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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.
| 201. |
A rectangular vessel when full of water takes 10 minutes to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with waterA. `9` minuteB. `7` minuteC. `5` minuteD. `3` minute |
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Answer» Correct Answer - D |
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| 202. |
A tube of uniform cross section is used to siphon water from a vessel `V` as shown in the figure. The pressure over the open end of water in the vessel is atmospheric pressure `(P_(0))`. The height of the tube above and below the water level in the vessel are `h_(1)` and `h_(2)`, respectively. Given `h_(1) = h_(2) = 3.0 m`, the gauge pressure of water in the highest level `CD` of the tube will beA. `3.0xx10^(4)N//m^(2)`B. `5.9xx10^(4)N//m^(2)`C. `5.9xx10^(4)N//m^(2)`D. `1.5xx10^(4)N//m^(2)` |
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Answer» Correct Answer - B Putting `h_(1)=h_(2)=3.0m` in eqn iii above `P_(1)=` pressure at level `CD` `=P_(atm)-(rhog)(3+3)=1.0xx10^(5)-6rhog` The gauge pressurre at level `CD=6rhog=6xx10^(3)xx9.8N//m^(2)` `=5.9xx10^(4)N//m^(2)` |
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| 203. |
The thickness of the ice layer on the surface of lake is 20 m. A hole is made in the ice layer. What is the minimum length of the rope required to take a bucket full of water out ? (Take density of ice `= 0.9 xx 10^3 kg//m^3)`A. 2 mB. 5 mC. 9 mD. 18 m |
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Answer» Correct Answer - A (a) Given that density of ice ` = (9)/(10)xx` density of water, so only `(1)/(10)` the of ice will be outside the water. This means that `20 xx (1)/(10) = 2m`of ice layer will be above the water surface. In other words, in the hole, the water level will rise up to 18 m of the ice layer. So minimum length of rope required = 2m. |
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| 204. |
A ball of weight `W` is supported on a vertical jet of water. If the stream of water flowing from the nozzle has a diameter `D` and velocity `u`, determine the value of `H`. Assume that no loss of energy takes place. |
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Answer» Correct Answer - `1/(2g)` Applying continuity equation at nozzle end at a top position `a_(1)u=a_(2)v` `v=(a_(1)u)/(a_(2))`…..i `v^(2)-u^(2)=-2gH` `implies H=1/(2g) {u^(2)-[W/(rhoa_(1)^(2)u)]^(2)}`…….ii `implies rhoa_(2)v^(2)=Wimplies v=W/(rhoa_(1)u)` `implies (rhoa_(2)u)/W=1/vimpliesv^(2)=W/(rhoa_(1)u)` `impliesH=1/(2g) {u^(2)-[W/(rhoa_(1)^(2)u)]^(2)}` `=1/(2g){u^(2)-[(4D)/(rhopiD^(2)u)]^(2)}` |
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| 205. |
Three liquids of equal masses are taken in three identical cubical vessels A, B and C. Their densities are `rho_(A), rho_(B) " and " rho_(C )` respectively but `rho_(A) lt rho_(B) lt rho_(C )`. The force exerted by the liquid on the base of the cubical vessel isA. maximum in vessel CB. minimum in vessel CC. the same in all the vesselsD. maximum in vessel A |
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Answer» Correct Answer - C Force exetred by the liquid on the base of the vessel is F = mg `" Here, "m_(A)=m_(B)= F_(C)` |
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| 206. |
A closed rectangular tank is completely filled with water and is accelerated horizontally with an acceleration towards the right. Pressure is i. maximum and ii. minimum at A. (i) B (ii) DB. (i) C (ii) DC. (i) B (ii) CD. (i) B (ii) A |
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Answer» Correct Answer - A Due to acceleration towards right, there will be a psedo force in a left direction. So the pressure will be more on rear side (Point `A` and `B` ) in comparison with front side pressure will be more at the bottom (point `B` and `C`) in compariosn with top (point `A` and `D`). so overall maximum pressure will be at point `B` and minimum pressure will be at point `D`. |
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| 207. |
A smooth gate is kept in equilibrium by applying a horizontal force. What is the value of y so that no horizontal reaction force acts at the pivot ? A. `(h)/(3)`B. `(h)/(6)`C. `(2h)/(3)`D. zero |
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Answer» Correct Answer - A (a) The centre of liquid force passes through `y = (h)/(3).` |
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| 208. |
Figure shows an open tube which contains some water and a less dense liquid poured ino the right hand side. If the density of the unknown liquid is `rho_(x)`, show that `rho_(x)=(rho_(w)h_(w))/(h_(x))` |
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Answer» Consider levels `D` and `A` in the tube. If the pressures at `D` and `A` are not equal, then water would flow. Since water does no flow, the pressures at `D` and `A` are equal. `P_(A)=P_(atm)=rho_(w)gh_(w)` and `P_(D)=P_(atm)+rho_(x)gh_(x)` Equating `P_(A)` and `P_(D)` we obtain `rho_(x)=(rho_(w)h_(w))/(h_(x))` |
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| 209. |
A vertical `U` tube of uniform cross section contains mercury in both of its arms. A glycerine `(d=1.3g//cm^(3))` column of length `10 cm` is introduced into one of the arms. Oil of density `0.8g//cm^(3)` is poured in the other arm until the upper surfaces of the oil and glycerine are in the same horizontal level. Find the length of oil column. Densit of mercury is `13.6g//cm^(3)`. |
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Answer» Let the length of oil column `=CD=xcm` Length of glycerine `=AB=10 cm` `:. DE=(10-x)cm` Equating pressure at horizontal level `BE` in both arms, we have `P_("oil")+P_(Hg)=P_("glycerine")` `xd_("oil")g+(10-x)rho_(Hg)g=10d_("glycerine")g` or `x(0.8)=(10-x)13.6=10(1.3)impliesx=(136-13)/12.8=9.6cm` |
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| 210. |
The cross sectional area of a horizontal tube increases along its length linearly, as we move in the direction of flow. The variation of pressure, as we move along its length in the direction of flow (x-direction), is best depicted by which of the following graphs.A. B. C. D. |
| Answer» Correct Answer - A | |
| 211. |
The ratio of the terminal velocities of two drops of radii R and `R//2` isA. `2`B. `1`C. `1//2`D. `4` |
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Answer» Correct Answer - D Terminal velocity, `v_(T) prop r^(2)` ` therefore (v_(T))_(1)/(v_(T))_(2)=((R )/(R//2))^(2)=4` |
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| 212. |
When a jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. `F=(rhoAV_(0))V_(0)-(rhoAV_(0))V_(0)costheta=rhoAV_(0)^(2)[1-costheta]` If surface is free and starts moving due to thrust of the liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let, at any instant, the velocity of surface be `u`. Then the above equation becomes ` F=rhoA(V_(0)-u)^(2)[1-costheta]` Based on the above concept, as shown in the following figure, if the cart is frictionless and free to move in the horizontal direction, then answer the following. Given that cross sectional area of jet `=2xx10^(-4)m^(2)` velocity of jet `V_(0)=10m//s` density of liquid `=1000kg//m^(3)` ,mass of cart `M=10 kg`. Initially (`l=0`) the force on the cart is equal toA. `20N`B. `40N`C. `80N`D. zero |
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Answer» Correct Answer - B `F-rho_(A)(V_(0)-0)^(2)[1-cos180^@]` `=2rhoAv^(2)=2xx1000xx10^(-4)xx10xx10=40N` |
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| 213. |
What is principle of floatation ? |
| Answer» Whenever a body floats (partly submerged ) in a liquid, the weight of the body is equal to the weight of the liquid displaced by the submerged part of the body. | |
| 214. |
Two `U-`tube manometers are connected to a same tube as shown in figure. Datermine difference of pressure between `X` and `Y`. Take specific gravity of mercury as `13.6. (g = 10 m//s^(2), rho_(Hg) = 13600 kg//m^(3))` |
| Answer» Correct Answer - If `g = 10m//s^(2), 253200 N//m^(2)` | |
| 215. |
Statement I: When spinning ball is thrown it deviates from its usual path in flight. Statement II: Time of flight will remain same if axis of rotation is vertical.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. |
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Answer» Correct Answer - A Due to pressure difference across two sides of the ball, it deviates from its usual path. Force on vertical direction will remain the same. |
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| 216. |
Assertion: A spinning cricket ball deviates from is trajectory as it moves through air. Reason: The ball is moving forward and relative to it the air is moving backward.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 - A A spinning ball deviates from its parabolic trajectory as it moves through air. Suppose ball is moving in forward direction and air is moving in backward relative to ball. Due to this there is dynamic lift on the ball and ball deviates from its parabolic trajectory. |
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| 217. |
Assertion : Hydrostatic pressure is a vector quantity. Reason : Pressure is force divided by area, and force is a vector quantity.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. If assertion ture but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D |
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| 218. |
Water is flowing in a pipe of diameter 4 cm with a velocity `3 m//s` . The water then enters into a tube of diameter 2 cm . The velocity of water in the other pipe isA. `3 m//s`B. `6 m//s`C. `12 m//s`D. `8 m//s` |
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Answer» Correct Answer - C |
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| 219. |
A sphere of radius R and made of material of relative density `sigma` has a concentric cavit of radius r. It just floats when placed in a tank full of water. The value of the ratio `R//r` will beA. `((sigma)/(sigma-1))^(1//3)`B. `((sigma-1)/(sigma))^(1//3)`C. `((sigma+1)/(sigma))^(1//3)`D. `((sigma-1)/(sigma+1))^(1//3)` |
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Answer» Correct Answer - A `B=W" " implies(4)/(3)piR^(3) xx 1 xx g =(4)/(3)pi(R^(3)-r^(3))sigma g` `R^(3)=sigmaR^(3)-sigmar^(3)" "impliessigmar^(3)=(sigma-1)R^(3)" "implies(R )/(r )=((sigma)/(sigma-1))^(1//3)` |
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| 220. |
Assume the density of brass weights to be `8 g cm^(-3)` and that of air to be `0.0012g cm^(-3)`. What fractional error arises from neglecting buoyancy of air in weighing and objecte of density `3.4 g cm^(-1)` on a beam balance?A. `2xx10^(-1)`B. `2xx10^(-2)`C. `2xx10^(-3)`D. `2xx10^(-4)` |
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Answer» Correct Answer - D Let `m gwt` be the weight of object in vacuum. Volume of object = `(m)/(3.4)` weight of air displaced by object = `m/3.4 xx 0.0012` Volume of brass weight =` m/8` weight of air displaced by brass weights `=m/8 xx 0.0012` Error=differecne in buouancy `=0.0012[1/3.4-1/8]` Fractional error = `(0.0012 xx 4.6)/(3.4xx8) = 2 xx 10^(-4)`. |
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| 221. |
The centre of buoyancy of a floating object is :A. At the centre of gravity of the objectB. At the centre of gravity of the submerged part of the objectC. At the centre of gravity of the remaining part outside the fluid of the objectD. At the centre of gravity of the fluid displaced by the submerged part of the object. |
| Answer» Correct Answer - B | |
| 222. |
The correct curve between the height or depression `h` of liquid in a capillary tube and its radius isA. B. C. D. |
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Answer» Correct Answer - B `h=(2T cos theta)/(r d g) therefore h prop (1)/(r ). " If " theta` is less than h will be more. |
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| 223. |
There is a hole of area `(1)/(25)cm^(2)` in the bottom of a cylindrical vessel containing fluid up to height `h`. The liquid flows out in time `t`. If the liquid were filled in the vessel up to height `4h`, then it would flow out in timeA. tB. 2tC. 4tD. `(t)/(2)` |
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Answer» Correct Answer - B Let `A` and `a` be the corss-sectional areas of the vesser and hole respectively. Let `h` be the height of water in the vessel at time. Let `(-(dh)/(dt))` represent the rate of fall of level. then `A(-(dh)/(dt)) = alpha v = asqrt(2gh)` or `-(dh)/(dt) = (alphasqrt(2g))/(A) dt` `-int_(A)^(0) (1)/(sqrt(h))dh = (asqrt(2g))/(A) int_(0)^(g) dt` `-(-2sqrt(h)) = (alphasqrt(2g))/(A)t` or `t=(A)/(alpha) (1)/(sqrt(2g)) xx 2sqrt(h) or t = A/(alpha) sqrt((2h)/(g))` Now `t prop sqrt(h)` where `h` is quarupled, `t` is doubled. |
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| 224. |
A water tank placed on the floor has two small holes, pinched in the vertical wall, one above the other. The holes are `3.3 cm` and `4.7cm` above the floor. If the jets of water issuing out from the holes hit the floor at the same point on the floor, then the height of water in the tank isA. 3cmB. 6cmC. 8cmD. 9cm |
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Answer» Correct Answer - C `R = 2sqrt(h(H-h))` Now, `3.3[H-3.30 = 4.7[H-4.7]` or (4.7-3.3)H = 4.7xx4.7 - 3.3xx3.3` `or H = (22.09-10.89(/(1.4) cm` `=(11.2)(1.4)cm = 8cm`. |
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| 225. |
A boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water ?A. It will remain unchangedB. It will riseC. It will fallD. First it will first rise and then fall |
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Answer» Correct Answer - C |
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| 226. |
A cylinder containing water up to a height of `25cm` has a hole of cross-section `1/4 cm^(2)` in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begin to flow out? A. Increase of `12.5 gm - wt`B. Increase of `6.25 gm-wt`C. Decrease of `12.5 gm-wt`D. Decrease of `6.25 gm-wt` |
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Answer» Correct Answer - C |
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| 227. |
A cylinder containing water up to a height of `25cm` has a hole of cross-section `1/4 cm^(2)` in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begin to flow out? A. Increase of `12.5 gm-wt`B. Increase of `6.25 gm-wt`C. Decrease of `12.5 gm-wt`D. Decrease of `6.25 gm-wt` |
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Answer» Correct Answer - C Let A = The area of cross section of the hole, `upsilon` = Initial velocity of efflux, d = Density of water, Initial mass of water flowing out per second = `Aupsilon` Initial mass of water flowing out per second = `Aupsilond` Rate of change of momentum `= Adupsilon^(2)" " therefore` Initial downward force on the out flowing water `=Adupsilon^(2)` So equal amount of reaction acts upwards on the cylinder. `therefore` Initial upward reaction `=Adv^(2) " " [As v =sqrt(2gh)]` `therefore` Initial decrease in weight `=Ad (2gh) = 2Adgh = 2xx((1)/(4))xx1xx980xx25 = 12.5 gm-wt`. |
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| 228. |
A spherical drop of water as `1mm` radius. If the surface tension of the the water is `50xx10^-3(N)/(m)`, then the difference of pressure between inside and outside the spherical drop is:A. `25 N//m^(2)`B. `10000 N//m^(2)`C. `100 N//m^(2)`D. `50 N//m^(2)` |
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Answer» Correct Answer - C `P_("excess")=(2T)/R=(2(50xx10^(-3)))/((10^(-3)))=100 N//m^(2)` |
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| 229. |
A cylinder containing water up to a height of `25cm` has a hole of cross-section `1/4 cm^(2)` in its bottom. It is counterpoised in a balance. What is the intial change in the balancing weight when water begin to flow up? A. increases of `12.5 gwt`B. increase of `6.25 gwt`C. Decrease of `12.5 gwt`D. Decrease of `6.25 gwt` |
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Answer» Correct Answer - C Let `A = the area of cross section of the hole `v` = Initial velocity of efflux `d`=density of water , Initial volume of water flowing out per second =`Av` Initial mass of water flowing out per second = `Avd` Rate of change of momentum = `Adv^(2)` Initial downward force on the flowing out water `=Adv^(2)` So equal amount of reaction actys upwards on the cylinder. `:. I nitial upward reaction = `Adv^(2) [As v = sqrt(2gh)]` `:. i nital decrease in weight = Ad(2gh)` `=2Adgh = 2 xx (1/4)xx1xx980xx25=12.5 gm-wt`. |
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| 230. |
What is the pressure drop (in mm Hg) in the blood as it passes through a capillary `1mm` long and `2mu m` in radius if the speed of the blood through the centre of the capillary is `0.66mm/s`? (The viscosity of whole blood is `4xx10^(-3) PI)` |
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Answer» Correct Answer - A `v_(max)=((p_(1)-p_(2))R^(2))/(4 eta L)` `:. (p_(1)-p_(2))=(4 eta Lv_(max))/(R^(2))` `=((4xx4xx10^(-3))(10^(-3))(0.66xx10^(-3)))/(2xx10^(-6))^(2)` `h=(2.64xx10^(3))/(rhog)` `=(2.64xx10^(3))/(13.6xx10^(3)xx9.81)m of Hg` `=0.0195m of Hg` `~~19.5 mm of Hg`. |
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| 231. |
The relative velocity of two parallel layers of water is 8 cm/sec. If the perpendicular distance between the layers is 0.1 cm, then velocity gradient will beA. `8sec`B. `80 sec`C. `0.8 sec`D. `0.08 cm` |
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Answer» Correct Answer - B |
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| 232. |
A closed container shown in figure is filled with water `(rho=10^(3) kg//M^(3))` This is accelerated in horizonal direction with an acceleration, `a=2m//s^(2)`.Find (a)`p_(C)-p_(D)` and (b) `p_(A)-p_(D)`. |
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Answer» Correct Answer - A::B::D (a) In horizontal direction, pressure decreases in the direction of acceleration. Thus, `p_(C) gt p_(D)` or, `p_(C)-p_(D)=+ rhoax` Substituting the values, we have `p_(C)-p_(D)=(10^(3))(2)(2)` or,`p_(C)-p_(D)=4.0xx10^(3) N//m^(2)` (b) In vertical direction, pressure increses with depth. `:. p_(C) gtp_(A)` or, `p_(A)-p_(C)=-rho gh` `=-(10^(3))(10)(6) (g_(e) =g)` `=-60xx10^(3) N//m^(2)` (b) Now, `p_(A)-p_(D)=(p_(A)-p_(C))+(p_(C)-p_(D))` `=(-60xx10^(3))+(4.0xx10^(3))` `=-56xx10^(3) N//m^(2)` `=-56xx10^(4) N//m^(2)`. |
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| 233. |
A U-tube having a liquid of density `rho` is accelerated at a `m//s^(2)` , so as to create be the height difference between two columns of l/2 (as shown if figure) . If/is the length of the base of U-tube the value of acceleration given to the system is A. `4.9 m//s^(2)`B. `9.8 m//s^(2)`C. `5.6 m//s^(2)`D. `6.4 m//s^(2)` |
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Answer» (a) Pressure difference between two column, `Deltap=pg(h_(1)-h_(2))=pg.(1)/(2)` `:.` Force on the liquid contained in the horizontal portional of the tube. `=(Deltap)xx"area"`. Now this force must be equal to the product of mass of liquid (in bas tube) and acceleration a lof the system `rArr " " pg.(1)/(2)xxarea=(volume xxdensity)xxa` `rArr " " pg.(1)/(2)xxarea=arealxxpxxa` `rArr" " a=g//2=4.9 m//s^(2)` |
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| 234. |
One end a U-tube of unifrom bore (area A) cpmtaining mercury is connected to sunction pump . Because of it the level of liquid of density `rho` falls in one limb. When the pump is removed, the restoring force in the other limb is A. `2xrhoAg`B. `xrhog`C. `Arhog`D. `xrhoAg` |
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Answer» (a) `:.` The force due to excess pressure = restoring force Or `rhoghA`=restoring force or `rhog(2x)A` = restoring force `:.` Restoring force `=2rhogxxA` |
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| 235. |
A spherical ball of density `rho` and radius` 0.003m` is dropped into a tube containing a viscous fluid up to the 0 cm mark as shown in the figure. Viscosity of the fluid `=1.26N-s//m^(2)` and its density `rho_(L)=(rho)/(2)=1260kg//m^(3)`. Assume that the ball reaches a terminal speed at 10cm mark. The time taken by the ball to travel the distance between the 10cm and 20cm mark is `(g=10m//s^(2))` |
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Answer» Correct Answer - 5s `{:(,V_(T)=(2)/(9)xx(r^(2)(rho-rho_(L))g)/(eta),impliesV_(T)=(2)/(9)xx((3xx10^(-3))^(2)xx1260xx9.8)/(1.260)),(,V_(T)=2xx10^(-2)m//s,impliesV_(T)=2cm//s),(,t=(10cm)/(V_(T)),impliest=5sec):}` |
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| 236. |
Water is conveyed through a horizontal tube 8 cm in diameter and 4 kilometer in length at the rate of 20 litre/s Assuming only viscous resistance , calculate the pressure required to maintain the flow . Coefficient of viscosity of water is 0.001 pa s |
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Answer» Here, 2r=8 cm =0.08 m or r=0.04 m, l=4 km =4000 m, `V=20 "litre"//s=20xx10^(-3) m^(3)s^(-1), eta=0.001 Pa-s, p= ?` As, `V=(pi p r^(4))/(8eta l) " or" p=(8V eta l)/(pi r^(4))` `=(8xx(20xx10^(-3))xx0.001xx4000)/(((22)/(7))xx(0.04)^(4)) =7.954xx10^(4) Pa` `therefore` Height of mercury column for pressure difference p will be, `h=(p)/(rho g)=(7.954xx10^(4))/((13.6xx10^(3))xx9.8)=0.5968 m=59.68 cm` |
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| 237. |
A container contains liquid upto height H and kept on a horizontal frictionless surface .At t=0, the container is given a constant acceleration `a_0` along positive x-axis.the pressure at point P depends upon : A. Only on the x-co-ordinate of the point PB. Only on the y-co-ordinate of the point PC. On both x and y co-ordinate of the point PD. None |
| Answer» Correct Answer - A | |
| 238. |
The weight of an aeroplane flying in air is balanced byA. Up thrust of the air which will be equal to the weight of the air having the same volume as the planeB. Force due to the pressure difference between the upper and lower surfaces of the wings, created by different air speed on the surfaceC. Vertical component of the thrust created by air currents strinking the lower surface of the wingsD. Force due to the reaction of gases ejected by the revolving propeller |
| Answer» Correct Answer - A | |
| 239. |
At what speed the velocity head of a stream of water be equal to 40 cm of HgA. `282.8 cm//sec`B. `432.6 cm//sec`C. `632.6 cm//sec`D. `832.6 cm//sec` |
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Answer» Correct Answer - A |
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| 240. |
A silver ingot weighing 2.1 kg is held by a string so s to be completely immersed in a liquid of relative density `0.8`. The relative density of silver is `10.5`. The tension in the string in `kg-wt` isA. `1.6`B. `1.94`C. `3.1`D. `5.25` |
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Answer» Correct Answer - B Apparent weight `=V(rho-sigma)g =(M)/(rho) (rho-sigma)g = M(1-(sigma)/(rho))g = 2.1(1-(0.8)/(10.5))g = 1.94 g` Newton `= 1.94 kg-wt` |
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| 241. |
A metallic sphere weighs `210g` in air, 180 g in water and 120 g in an unknown liquid. Find the density of metal and of liquid. |
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Answer» Correct Answer - C Relative density of metal `=("weight in air")/("change in weight of water")=(210)/(210-180)=7` `:.` Density of metal`=7g//cm^(2)` Change in weight in a liquid =upthrust in liquid `=(V_("solid")) (rho_("liquid"))g` or, `Delta w prop rho_(liquid)` `:. (Delta w_(l))/(Delta w_(w))=(rho_(l))/(rho_(w))` `rho_(l)=(Deltaw_(l))/(Deltaw_(w)) rho_(w)` `((210-120)/(210-180))(1)gm//cm^(3)` `=3g//cm^(3)` |
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| 242. |
A sample of metal weights 210 grams in air 180 grams n water and 120 grams in an unknown liquid thenA. Metal of 3B. Metal is 7C. Liquid of 3D. Liquid is `(1)/(3)` |
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Answer» Correct Answer - B::C Let the density of metal is p and density of liquid is `sigma`. If V is the volume of sample then according to problem `210 = Vrhog " "…….(i)` `180 = V(rho-1)g " "……(ii)` `120 = V(rho-sigma)g " "……..(iii)` By solving (i),(ii) and (iii) we get `rho = 7` and `sigma = 3`. |
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| 243. |
Coatings used on raincoat are waterproof because theyA. increases angle of contactB. decreases angle of contactC. does not alters angle of contactD. forms a smooth surface |
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Answer» Correct Answer - A Water proof surface coatings are made up of materials which increases angle of contact and water does not wets the surface. |
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| 244. |
A cubical closed vessel of side 5m filled with a liquid is accelerated with an acceleration a. Find the value of a so that pressure at mid point` M` of `AC` is equal to pressure at `N`. . |
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Answer» Correct Answer - B `p_(N)+rho_(g)h-rho a((h)/(2))=p_(M)` But `p_(N)=p_(M)` `:. a=2g`. |
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| 245. |
Statement I: Imagine holding two identical bricks under water. Brick `A` is completely submerged just below the surface of water, while brick `B` is at a greater depth. The magnitude of force exerted by the person (on the brick) to hold brick `B` in place is the same as magnitude of force. exerted by the person (on the brick) to hold brick `A` in place. Statement II: The magnitude of buoyant force on a brick completely submerged in water is equal to the magnitude of weight of water it displaced and does not depend on the depth of the brick in water.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. |
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Answer» Correct Answer - A Since the net buoyant force on the brick completely submerged, in water is independent of its depth below the water surface, the man will have to exert the same force on both the bricks. Hence, statement I is true, Statement II is true, statement II is a correct explanation for Statement I. |
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| 246. |
We have two different liquids `A` and `B` whose relative densities are `0.75` and `1.0`, respectively. If we dip solid objects `P` and `Q` having relative densities `0.6` and `0.9` in these liquids, thenA. `P` floats in `A` and `Q` sinks in `B`B. `P` sinks in `A` and `Q` floats in `B`C. `P` floats in `B` and `Q` sinks in `A`D. `P` sinks in `B` and `Q` floats in `A` |
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Answer» Correct Answer - C The relative density of object `P` is less than the relative density of both the given liquids. So , `P` will float in both the liquids .The relative density of `Q` is higher than the relative of liquid `A`. So `Q` sinks in `A`. |
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| 247. |
An unsymmetrical sprinkler the top view of the setup has frictionless shaft and equal fluid flows through each nozzle with a velocity of 10 m/sec relative to nozzle. If the shaft is rotating at constant angular speed then its angular speed of rotation is : A. 3 rad/sB. 4 rad/sC. 2 rad/sD. 10 rad/s |
| Answer» Correct Answer - D | |
| 248. |
A body is just floating in a liquid whose density is equal to the density of body. What happens to the body if it is slightly pressed and released?A. It will slowly come back to its earlier positionB. It will remain submerged, where it is leftC. it will stockD. it will come out violently |
| Answer» Correct Answer - B | |
| 249. |
Which of the following is correct.A. Gauge pressure=absolute pressure+atmospheric pressureB. Absolute pressure=gauge pressure-atmospheric pressureC. Gauge pressure=absolute pressure-atmospheric pressureD. Absolute pressure=atmospheric pressure -gauge pressure |
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Answer» Correct Answer - C Gauge pressure = absolute pressure-atmospheric pressure |
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| 250. |
A tank is filled with water up to a height `H`. Water is allowed to come out of a hole `P` in one of the walls at a depth `D` below the surface of water. Express the horizontal distance `x` in terms of `H` and `D` .A. `x = sqrt(D(H - D))`B. `x = sqrt((D(H - D))/(2))`C. `x = 2sqrt(D(H - D))`D. `x = 4sqrt(D(H - D))` |
| Answer» Correct Answer - C | |