid of high viscosity and high density flowing through a pipe of small radiusB. Liquid of high viscosity and low density flowing through a pipe of small radiusC. Liquid of low viscosity and low density flowing through a pipe of large radiusD. Liquid of low viscosity and high density flowing through a pipe of large radius |
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Answer» Correct Answer - B |
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| 72. |
Assertion : A solid is floating in a liquid of density `rho_(1)`. When the solid melts its density becomes `rho_(2)` in liquid state. If `rho_(1) gt rho_(2)` level of liquids will increase after melting. Reason : In liquid state volume always increases after a solid melts.A. If both Assertion and Reason are correct 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 is false but Reason is true. |
| Answer» Correct Answer - C | |
| 73. |
The excess pressure inside one soap bubble is three times that inside a second bubble. The ratio of the volume of first bubble to that of the secondA. `1 : 27`B. `27 : 1`C. `1 : 9`D. `9 : 1` |
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Answer» Correct Answer - A We know that for soap bubble, `p=(4T)/(r )` `"Given" p_(1)=3p_(1)` `"Here" (4T)/(r_(1))=3((4T)/(r_(2))) Rightarrow r_(2)=3r_(1)....(i)` `V_(1)=(4)/(3)=3pir_(1)^(3) and V_(2)=(4)/(3)pir_(2)^(3)=(4)/(3)pi(3r_(2)^(2))=(4)/(3)pi(27r_(1)^(3))` `Rightarrow (V_(1))/(V_(2))=((4//3)pir_(1)^(3))/((4//3)pi27r^(3))=(1)/(27)` |
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| 74. |
If a block of iron (density `5g cm^(-3))` is size 5 cm xx 5 cm xx 5 cm was weight while completely submerged in water, what would be the apperent weight ?A. 5xx5xx5xx5 gfB. 4xx4xx4xx5 gfC. 3xx5xx5xx5 gfD. 4xx5xx5xx5 gf |
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Answer» Correct Answer - D (d) Upthrust = 5xx5xx 5xxx1 gf Weight = 5xx5xx5xx5 gf Apparent weight = 5xx5xx5(5 -1) gf = 5xx5xx5xx4 gf |
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| 75. |
If a block of iron (density `5 g cm^(-3))` is size 5 cm x 5 cm x 5 cm was weight while completely submerged in water, what would be the apparent weight ?A. `5 xx 5 xx 5 xx 5 gf`B. `4 xx 4 xx 4 xx 4 gf`C. `5 xx 4 xx 4 xx 4 gf`D. `4 xx 5 xx 5 xx 5 gf` |
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Answer» Correct Answer - D |
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| 76. |
A body `B` is capable of remaining stationary inside a liquid at the position shown in Fig. (a). If the whole system is gently placed on smooth inclined plane and is allowed to slide down, then `(0 lt 0 lt 90^@`) A. The body will move up (relative to liquid)B. The body will move down (relative to liquid)C. The body will remain stationary (relative to liquid).D. The body will move up for some inclination `theta` and d move down for another inclination `theta` |
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Answer» Correct Answer - D Due to acceleration down the plane, both body `B` and liquid will received the same component of acceleration down the plane. |
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| 77. |
A stream-lined body falls through air from a height `h` on the surface of a liquid . Let `d` and `D` denote the densities of the materials of the body and the liquid respectively, if `D gt d`, then the time after which the body will be intantaneously at rest, is:A. `sqrt((2h)/(g))`B. `sqrt((2h)/(g).(D)/(d))`C. `sqrt((2h)/(g).(d)/(D))`D. `sqrt((2h)/(g)) ((d)/(D-d))` |
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Answer» Correct Answer - D |
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| 78. |
A stream-lined body falls through air from a height `h` on the surface of a liquid . Let `d` and `D` denote the densities of the materials of the body and the liquid respectively, if `D gt d`, then the time after which the body will be intantaneously at rest, is:A. `sqrt((2h)/(g))`B. `sqrt((2h)/(g)(D)/(d))`C. `sqrt((2h)/(g)(d)/(D))`D. `sqrt((2h)/(g)) ((d)/(D-d))` |
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Answer» Correct Answer - D Velocity `u` of the body when it enters the liquid is given by `mgh=1/2mu^(2) or u = sqrt(2gh)` Let Volume of the body = `V` Mass of the body = `Vd` weight of the body = `V dg` Mass of liquid displaceed = `VD` weight of liquid displaced = `VDg` Net upward force = `VDg-VDg` `=V_(g)(D-d)` Retardation = `(n et weight)/(mass)` `=(V(D-d)g)/(Vd) = ((D-d)/(d))g` Acceleration `a = -((D-d)/(d))g` Final velocity , `v` in the liquid when the body is instantaneously at rest is zero. let the time taken be `t`. `v = u+at` `0=sqrt(2gh)-((D-d)/(d))g t * ((D-d)/(d))g t =sqrt(2gh)` `:. t= [(d)/(D-d)] sqrt((2h)/(g))`. |
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| 79. |
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 assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. if assertion is true but 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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| 80. |
In turbulent flow the velocity of the liquid molecules in contact with the walls of the tube.A. ZeroB. MaximumC. Equal to critical velocityD. May have any value |
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Answer» Correct Answer - B |
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| 81. |
While measuring surface temsoion of water using capillary rise method, heihgt of the lowr meniscus from free surface of water is 3 cm while inner radius of capillary tube is found to be `0.5` cm. Then compute tension of water using this data. (Take contact angle between glass and water and a 0 and `g=9.81 m//s^(2)`A. `0.72 N//m`B. `0.77 N//m`C. `1.67 N//m`D. None of the above |
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Answer» (b) As , T `(r(h+(r)/(3))rhog)/(2 costheta)` `=(0.5xx10^(-2)[3+(0.5)/(3)]xx10^(-2)xx10^(3)xx9.81)/(2)=0.77N//m` |
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| 82. |
A capillary tube of radius r is placed in a liquid I the angle of contact is `theta` , the radius of curvature R of the meniscus in the capillary isA. B. C. D. |
| Answer» Correct Answer - B | |
| 83. |
A U-tube containing a liquid is accelerated horizontally with a constant acceleration `a_0`. If the separation between the vertical limbs is l find the difference in the heights of the liquid in the two arms. |
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Answer» Correct Answer - A Since the U tube is accelerated horizotaly inertia force will be experience by the horizontal part. `If P_ararr atm pressure Ararr area of cross section `hrarr ` increase in height `P_aA+AxxLxxPxxa_0` `=P_a A+hrhogxxA` `rarr hg=a_0L` `rarr h=(a_0L)/g` |
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| 84. |
A barometer kept in an elevator reads 76 cm when it is at rest. If the elevator goes up with increasing speed, the reading will beA. zeroB. 76cmC. lt76cmD. gt76cm |
| Answer» Correct Answer - C | |
| 85. |
Fig, shows a U-tube of uniform cross-sectional area `A` accelerated with acceleration a as shown. If `d` is the seperation 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 Mass of liquid in horizontal portion of U-tube =`A d rho` Pseudo force on this case = `Adrhoa` force due to pressrue difference in the two limbs `=(h_(1)rho g - h_(2)rho g)A` Equating, `(h_(1)-h_(2))rho g A = A d rho a` Equanting , `(h_(1)-h_(2))rhogA = A d rho a` or `h_(1) - h_(2) = (Adrhoa)/(rho g A ) = (ad)/(g)`. |
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| 86. |
A barometer kept in a stationary elevator reads `76 cm`. If the element starts accelerating up the reading will beA. ZeroB. Equal to `76 cm`C. More than `76 cm`D. Less than `76 cm` |
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Answer» Correct Answer - D |
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| 87. |
A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?A. B. C. D. |
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Answer» Correct Answer - C Soap solution has lower surface tension, T as compared to pure water and capillary rise `h=(2Tcostheta)/(rho rg)`, so h is less for saop solution. |
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| 88. |
Assertion : Small water drops are sperical while bigger water drops are not. Reason : In small water drops surface tension forces dominate while in bigger water drops gravity forces dominate.A. If both Assertion and Reason are correct 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 is false but Reason is true. |
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Answer» Correct Answer - B In small drops force of surface tension is much larger. So, they colasces to a smaller radii spheres. |
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| 89. |
The pressure inside a small air bubble of radius `0.1mm` situated just below the surface of water will be equal to `("Take surface tension of water" 70xx10^(-3)1^(J)m^(-1) "and atmospheric pressure" =1.013xx10^(5)Nm^(-2))`A. `2.054xx10^(3)Pa`B. `1.027xx10^(3)Pa`C. `1.027xx10^(5)Pa`D. `2.054xx10^(5)Pa` |
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Answer» Correct Answer - C Pressure inside a bubble when it is in a liquid `P_(o)+(2T)/(R ) = 1.013xx10^(5)+2xx(70xx10^(-3))/(0.1xx10^(-3)) = 1.027xx10^(5) Pa`. |
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| 90. |
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)` |
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Answer» Surface tension of water `T=7.2xx10^(-2) Nm^(-1)` Radius of air bubble `R=0.1 mm = 10^(-4) m` The excess pressure inside the air bubble is given by, `p_(2)-p_(1)=(2T)/(R )` `therefore` Pressure inside the air bubble, `p_(2)=p_(1)+(2T)/(R )` Substituting the values, we have `p_(2)=(1.013xx10^(5))+((2xx7.2xx10^(-2)))/(10^(-4))=1.027xx10^(5)Nm^(-2)` |
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| 91. |
Suppose the tube in the previous problem is kep vertical with A upward but the other conditions remain the same. The separation between the cross sections at A and B is 15/16 cm. Repeat parts a., b, and c of the previous problem. `take g=10ms^-2`. |
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Answer» Correct Answer - A::B::C::D a. `v_A =25cm/sec` `b. v_B=50cm/sec` c. From Bernoullis equation `1/2pv_A^2+rhogh_A+pA=1/2pv_B^2=rhogh_B+pB` `rarr P_A-P_B=(1/2)p(v_B^2-v_A^2)+rhog(h_B-h_A)` `[Given h_A-h_B=(15/16)cm]` `rarr P_A-P_B=1/2(2500-625)-1000x15/16` |
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| 92. |
Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. The separation between the cross sections at A and B is 15/16 cm. Repeat parts a., b, and c of the previous problem. `take g=10ms^-2`. |
| Answer» Correct Answer - `(i) 25 cm//s, (ii) 50 cm//s, (iii)` zero | |
| 93. |
Suppose the tube in the previous problem is kept verticla with B upward. Water enters through B at the rate of `1cm^3s^-1`. Repeat parts a, b, and c. Note that the speed decreases as the water falls down. |
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Answer» Correct Answer - `187.5 N//m^(2)` `A_(1)v_(1) = A_(2)v_(2)` `P_(y) + (1)/(2)rhov_(y)^(2) + rhogh = P_(x) + (1)/(2) rhov_(x)^(2) + 0`. `P_(x) - P_(y) = (1)/(2)rho(v_(y)^(2) - v_(x)^(2)) + rhogh` `P_(x) - P_(y) = (1)/(2) 1000[((1)/(2))^(2) - ((1)/(4))^(2)] + 1000 xx 10 xx (15)/(1600)` `P_(x) - P_(y) = 187.5 N//m^(2)` |
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| 94. |
An ornament weighing `36 g` in air weighs only `34 g` in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. Specific gravity of gold is `19.3` and that of copper is `8.9`A. `2.2g`B. `4.4g`C. `1.1g`D. `3.6g` |
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Answer» Correct Answer - A Given that `m_("real")=36g, m_(app)=34g`, Density of gold `rho_(Au)=19.3g//"cc"` Density of copper `rho_(Cu)=8.9g//"cc"` We know that loss of weight `=` weight of displaced water `=36-34=2g=` Buoyant force `=B` Here, `m_("real")=m_(Aw)+m_(Cu)=36g` ...........i Let `v` be the volume of the ornament in centimetres. Then `B=vxxrho_(w)xxg=2xxg` `implies((m_(Au))/(rho_(Au))+(mu_(Cu))/(rho_(Cu)))rho_(w)xxg=2xxg` `m_("Au")rho_("Cu")+m_("Cu")rho_("Au")=2rho_("Acc")rho_("Cu")` `8.9m_(Au)+19.3m_(Cu)=2xx19.3xx8.9=343.54`..........ii `implies8.9(m_(Au)+m_(Cu))+10.4m_(Cu)=343.54` `implies 8.9xx36+10.4m_(Cu)=343.54` `m_(Cu)=2.225g` So the amount of copper in the ornament is `2.2g`. |
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| 95. |
A matallic block weighs `15N` in air. It weights `12N` when immersed in water and `13 N` when immersed in another liquid. What is the specific gravity of the liquid?A. `1//3`B. `2//3`C. `12//13`D. `13//15` |
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Answer» Correct Answer - B `(sigma)/(rho_w) = (15)/(15-12) , (sigma)/(rho_1) = (15)/(15-13)` `implies (rho_l)/(rho_w) = 2/3`. |
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| 96. |
A liquid drop of radius R breaks into N smaller droplets of radii r, If liquid has density `rho,` specific heat s and surface tension, T, than the drop in temeperature is givenA. `(NT)/(rhos)((1)/(R)-(1)/(r))`B. `(3NT)/(rhos)((R)/(r)-1)`C. `(3)/(4)(NT)/(rhos)((1)/(R)-(1)/(r))`D. `(3NT)/(rhos)((1)/(R)-(1)/(r))` |
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Answer» Volume will remain same, hence we can write `(4)/(3)piR^(3)=Nxx(4)/(3)pir^(3)rArrN(R^(3))/(r^(3))` Increase in surface energy due to breaking of drop inot N driplets. `" " `DeltaUtheta` `ms Deltatheta=DeltaU` `Deltatheta=(DeltaU)/(ms)=(4piT(R^(2)-Nr^(2)))/(((4)/(3)piR^(3)rho)s)` `rArrDeltatheta=(3T)/(rhos)((1)/(R)-(r^(2))/(R^(3))N)=(3T)/(rhos)((1)/(R)-(1)/r)" " ( :. N=(R^(3))/(r^(3)))` |
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| 97. |
When equal volumes of two substance are mixed, the specific gravity of the mixurie is 4. When equal weights of the same substance are mixed, the specific gravity of the mixture is 3. The soecufuc gravities of the two substance could beA. 6 and 2B. 3and 4C. `2.5` and `3.5`D. 5 and 3 |
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Answer» Correct Answer - A `4=(Vd_(1)+Vd_(2))/(2V)` or, `d_(1)+d_(2)=8`…(i) `3=(m+m)/((m//d_(1))+(m//d_(2)))` or` (2 d_(1)d_(2))/(d_(1)+d_(2))=3` …(ii) Solving eqs. (i) and (ii), we get `d_(1)=6` and `d_(2)=2`. |
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| 98. |
In making an alloy, a substance of specific gravity `s_(1)` and mass `m_(1)` is mixed with another substance of specific gravity of the alloy isA. `((m_1 + m_2)/(s_1 + s_2))`B. `((s_1 s_2)/(m_1 + m_2))`C. `(m_1 + m_2)/((m_1)/(s_1) + (m_2)/(s_2))`D. `((m_1)/(s_1) + (m_2)/(s_2))/(m_1 + m_2)` |
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Answer» Correct Answer - C `S = (m_1+ m_2)/(V_1 + V_2) = (m_1 + m_2)/((m_1)/(s_1) + (m_2)/(s_2))`. |
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| 99. |
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` of the water `=` Pressure energy of the water at that layer `1/2mV^(2)=mxxP/(rho)` `V^(2)=(2P)/rho=2/(rho)[P_(0)+rhog(h_(1)-h_(2))]` `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)` metre below the top or `X` is `-h_(1)` above the top. |
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| 100. |
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 of the water in the long tube above the top when the water stops coming out of the hole.A. `-2h_(0)`B. `h_(0)`C. `h_(2)`D. `-h_(1)` |
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Answer» Correct Answer - D `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` of the water `=` Pressure energy of the water at that layer `1/2mV^(2)=mxxP/(rho)` `V^(2)=(2P)/rho=2/(rho)[P_(0)+rhog(h_(1)-h_(2))]` `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)` metre below the top or `X` is `-h_(1)` above the top. |
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