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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.
| 401. |
For the system shown in the figure the cylinder on the left at L has a mass of 600 kg and a cross sectional area of `800cm^(2)` the piston on the right at S has cross sectional area `25cm^(2)` and negligible weight if the apparatus is filled with oil `(rho=0.75gm//cm^(3))` find the force F required to hold the system in equilibrium. |
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Answer» Correct Answer - `37.5m` `(F)/(A_(1))+hrhog=(600g)/(A_(2)) implies (F+hrhogA_(1))/(A_(1)) = (600xxg)/(A_(2)) implies (F+hrhogA_(1))/(25) = (600g)/(32)` `F + 8 xx 750 xx g xx 25 xx 10^(-4) = (1500g)/(8)` `F = 187.5-150 " "implies F=37.5 N` |
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| 402. |
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 |
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Answer» Correct Answer - C `hrhog=(12g)/(800xx10^(-4))" "implies hxx1000=(12xx10^(4))/(800)` `h=(120)/(800)m" "impliesh=15cm` |
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| 403. |
Figure shown a hydraulic press with the larger piston if diameter 35 cm at a height of 1.5 cm at a height of 1.5 m relative to the smaller piston of diameter 10cm. The mass on the smaller piston is 20 kg. What is the force exerted on the load by the larger piston? The density of oil in the press is `750 kh//m^(3)` . (Take `g=9.8 m//s^(2))` . |
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Answer» Correct Answer - A::C Pressure on the smaller piston `= (20xx9.8)/(pi xx(5xx10^(-2))^(2))N//m^(2)` Pressure on the larger piston`=(F)/(pixx(17.5xx10^(-2))^(2)) N/m^(2)` The difference between the two pressure `=h rho g` where, `h=1.5 m` and ` rho=750 kg//m^(3)` Thus, `(20xx9.8)/(pi xx(5xx10^(-2))^(2))-(F)/(pixx(17.5xx10^(-2))^(2))=1.5xx750xx9.8` which gives, `F=1.3xx10^(3)N`. |
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| 404. |
Figure shows a liquid of densith `1200kgm^-3` flowing steadily in a tube of varying cross section. The cross section at a point A is `1.0cm^2` and that at B is `20mm^2`, the points A and B are in the same horizontal plane.The speed of the liquid at A is 10 `cms^-1`. Calculate the difference in pressure at A and B. |
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Answer» From equation of continuity the speed `v_2` at B is given by `A_1v_1=A_2v_2` ltbrrarr `(1.0cm^2)(10cms^-1)=(20mm^2)v_2` `or v_2=(1.0cm^2)/(20mm^2)xx10cms^-1=50cms^-1` By Bernoullil equation `P_1+rhogh_1+1/2 rhov_1^2=P_2+rhogh_2+1/2rhov_2^2`. here `h_1=h_2` Thus, `P_1-P_2=1/2rhov_2^2-1/2rhov_1^2` `=1/2xx91200kgm^-3)(2500cm^2s^-2-100cm^2^-2)` `=600kgm^-3xx2400cm^2s^-2=144Pa` |
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| 405. |
A sample of metal weights 210 grams in air 180 grams n water and 120 grams in an unknown liquid thenA. Metal is `3`B. Metal is `7`C. Liquid is `3`D. Liquid is `1/3` |
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Answer» Correct Answer - B::C |
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| 406. |
The weight of an aeroplane flying in air is balanced byA. Upthrust 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 speeds on the surfaceC. Vertical component of the thrust created by air currents striking the lower surface of the wingsD. Force due to the reaction of gases ejected by the revolving propeller |
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Answer» Correct Answer - B |
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| 407. |
Water is moving with a speed of `5.18 ms^(-1)` through a pipe with a cross-sectional area of `4.20 cm^2`. The water gradually descends `9.66 m` as the pipe increase in area to `7.60 cm^2`. The speed of flow at the lower level isA. `2.86 ms^(-1)`B. `3.0 ms^(-1)`C. `3.82 ms^(-1)`D. `5.7 ms^(-1)` |
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Answer» Correct Answer - A From the principle of continuity, `A_(1)v_(1)=A_(2)v_(2)` `v_(2)=(A_(1)v_(1))/(A_(2))=(4.20xx5.18)/(7.60)=2.86ms^(-1)` |
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| 408. |
Water is moving with a speed of `5.18 ms^(-1)` through a pipe with a cross-sectional area of `4.20 cm^2`. The water gradually descends `9.66 m` as the pipe increase in area to `7.60 cm^2`. The speed of flow at the lower level isA. `3.0 ms^(-1)`B. `5.7 ms^(-1)`C. `3.82 ms^(-1)`D. `2.86 ms^(-1)` |
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Answer» Correct Answer - D `a_(1)v_(1) = a_(2)v_(2) implies 4.20xx5.18 = 7.60xxv_(2) implies v_(2) = 2.86 m//s` |
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| 409. |
An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters `2.5 cm` and `3.75 cm`. The ratio of the velocities in the two pipes isA. `9 : 4`B. `3 : 2`C. `sqrt(3) : sqrt(2)`D. `sqrt(2) : sqrt(3)` |
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Answer» Correct Answer - A As given `d_(1)=`Diameter of Ist pipe is 3.75. `d_(2)=`Dimeter of IInd pipe is 3.75. Applying equation of continuty for cross-sections `A_(1)" and "A_(2)`. `rArr" "A_(1)v_(1)=A_(2)v_(2)rArrv_(1)/v_(1)=A_(2)/A_(2)=(pi(r_(2)^(2)))/(pi(r_(1)^(2)))=(r_(2)/r_(1))^(2)` `=((3.75/2)/(2.5/2))^(2)=(3.75/2.5)^(2)=9/4[{:(r_(2)= (d_(2))/(2)),(r_(1) = (d_(1))/(2)):}]` |
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| 410. |
Water enters through end A with a speed `v_1` and leaves through end B with a speed `v_2` of cylindrical tube AB. The tube is always completely filled with water. In case I the tube is horizontal, in case II it vertical with the end A upward and in case III it is vertical with the end B upward. We have `v_1=v_2`forA. Case IB. Case IIC. Case IIID. Each case |
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Answer» Correct Answer - D This happens in accordance with equation of continuity and this equation was derived on the principal of conservation of mass and it is true in every case, either tube remain horizontal or vertical. |
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| 411. |
Streamline flow is more likely for liquid withA. high density and high viscosityB. low density and low viscosityC. high density and low viscosityD. low density and high viscosity |
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Answer» Correct Answer - D Streamline flow is more likely for liquids with low density and high viscosity. |
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| 412. |
An ideal fluid through a pipe of circular cross-section made of two sections with diameters `2.5 cm` and `3.75 cm`. The ratio of the velocities in the two pipes isA. `9:4`B. `3:2`C. `sqrt(3):sqrt(2)`D. `sqrt(2) : sqrt(3)` |
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Answer» Correct Answer - A According to equation of continuity `A_(1)v_(1) = A_(2)v_(2)` `(v_1)/(v_2) = (A_2)/(A_1) = (pi D_(2)^(2)//4)/(pi D_(1)^(2)//4) = ((D_2)/(D_1))^(2)` Here, `D_(1) = 2.5 cm, D_(2) = 3.75 cm` `:. (v_1)/(v_2) = (3.75/2.5)^(2) = (3/2)^(2) = 9/4`. |
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| 413. |
Water enters through end A with a speed `v_1` and leaves through end B with a speed `v_2` of cylindrical tube AB. The tube is always completely filled with water. In case I the tube is horizontal, in case II it vertical with the end A upward and in case III it is vertical with the end B upward. We have `v_1=v_2`forA. caseIB. caseIIC. caseIIID. each case |
| Answer» Correct Answer - D | |
| 414. |
Consider the situations of the previous proble. Let the water push the left wass by a force `F_1` and the right wall byi as force `F_2`.A. `F_1=F_2`B. `F_1gtF_2`C. `F_1ltF_2`D. the information in insufficient to know the relation between `F_1 and F_2` |
| Answer» Correct Answer - B | |
| 415. |
Suppose the glass of the previous problem is covered by a jar and the air inside the jar is completely pumped out. (a) What will be the answer to the problem ? (b) Show that the answers do not change if a glass of different shape is used provided the height, the bottom area and the volume are unchanged. |
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Answer» Now `P_0 = 0` (a) (i) Force at bottom `F_0 = rho gh A = 1000 xx 10 xx 0.2 xx 20 xx 10^-4` `= 4 N (downward)` (ii) Force at top `F_1 = 0` Weight of water `W = rho V g` `1000 xx (1)/(2) xx 10^-3 xx 10 = 5 N (downward)` Force by bottom on water `= 4 N (upward)` Since water is in equilibrium, Force by glass sides on water =`5 - 1 = 1 N (upward)` (b) Answer will not change because for solving problems, height, bottom area and volume are required, shape does not matter. |
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| 416. |
Assertion: Pressure is a vector quantity. Reason: Pressure `P=(F)/(A)`. Here `F`, the force is a vector quantity.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D Direction of force is `p=(F)/(A)` is always perpendicular to the surface or it it always specific. Hence, it is not a vector. |
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| 417. |
A piece of ice is floating in a beaker containing water when ice melts, the temperature falls from `20^@C to 4^C` When ice melts, the temperature falls from `20^@C to 4^@C` and the level of water :A. remains uchangedB. fallsC. risesD. changes erratically |
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Answer» Correct Answer - B (b) As the density of water is maximum at `4^@C`, hence volume of liquid decreases which makes the level fall. |
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| 418. |
Ice pieces are floating in a beaker `A` containing water and also in a beaker `B` containing miscible liquid of specific gravity `1.2` When ice melts, the level ofA. water increases in `A`B. water decreases in `A`C. liquid in `B` decreasesD. liquid in `B` increases |
| Answer» Correct Answer - D | |
| 419. |
Two lock gates of `7.5 m` height are provided in a canal of `16 m` width meeting at an angle of `120^@`. Calculate the force acting on each gate when the depth of water in upstream side is `5 m`. |
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Answer» Height of lock gates `=7.5m` Width of canal `=16m` Inclination of gates `=120^@` and `H=5m` From the geometry of the lock gates, we find that inclination of the lock gates with the walls `a=(180^@-120^@)/2=30^@` and width of each gate `=(16/2)/(cosalpha)=8/(cos30^@)=8/0.866=9.24m` Therefore, wetted area of each gate `A=5xx9.24=46.2m^(2)` and force acting on each gate is `F=gAxx(H/2)=9.81xx46.2xx(5/2)=1133kN` |
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| 420. |
An isosceles triangular plates of base 3 m and altitude 3 m is immersed in oil vertically with its base coinciding with the free surface of the oil of relative density 0.8. Determine the total thrust. |
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Answer» Area of the plate `=1/2xx"base"xx"altitude"` `=1/2xx3xx3=4.5m^(2)` Density of oil` =0.8xx1000-800kgm^(-3)` Depth of `CG` of the plate from the free surface `=(1/3)3=1m` Now, total thrust on the plate `=("plate area" )xx("pressure at the CG of the plate")` `=(4.5)(800x10xx1)=36kN` |
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| 421. |
A triangular lamina of area `A` and height `h` is immersed in a liquid of density `rho` in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is.A. `1/2 Arhogh`B. `1/3 Arhogh`C. `1/6 Arhogh`D. `2/3 Arhogh` |
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Answer» Correct Answer - B |
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| 422. |
To what depth must a rubber ball be taken in deep sea so that its volume is decreases by `0.1%` (The bulk modulus of rubber is `9.8xx10^(8) N//m` , and the density of sea water is `10^(3) kg//m^(3))`A. 100 mB. 60 mC. 75 mD. 65 m |
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Answer» (a) From free body diagram of the wooden block, `Vsigmag=mg=(0.5xx10^(-2)[3+(0.5)/(3)]xx10^(-2)xx10^(3)xx9.81)/(2)=0.77N//m` |
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| 423. |
A solid sphere of mass `m=2 kg` and density `rho=500kg//m^(3)` is held stationary relative to a tank filled with water. The tank is acceerating upward with acceleration `2m//s^(2)`. Calculate (a) Tension in the thread connected between the sphere and the bottom of the tank. (b) If the thread snaps, calculate the acceleration of sphere with respect to the tank. (Density of water`=1000kg//m^(3),g=10m//s^(2))` |
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Answer» Correct Answer - A::B::D (a) Upthrust-weight -T=ma `:. T=Upthrust - Weight-ma` `=((2)/(500))(1000)(10+2)-20-4` `=48-20-4=24N` (b) Downward force T suddenly becomes zero. Therefore `a=("upthrust-weight")/(m)=(48-20)/(2)` `=14mm//s^(2)` `:.` Acceleration w.r.t. tank `=14-2` `=12m//s^(2)` |
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| 424. |
The rate of flow of liquid ina tube of radius r, length l, whose ends are maintained at a pressure difference P is `V = (piQPr^(4))/(etal)` where `eta` is coefficient of the viscosity and Q isA. `8`B. `1/8`C. `16`D. `1/16` |
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Answer» Correct Answer - B |
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| 425. |
A cubical vessel of height `1` m is full of water. What is the workdone in pumping water out of the vessel?A. 1250 JB. 5000 JC. 1000 JD. 2500 J |
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Answer» Correct Answer - B The volume of water contained in the vessel would be equal to `V=I^(3)=(1m)^(3)`. Assuming water to be pure so the density of water would be equal to `d = 1000 kgm^(-3)`, hence the mass of water containded in the vessel would be equal to m = Vd = 1000 kg. Thus, the work done would be equal to `W=mgh=1000xx10xx1/2=5000 j` |
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| 426. |
A capillary glass tube records a rise of `20cm` when dipped in water. When the area of cross-section of the tube is reduced to half of the former value, water will rise to a height ofA. `10sqrt(2)cm`B. `10cm`C. `20cm`D. `20sqrt(2)cm` |
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Answer» Correct Answer - D Area is halved, means radius of tube is made `(1)/sqrt(2)` times. `h=(2T cos theta)/(rrhog)` or `h prop (1)/(r)` Hence, h will become `sqrt(2)` times. |
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| 427. |
An ice cube of side `1 cm` is floating at the interface of kerosene and water in a beaker of base area `10 cm^(2)`. The level of kerosene is just covering the top surface of the ice cube. a. Find the depth of submergence in the kerosene and that in the water. b. Find the change in the total level of the liquid when the whole ice melts into water. |
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Answer» Correct Answer - `0.1mm` According to the condition of floating `0.8rho_(w)gh_(k)+rho_(w)gh_(w)=0.9rho_(w)gh`……….i where `h_(k)` and `h_(w)` are the submerged depth of the ice in the kerosene and water, respectively `h_(k)+h_(w)=h`…….ii Solving eqn i and ii we get `h_(k)=0.5cm, h_(2)=0.5cm` b. `1cm^(3)` ice `overset("melts") (rarr)0.9cm^(3)` water After melting the ice `0.5cm^(3)` of water will be occupied by the kerosene. Therefore the fall in the level of kerosene is `/_h_(k)=0.5/A`. Out of `0.9cm^(3)` of water formed `0.5cm^(3)` of water will be filled in the vacant part of ice and remaining `(0.9-0.5)=0.4cm^(3)` of water will be available to increase the level of water. Rise in the level of water is `/_h_(w)=(0.9-0.5)/A=0.4/A` Hence, the net fall in the overall level is `/_h=0.1/A=0.1/10=0.01mm` |
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| 428. |
One small and one big piece of cork are pushed below the surface of water. Which will have greater tendency to rise swiftly? |
| Answer» The upthrust on cork is equal to the weight of the displaced water. Therefore, upthrust will be greater in the case of bigger cork and hence it has greater tendency to rise swiftly. | |
| 429. |
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. `2 mum`B. `3 mu m`C. `5 mu m`D. `1 mu m` |
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Answer» Correct Answer - D `rhog h pi r^(2)=2pi r S cos theta` `rArr r=(2S cos theta)/(rho g h) =(2xx1xx0.5)/(10^(3)xx10xx10)=10^(-6) m` |
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| 430. |
Choose the correct statements from the following:A. A body will sink in a liquid if its weight is equal to or greater than the weight of the liquid displaced by it.B. A body will float in a liquid if its weight is equal to or less than the weight of liquid displaced by it.C. When a body floats in a liquid, the portion of the body above the surface of the liquid is independent of the density of the body relative to that of the liquid.D. In still air, a hydrogen filled balloon rises up to a certain height and then stops rising. |
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Answer» Correct Answer - B::D The option a is incorrect. A body will sink if its weight is greater than the upthrust. Option b is correct. Option c is incorrect. Option d is correct. The density of hydrogen is less than that of air. Therefore, initially, the balloon rises because the upthrust due to air is greater than the weight of the hydrogen filled balloon. The density of air decreases, as we go up. Therefore, the upthrust on the balloon also decreases. It will stop rising when it attains a height at which the upthrust becomes equal to its own weight. |
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| 431. |
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 eject out as fountain from capillary tube.B. Water will remain at 12 cm is in capillary tubeC. Height of water in capillary tube will be 10.3 cmD. Level of water will fall down from capillary tube |
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Answer» Correct Answer - B As if length of capillary tube is insufficient then water will rise in capillary tube upto its upper end and radius of meniscus will adjust automatically. |
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| 432. |
Water rises to a height of `10 cm` in a capillary tube and mercury falls to a depth of `3.42 cm` in the same capillary tube. If the density of mercury is `13.6 g//c.c.` and the angles of contact for mercury and for water are `135^@` and `0^@`, respectively, the ratio of surface tension for water and mercury isA. `1 : 0.15 `B. `1 : 3`C. `1 : 6.5`D. `1.5 : 1` |
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Answer» `h=(2Tcostheta)/(rrhog)` or `s=(hrrhog)/(2costheta)` or `S prop(hrho)/(costheta)` `(S_(w))/(S_(Hg))=(h_(1))/(h_(2))xx(costheta)/(costheta_(1))xx(rho_(1))/(rho_(2))` Putting the values, we obtain `1 : 6.5` |
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| 433. |
A block of volume `V` and density `sigma_(b)` is placed in liquid of density `sigma_(1)(sigma_(1) gt singma_(b))`, then block is moved upward upto a height `h` and it is still in liquid. The increase in gravitational energy of the block is :A. `sigma_(b) Vgh`B. `(sigma_(b) + sigma_(1))Vgh`C. `(sigma_(b) - sigma_(1))Vgh`D. none of these |
| Answer» Correct Answer - A | |
| 434. |
The density of ice is `xgm//c c` and that of water is `y gm//c c`. What is the change in volume in `c c`, when `m gm` of ice metls?A. `m(y-x)`B. `(y-x)//m`C. `mxy(x-y)`D. `m(1//y-1//x)` |
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Answer» `m=` volume `xx` density Mass remain constant. `V_(ice)xxrho_(ice)=V_(water)xxrho_(water)` `V_(water)=(V_(ice)xxrho_(ice))/(rho_(water))=(m)/(y)` `DeltaV=V_(water)-V_(ice)=(m)/(y)-(m)/(x)=m((1)/(y)-(1)/(x))` |
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| 435. |
A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal toA. `L//sqrt(2pi)`B. `2piL`C. `L`D. `1//2pi` |
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Answer» Correct Answer - A Equating the rate of flow, we have `sqrt((2gy))xxL^(2)=sqrt((2gxx4y))piR^(2)` `impliesL^(2)=2piR^(2)impliesR=L/sqrt((2pi))` Flow `=("area")xx("velocity")`. Here `Vel=sqrt(2gx)` where `x=ht` from top |
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| 436. |
The density of ice is `xgm//c c` and that of water is `y gm//c c`. What is the change in volume in `c c`, when `m gm` of ice metls?A. `M (y -x)`B. `(y - x)//m`C. `mxy (x - y)`D. `m (1//y - 1//x)` |
| Answer» Correct Answer - D | |
| 437. |
The reading of a spring balance when a block is suspended from it in air is 60 N. This reading is changed to 40 N when the block is submerged in water. The relative density of the block is:A. `3`B. `2`C. `6`D. `3//2` |
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Answer» Correct Answer - A `mg = 60` …..(i) `mg - rho_(i)vg = 40` …(ii) `(mg - rho_(l)vg)/(mg) = (2)/(3)` or `(rho_(0))/(rho_(l)) = 3` where `rho_(0) =` density of the block and `rho_(l) =` density of the liquid. |
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| 438. |
Consider ideal flow of water through a pipe with its axis horizontal. `A` and `B` are the two point in the pipe at the same horizontal level (`A` lies on the upstream), thenA. The pressure at `A` and `B` are equal for any shapes of the pipeB. the pressure are never equalC. The pressures are equal if the pipe has a uniform cross-sectionD. the pressure at `A` is always more than that at `B`. |
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Answer» Correct Answer - C Since water flows from `A to B`, so `A` is at higher pressure than `B`. |
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| 439. |
Water is flowing through two horizontal pipes of different diameters which are connected together. The diameters of the two pipes are 3 cm and 6 cm respectively. If the speed of water in the narrower tube is `4 ms^(-1)`. Then the speed of water in the wider tube isA. `16 ms^(-1)`B. `1 m s^(-1)`C. `4 m s^(-1)`D. `2 m s^(-1)` |
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Answer» Correct Answer - B From continutiy equation `v prop 1/A" or v "prop1/d^(2)` (d = diameter of pipe) `v_("narrow")/v_("Wider")d_(w)/d_(N)=(6/3)=rrArrv_("Wider")=v_("narrow")/4=4/4=1 ms^(-1)` |
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| 440. |
A solid floats in a liquid of different material. Carry out an analysis to see whether the level of liquid in the container will rise or fall when the solid melts. |
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Answer» Let `M=` Mass of the floating solid `rho_(2)`=density of liquid formed by the melting of the solid `rho_(1)`=density of the liquid in which the solid is floating The mass of liquid displaced by the solid is M. Hence, the volume of liquid displaced is `(M)/(rho_(1))`. When the solid melts, the volumes occupied by it is `(M)/(rho_(2))`. Hence the level in container will rise or fall according as `(M)/(rho_(2)) gt` or `lt(M)/(rho_(1))` i.e. `rho_(2)lt `or` gtrho_(1)` There will be no change in the level if `rho_(1)-rho_(2)`. In case of ice, floating in water `rho_(1)=rho_(2)` and hence, the level of water remains unchanged when ice melts. |
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| 441. |
From a horizontal tube with area of cross-section `A_(1)` and `A_(2)` as shosn in fiugre liquid is flowing in the level of the liauid in the two veritcal tunes is h. A. The volume of the liquid flowing through the tube in time is `A_(1)v_(1)`B. `v_(2)-v_(1)=sqrt(2gh)`C. `v_(2)^(2)-v_(1)^(1)=2gh`D. The energy per unit mass of the liquid is the same in both sections of the tube |
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Answer» (c) `(p_(1))/(rho)+(v_(1)^(2))/(2)=(p_(2))/(rho)+(v_(2)^(2))/(2)` or` " " p_(1)-p_(2)=(rho)/(2)(v_(2)^(2)-v_(1)^(2))` But `" " p_(1)-p_(2)=rhogh=(rho)/(2)(v_(2)^(2)-v_(1)^(2))` or `" " v_(2)^(2)-v_(1)^(2)=2gh` |
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| 442. |
Vessel contains oil `(density 0.8 g//c c)` over mercury (density `13.6 g//c c)` A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the sphere in g/c c isA. `6.4`B. `7.2`C. `12.8`D. `12.8` |
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Answer» (c) For equilibrium , the total upaward pull be equal to the downward. Pull. If V is the volume of he sphere, we have `((V)/(2))(13.6)g+((V)/(2))(0.8)g=Vrhog` `:. rho=((13.6+0.8)/(2))gcm^(-3)=7.2 gcm^(-3)` |
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| 443. |
Water is flowing in a pipe of diameter 6 cm with an average velocity `7.5 cm^(-1).s^(-1)` and its density is `10^(3) kg m^(-3)`. What is the nature of flow ? Given coefficient of viscosity of water is `10^(-3) kgm^(-1) s^(-1).` |
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Answer» Reynolds number for the given situation is given as `R_(e )=(rho upsilon )/(eta)` Here `rho`, density `=10^(3) kgm^(-3)` Coefficient of visosity , `eta=10^(-3) kgm^(-1)s^(-1)` Average velocity of water, `upsilon =7.5 cms^(-1) = 0.075 ms^(-1)` Diameter of pipe, D=0.075=6 cm = 0.06 m Hence, `R_(e )=(10^(3)xx0.075xx0.06)/(10^(-3))` `=10^(6)xx0.0045=4500` `because " " R_(e ) gt 2000` Therefore, the flow is turbulent |
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| 444. |
A cylindrical object of outer diameter 20 cm height 20 cm and density 8000 `kgm^-3` is supported by a vertical spring and is half dipped in water as shown in figue. a. Figure the elongation of the spring in equilibrium condition. b. If the object is sllightlly depressed and relation, find the time period of resulting oscillations of the objcet. The spring constant `=500 Nm^-1`. |
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Answer» Correct Answer - A::B::C `d=10cm` `r=5cm` `rarr h=20cm` `rho_b=8000kg/m^3=8gm/cc` `k=500N/m=500xxx10^3dyne/cm` a. Here `F+U=mg [whee Fkx]` `rarr kx+Vrho_ug=mg` `rarr 500xx10^3xx(x)+(pir^2)xx(h/2)xx1xx1000` `=pir^2xhxxrho_bxx1000 `rarr 500xx10^3xx(x)` brgt`=pir^2xxhxx1000(rho-1/2)` `=pixx(5)^2xx20xx1000(rho_b-1/2)` `rarr 50x=pixx25xx2xx(rho_b-1/2)` `=x=pi(8-0.5)` or `x=pixx7.5=23.5cm` b. If Xrarr displacement of the block form the equilibrium position, Driving force `F=kX+Vrho_wxxg` `rarr ma=kX+pir^2xx(X)xxrho_wxxg` `=(k+pir^2xxrho_wxxg)X` `rarr omega^2xx(X)=((k+pir^2+rho_wxxg))/m xx(X)` `[because a=w^2XinSHM]` `rarr T=2i m/(k+pir^2+rho_wxxg) `, `=2pi(pixx25xx20xx8)/(500+10^3+pix25xx1xx1000)` `=0.935s. |
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| 445. |
At a depth of 500m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? Density of sea water is `1.03 xx 10^(3) kg//m^(3), g=10ms^(-2)`. Atmospheric pressure = ` 1.01 xx 10^(5)Pa`.A. 40 atmB. 52 atmC. 32 atmD. 62 atm |
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Answer» Correct Answer - B Absolute pressure `p=p_(a)+ rho gh` here `p_(a)=1.01xx10^(5) Pa` `rho=1.03xx10^(3) kgm^(-3)` `therefore " " p=1.01xx10^(5) Pa+ 51.5xx10^(5) Pa` `=1.01xx10^(5) Pa + 51.5xx10^(5) Pa` `=52.5xx10^(5) Pa=52 atm` |
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| 446. |
A cylindrical object of outer diameter 20 cm and mass 2 kg flots in wter with its axis vertical. If it si lightly depressed and then released, find the time peiod of the resulting simple harmonic motion of the object. |
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Answer» Given d=20` `rarr r=d/2=10cm` When depressed downward the net unbalanced force will cause SHM. Let xrarr displacement of the block from the equilibrium positoinso, Driving force `=U=V(rho_ug)` `rarr ma=pir^2(X)xxrho_ug)` `a=(pir^2rho_w9x)/(2xx10^3)` `[because m=2kg=2xx10^3g]` `T=2pisqrt(displacement/AScceleration)` `=2pisqrt((x)xx2xx10^3)/((pir^2rho_wg(x)))` `=2pisqrt(2xx10^3)/(pixx(10)62xx1xx10))` `=2pisqrt(2/(pixx10))=0.5sec` |
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| 447. |
A cylindrical block of wood of mass M is floating n water with its axis vertica. It is depressed a little and then released. Show that the motion of the block is simple harmonic and find its frequency. |
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Answer» Suppose a height `h` of the block is dipped in the water in equilibrium position. If `r` be the radius of the cylindrical block, the volume of the water displaced `= pir^(2)h`. For floating in equilibrium, `pir^(2)hrhog = W` ……..(i) where `rho` is the density of water and `W` the weight of the block. Now suppose during the vertical motion, the block is further dipped through a distance `x` at some Instant. The volume of the displaced water `pir^(2)(h + x) rhog` vertically upward. Net force on the block at displacement `xx` from the equilibrium position is `F = W - pir^(2)(h + x)rhog = W - pir^(2)hrhog - pir^(2)rho xx g` Using `(i) F = -pir^(2) rhogx = -kx`, where `k = pir^(2)rhog`. Thus, the block executes `SHM` with frequency. `v = (1)/(2pi) sqrt((k)/(m)) = (1)/(2pi) sqrt((pir^(2)rhog)/(m))` |
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| 448. |
A piece of ice having a stone frozen in it floats in a glass vessel filled with water. How will the level of water in the vessel change when the ice melts? |
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Answer» Correct Answer - A::C::D Let,` m_(1)=`mass of ice,` m_(2)=`mass of stone `rho_(s)=`density of stone and `rho_(w)=` density of water In equilibrium, when the piece of ice floats in water, weight of (ice+stone)=upthrust or, `(m_(1)+m_(2))=V_(i)rho_(w)g` `:. V_(i)=(m_(1))/(rho_(w))+(m_(2))/(rho_(w))` ...(i) Here, `V_(i)=` Volume of ice immersed When the ice melts, `m_(1)` mass of ice converts into water and stone of mass`m_(2)` is completely submerged. Volume of water formed by `m_(1)` mass of ice, `V_(1)=(m_(1))/(rho_(w))` Volume of stone (which is also equal to the volume of water displaced) `V_(2)=(m_(2))/(rho_(s))` Since, `rho_(s)gtrho_(w)` therefore, `V_(1)+V_(2)ltV_(1)` or the level of water will decrease. |
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| 449. |
A sphere of mass m and radius r is projected in a gravity free space with speed v. If coefficient of viscosity of the medium in which it moves is `1/(6pi)` , the distance travelled by the body before it stops isA. `(mv)/(2r)`B. `(2mv)/r`C. `(mv)/r`D. `(mv)/(4r)` |
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Answer» Correct Answer - C The only force acting on the body is the viscous force Here, `.m(vdv)/(dx) =-6 pi eta r v=-rv` `rArr int_(v)^(0)mdv-int_(0)^(x)-rdx rArr x=(mv)/f` |
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| 450. |
The two femurs each of cross-sectional area `10 cm^(2)` support the upper part of a human body of mass `40 kg. the average pressure sustained by the femurs is (take `g=10 ms^(-2))`A. `2xx10^(3)Nm^(-2)`B. `2xx10^(4)Nm^(-2)`C. `2xx10^(5)Nm^(-2)`D. `2 xx10^(6)Nm^(-2)` |
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Answer» Correct Answer - C Total cross-sectional area of the femurs is , `A = 2 xx 10 cm^(2) = 2 xx 10 xx 10^(-4)m^(2)=20xx10^(-4)m^(2)` Force acting on them is `F = mg = 40 kg xx 10ms^(-2)=400N` `:.` Average pressure sustained by them is `P= F/A = (400N)/(20xx10^(-4)m^(2)) = 2 xx 10^(5)Nm^(-2)`. |
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