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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.
| 251. |
Density of ice is `900 kg//m^(3)`. A piece of ice is floating in water of density `1000kg//m^(3)`. Find the fraction of volume of the picec of ice outside the water. |
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Answer» Let V be the total volume and `V_(1)` the volume of ice piece immersed in water. For equilibrium of ice piece, weight =upthrust `therefore " " V rhoig=V_(i) rho_(w)g` Here, `rho_(i)`=density of water`=900 kgm^(-3)` and `rho_(w)`=density of water `=1000kgm^(-3)` Substituting in Eq. (i), we get `(V_(i))/(V)=(rho_(i))/(rho_(w))=(900)/(1000)=0.9` i.e., the fraction of volume outside the water, f=1-0.9=0.1 |
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| 252. |
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`. Velocity of cart at `t = 10 s` is equal toA. `4m//s`B. `6m//s`C. `8m//s`D. `5m//s` |
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Answer» Correct Answer - C `F=2rhoA(V_(0)-u)^(2)` `u=` speed of the cart `m(du)/(dt)=2rhoA(V_(0)-u)^(2)` `int_(0)^(u)(du)/((V_(0)-u)^(2))=(2rhoA)/mint_(0)^(1)dt` `[1/(V_(0)-u)]_(0)^(u)=(2rhoAt)/m` `1/(V_(0)-u)=-1/(V_(0))=(2rhoAt)/m=(4t)/100` ...........i `((2rhoA)/m=(2xx10^(3)xx2xx10^(-4))/10=4/100)` At `t=10s` `1/(V_(0)-h) =4/10+1/10=1/2` `impliesV_(0)-u=2` `u=8m/s` |
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| 253. |
Statement I: A liquid will flow faster and more smoothly from a sealed can when two holes are punched in the can than when one hole is punched. Statement II: The flow becomes streamlined with two holes rather than with one hole.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 - B When two holes are made, the velocity of flow through each holes is much less tan the velocity of flow when there is only one hole. Hence, the flow becomes streamline with two holes. This is the reason why a liquid flows more smoothly and faster with two holes. |
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| 254. |
Which of the following statements are ture in case when two water drops coalesce and make a bigge drop:A. energy is releasedB. energy is absorbedC. the surface area of the bigger drop is greater than the sum of the surface areas of both the dropsD. the surface area of the bigger drop is same that of the sum of the surface areas of both the drops |
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Answer» Correct Answer - A Energy is reased in process. |
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| 255. |
A beaker containing water is placed on the pan of a balance which shows a reading of `M`. A lump of sugar of mass `m` ad volume `v` is now suspend by a thread (from an independent support) in such a way that it is completely immersed in water without touching the beaker and without any overflow of water. How will the reading change as time passes on? |
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Answer» Initially, the balance will show the weight of `M` plus thrust. But we know that Thrust `-= vsigmag=m/(rho)xx1xxg` [as `v=m/(rho)` and `sigma=1`] `:.W_(1)=(M=m/(rho))g=(M+m/(2rho)+m/(2rho))g`…………i Now as the sugar dissolves, the thrust and hence the reading of the balance should decrease. However, as dissolved sugar comes is solution, the reading of the balance should increase. To find which half the sugar has dissolved. In this situation the thrust reduces by `(m//2rho)` from `(mg//rho)` while weight of the solution increases by `(m//2)g` (due to sugar dissolved). So reading of the balance will become `W_(2)=(M+m/(2rho)+m/(rho))g`..........ii Finally, when all the sugar is dissolved, thrust will become zero and weighht of the solution will increase by `mg`. So the reading will become. `W_(3)=(M+m)g=(M+m/2+m/2)g`..........iii Nos as `rhogt1,m//2gtm//2rho`. So comparing eqs. i, ii and iii, we find that the reading of the balancing will gradually increase till all the sugar dissolves in water and finally will become constant equal to `(M+m)g`. |
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| 256. |
A block of mass m is kept over a fixed smooth wedge. Block is attached to a sphere of same mass through fixed massles pullies `P_(1)` and `P_(2)`, sphere is dipped inside the water as shown. If specfic gravity of material of sphere is 2. Find the acceleration of sphere. |
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Answer» Writing equation fo motion for the block `T=mg sin 30^(@)=ma` for the sphere Weight-Buoyant force `-T=ma` or, `mg-(mg)/(2)-T=ma` solving we get `a=0`. |
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| 257. |
Three liquids of densities `d, 2d` , and `3d` are mixed in equal proportions of weights. The relative density of the mixture isA. `(11d)/(7)`B. `(18d)/(11)`C. `(13d)/(9)`D. `(23d)/(18)` |
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Answer» Correct Answer - B |
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| 258. |
the correct observation is:A. The pressure on the bottom of tank (a) is greater than at the bottom of (b).B. The pressure on the bottom of the tank (a) is smaller than at the bottom of (b)C. The pressure depend on the shape of the containerD. The pressure on the botton of (a) and (b) is the same |
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Answer» Correct Answer - D |
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| 259. |
Two soap bubbles coalesce to form a single large drop. Match the following. `|{:(,"Table-1",,"Table-2"),((A),"Surface energy in the",(P),"increase"),(,"process will",,),((B),"Temperature of the",(Q),"decrease"),(,"drop will",,),((C),"Pressure inside the",(R),"remain same"),(,"soap bubble will",,):}|` |
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Answer» Correct Answer - A::B::C When two soap bubbles coalesce to form a larger drop, radius increases. Hence, the excess pressure `4T//R` will decrease. Further, surface are will decrease. Hence, the surface energy will decrease or the temperature will increases. |
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| 260. |
A hollow cylinder of mass `m` made heavy at its bottom is floating vertically in water. It is tilteed from its vertical position through an angle `theta` and is left. The respecting force acting on it isA. `mg cos theta`B. `(mg)/(cos theta)`C. `mg[(1)/(cos theta)-1]`D. `mg[(1)/(cos theta)+1]` |
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Answer» Correct Answer - C Let `l` be the length of the cylinder, when verticlal in water. Let `A` be the cross-sectional area of the cylinder. Equanting weight of the cylinder with the upthrust, we get `Mg = A l rho g or m=Alrho` When the cylinder is titled through an angle `theta`, length of cylinder in water =`l=(cos theta)` weight of water displaced = `(l)/(cos theta) A rho g` Restoring force = (lA rho g)/(cos theta) -lArhog` `=lArhog [(1)/(costheta)-1]=mg[(1)/(costheta)-1]`. |
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| 261. |
Mass of solid floating in liquid is `m_(1)` and mass of liquid is `m_(2)`. Base area is A. Then pressure at bottom is `p_(0) + ((m_(1)+m_(2))g)/(A)` Reason : Upward force on liquid from base of vessel is pA, where p is pressure at bottom.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 - B | |
| 262. |
The pressure gauge shown in figure has a spring for which `k=60N//m` and the area of the piston is `0.50cm^(2)`. Its right end is connected to a closed container of gas at a gauge pressure of `30kPa`. How far will the spring be compressed if the region containing the spring is (a) in vacuum and (b) open to the atmosphere? Atmospheric pressure is `101kPa`. |
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Answer» Correct Answer - A::B::C (a) Force from left hand side =force from right hand side `:. Kx=DeltapA` `or, x=((Delta p)A)/k` `=((30+101)xx10^(3)xx0.5xx10^(-4))/(60)` `=0.109m or 10.9 cm` (b) `x=((Delta p)A)/(k)` `=(30xx10^(3)xx0.5xx10^(-4))/(60)` `=0.025m` or `2.5 cm`. |
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| 263. |
A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance l and h are shown here. After some time the coin falls into water. Then A. `l` decreases and h increasesB. l increases and h decreasesC. Both and h increaseD. Both l and h decrease |
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Answer» Correct Answer - D |
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| 264. |
A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance l and h are shown here. After some time the coin falls into water. Then A. `l` decreases and `h` increaseB. `l` increases and `h` decreasesC. both `l` and `h` increasesD. both `l` and `h` decrease |
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Answer» Correct Answer - D `l` will decrease because the block moves up. `h` will decrease because the coin will displace the volume of water `(V_(1))` equal to its own volume when it is in the water whereas when it is on the block it will displace the volume of water `(V_(2))` whose weight is equal to weight of coin and since density of coin is greater than the density of water `V_(1) lt V_(2)` |
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| 265. |
Water flows through a tube shown in figure. The areas of cross section at A and B are `1cm^2 and 0.5 cm^2` respectively. The height difference between A and B is 5m. If the speed of water at A is 10 `cms^-1` find a the speed at B and b the difference in pressures at A and B. |
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Answer» Correct Answer - A::B::C::D a. `vecV_Axxa_A=vecV_B=a_B` `rarr 10xx1=vecV_Bxx0.5` `rarr vecV_B=20cm/sec` `b. 1/2rhov_A^2+rhogh_A+p_A` `=1/2rhov_B^2+rhogh_A+p_B` `rarr p_B-p_A=1/2rho(v_A^2-v_B^2)+rhog(h_A-h_B)` `=1/2xx1(100-400)+1xx1000(5.0)` `=-150+5000=4850Dyne/cm^2` `485N/m^2` |
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| 266. |
Assertion: Force of bouyancy due to atmosphere on a small body is almost zero(or negligible). Reason: If a body is completely submerged in a fluid, then buoyant force is zero.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 - C Force of buoyancy, `U=Vrho_(air)g` since, `rho_(air)` is negligible. Hence, `U` is negligible. |
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| 267. |
Bernouli theorem is based on conservtion ofA. Law of Conservation of EnergyB. Law of Conservation of MassC. Law of Conservation of MomentumD. Law of Conservation of Angular Momentum |
| Answer» Correct Answer - C | |
| 268. |
Bernouli theorem is based on conservtion ofA. momentumB. massC. energyD. angular momentum |
| Answer» Correct Answer - C | |
| 269. |
A small block of wood of density `0.4xx10^(3)kg//m^(3)` is submerged in water at a depth of `2.9m`. Find (a) the acceletation of the block towards the surface when the block is released and (b) the time for the block to reach the surface, Ignore viscosity. |
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Answer» Correct Answer - A::B::C::D (a) `a=("upthrust-weight")/(mass)` `=((V)(1000)(9.8)-(V)(0.4xx10^(-3))(9.8))/((V)(0.4xx10^(3))` `=14.7m//s^(2)` (b) `t-sqrt((2s)/(a))` `=sqrt((2xx2.9)/(14.7))` `=0.63s`. |
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| 270. |
A block of wood floats in water with `(4//5)th` of its volume submerged. If the same block just floats in a liquid, the density of liquid in `(kg m^(-3))` isA. 1250B. 600C. 400D. 800 |
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Answer» Correct Answer - D Submerge part = Replaced water `4/5hrho_(w)=hxxrho_(w)hxxrho_(lamda)` `therefore" "rho_(lambda)=4/5xxrho_(w)=4/5xx1000` `rho_(w)=800 kgm^(-3)` |
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| 271. |
A trolley containing a liquid slides down a smooth inclined plane of angle `alpha` with the horizontal. Find the angle of inclination `theta` of the free surface with the horizontal. |
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Answer» The acceleration of the trolley is `a=gsintheta` down the inclined plane `a_(x)=acosalpha=(gsinalpha)cosalpha` `a_(y)=asinalpha=(gsinalpha)sinalpha=gsin^(2)alpha` Let `theta` be the angle made by free surface of liquid, then `tantheta=(a_(x))/(g-a_(y))=(gsinalphacosalpha)/(g-gsin^(2)alpha)=tanalpha` `theta=alpha` |
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| 272. |
The blood pressure in humans is usually taken using the arm. How ever suppose the pressure reading were taken on the calf on the leg of a standing person. Would there be a difference ? |
| Answer» Yes, blood pressure taken on the calf of the leg will be higher than that taken from the arm. It is because, the height of the blood column is quite large at than that at the arm. | |
| 273. |
Assertion : The viscosity of liquid increases rapidly with rise of temperature. Reason : Viscosity of a liquid is the property of the liquid by virtue of which it opposes the relative motion amongst its different layers.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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| 274. |
A non-uniform rod of mass `m`, length `l` and radius `r` is having its centre of mass at a distance `l//4` from the centre and lying on the axis of the cylinder. The cylinder is kept in a liquid of uniform density `rho`. The moment of inertia of the rod about the centre of mass is `I`. The angular acceleration of point `A` relative to point `B` just after the rod is released from the position as shown in the figure is A. `(pirhogl^(2)r^(2))/(I)`B. `(pirhogl^(2)r^(2))/(4I)`C. `(pirhogl^(2)r^(2))/(2I)`D. `(3pirhogl^(2)r^(2))/(4I)` |
| Answer» Correct Answer - B | |
| 275. |
`A` liquid film is formed over a frame `ABCD` as shown in figure. Wire `CD` can slide without friction. The mass to be hung from `CD` to keep it in equilibrium is A. `(Tl)/(g)`B. `(2Tl)/(g)`C. `(g)/(2Tl)`D. `Txxl` |
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Answer» Correct Answer - B Weight of the body hung from wire `(mg)` = upward force due to surface tension `(2Tl) implies m=(2Tl)/(g)` |
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| 276. |
A good lubricant must haveA. High viscosityB. Low viscosityC. Moderate viscosityD. High density |
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Answer» Correct Answer - A |
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| 277. |
What is one bar of pressure ? |
| Answer» `1` bar = 10^(5)Nm^(-2)` | |
| 278. |
A cylindrical drum, open at the top, contains `30` litres of water. It drains out through a small opening at the bottom. `10` litres of water comes out in time `t_1`, the next `10`litres in a further time `t_1` and the last `10` litres in a further time `t_3` Then,A. `t_1 = t_2 = t_3`B. `t_1 gt t_2 gt t_3`C. `t_1 lt t_2 lt t_3`D. `t_2 gt t_1 = t_3` |
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Answer» Correct Answer - C Velocity of efflux `v = sqrt(2 gh)`, `h` : height of liquid As water drains out, `h` decreases and hence `v` reduces. This results in decrease in rate of drainage. `t_1 lt t_2 lt t_3`. |
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| 279. |
Figure shows how the stream of water emerging from a faucet necks donw as it falls. The area changes from `A_(0)` to A through a fall of h. At what rate does the water flow from the tap ? A. `A_(0) sqrt((2ghA^(2))/(A_(0)^(2)-A^(2)))`B. `2A_(0) sqrt((ghA^(2))/(A_(0)^(2)-A^(2)))`C. `A_(0)sqrt((gh)/(2))`D. `2A sqrt((ghA_(0)^(2))/(A_(0)^(2)-A^(2)))` |
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Answer» Correct Answer - A As,`A_(0)v_(0)=Av implies v^(2)=v_(0)^(2)+2gh` `implies v_(0) =sqrt((2ghA^(2))/(A_(0)^(2)-A^(2))) " or" R=A_(0)v_(0)=A_(0)sqrt((2ghA^(2))/(A_(0)^(2)-A^(2)))` |
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| 280. |
The work done in blowina siap bubble of surface tension `0.06 N,^(-1)` from 2 cm radius to 5 cm radiu isA. `0.004168 J`B. `0.003168 J`C. `0.003158 J`D. `0.004158 J` |
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Answer» As given, `s=0.06 Nm^(-1)` `r_(1)=2 cm =0.02 m,r_(2)=5 cm =0.05` Since, bubble has two surface Initial surface areqa of the bubble `=2xx4pir_(2)^(2)` `=2xx4pixx(0.2)^(2)` `=32pixx10^(-4)m^(2)` Final surface area of the bubble `=2xx4pir_(2)^(2)` `=2xx4pixx(0.05)^(2)` `=200pixx10^(-4)m^(2)` So, work done `=s xx` increase in surface `=0.06xx(200pi10^(-4)-32pixx10^(-4))` `0.06xx168pi10^(-4)` `0.0031168 J` |
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| 281. |
A wide cylindrical tank with a small opening in the bottom has a water column of height `h_(1)` and above the water column, there is a layer of kerosene oil of thickness `h_(2). The velocity of efflux through the opening isA. `sqrt(2gh_(1))`B. `sqrt(2gh_(2))`C. `sqrt(2g(h_(1)+h_(2)))`D. Data is not sufficient. |
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Answer» Correct Answer - C The velocity of efflux does not depend on the density of liquid. |
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| 282. |
In the absence of intermoecular forces of attraction . The observed pressure p will beA. ramin sameB. decreasC. IncreasesD. zero |
| Answer» In the absence of intermolecular forces, there will be no stickneee of molecules. Hence pressure will increases. | |
| 283. |
Water rises in a capillary tube to a height 2.0 cm. In an another capillary tube whose radius is one third of it, how much the water will rise ? If the first capillary tube is inclined at an angle of `60^@` with the vertical then what will be the position of water in the tube.A. `2.0cm`B. `4.0cm`C. `(4)/sqrt(3)cm`D. `2sqrt(2)cm` |
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Answer» Correct Answer - B The height upto which water will rise `l=(h)/(cos alpha) = (2cm)/(cos 60) = 4cm`. `[h = "vertical height", alpha = "angle with vertical"]` |
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| 284. |
The density of ice `x cm^(-3)` and that of water is `y gcm^(-3)`. What is the change in volume when `mg` of ice melts?A. `m(y-x)cm^(3)`B. `(y-x)/(m) cm^(3)`C. `mxy(x-y)cm^(3)`D. `m((1/y-1/x) cm^(3)`. |
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Answer» Correct Answer - D Volume of ice = `M/(x)` volume of water = `M/y` `:. Change in volume =`(M)/(x) - (M)/(y) = M((1)/(y)-(1)/(x)) cm^(3)` . |
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| 285. |
A boy carries a fish in one hand and a bucket (not full) of water in the other hand. If he places the fish in the bucket, the weight now carried by him (assume that water does not spill) :A. is less than beforeB. is more than beforeC. is the same as beforeD. depends upon his speed |
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Answer» Correct Answer - C In either case he carries same mass and hence same weight. (Buoyant force is internal force of bucket and fish system). |
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| 286. |
An empty glass jar is submerged in tank of water with open mouth of the jar downwards, so that air inside the jar is trapped and cannot get out. As the jar is pushed down slowly, the magnitude of net buoyant force on the system of volume of gas trapped in the jar and the jar. A. increasesB. decreasesC. remain sameD. Information is insufficient to draw inference. |
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Answer» Correct Answer - B As the jar is pushed down, due to increase in hydrostic pressure volume of gas trapped decreases. Hence net buoyant force decreases. |
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| 287. |
What is meant by one torr of pressure ? |
| Answer» `1` Torr `=` pressure exerted by `1 mm` of mercury column `= 10^(-3) xx 13,600 xx 9.8 = 133.3 Nm^(-2)` | |
| 288. |
Water flows along horizontal pipe whose cross-section is not constant. The pressure is 1 cm of Hg, where the veloecity is 35 cm/s . At a point where the velocity is 65cm/s then pressure will beA. `0.89 cm of Hg`B. `89 cm of Hg`C. `0.5 cm of Hg`D. `1 cm of Hg` |
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Answer» (a) In horizontal pipe, `p_(1)+(1)/(2)rhov_(1)^(2)=o_(2)+(1)/(2)rhov_(2)^(2) " " ....(i)` Here, " " `p_(1)=rho_(m)gh_(1)=13600xx9.8xx10^(-2)` `p_(2)13600xx9.8xxh` `rho=1000 kg//m^(3)` `v_(1)=35xx10^(-2)m//s` `v_(2)=65xx10^(-2)m//s` `:.` From Eq.(ii) , we get `13600xx9.8xx10^(-2)+(1)/(2)xx1000xx(0.35)^(2)` `=13600xx9.8xxh+(1)/(2)xx1000xx(0.65)^(2)` After solving `0.89` cn of Hg |
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| 289. |
The liquid inside the container has density `rho`. Choose the correct option. A. `p_(A)-p_(C )=2 rho gL`B. `p_(C )-p_(B)=sqrt(2)rhogL`C. `p_(C )-p_(D)=rho gL`D. `p_(A)-p_(D)=0rho gL` |
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Answer» Correct Answer - C `(dp)/(Delta x)= rho a= rho g` (in horizontal direction) Pressure decreases in the direction of acceleration in horizontal. `therefore " " p_(D) lt p_(C ) implies p_(C )-p_(D) = rho gh` `(Delta p)/(Detla x)=rho g` |
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| 290. |
A solid is completely immersed in a liquid. The force exerted by the liquid on the solid willA. increase if it is pushed deeper inside the liquidB. change if its orientation is changedC. decrese if it is taken partially out of the liquidD. be in the vertially upward direction |
| Answer» Correct Answer - C::D | |
| 291. |
Find the force acting on the pistion of `3 cm^2` at point 2 due to the water coluum of height 10 m. A. 10 NB. 20 NC. 30 ND. 40 N |
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Answer» Correct Answer - C (c ) `F = PA = rho gh A` `= 10^3 xx 10xx 10 xx 3 xx 10^4 = 30 N` |
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| 292. |
A solid is completely immersed in a liquid. The force exerted by the liquid on the solid willA. (i),(ii)B. (ii),(iii)C. (iii),(iv)D. (i),(iv) |
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Answer» Correct Answer - C Upthrust act in upward direction (iv) is O.K. When some part is outside, immersed volume decreases. |
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| 293. |
A uniform rod of density `rho` is placed in a wide tank containing a liquid of density `rho_0 (rho_0 gt rho)`. The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position the rod makes an angle `theta` with the horizontal.A. `sintheta = 1/2 sqrt(rho_(0)//rho)`B. `sintheta = 1/2.(rho_(0))/(rho)`C. `sintheta = sqrt(rho//rho_(0))`D. `sintheta = rho_(0)//rho` |
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Answer» Correct Answer - A |
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| 294. |
A container shown in figure contains a liquid to depth H, and of density `rho` The gauge pressure at point P is : A. `(rho gH)/(2 + P_0)`B. `(rho g H)/(2)`C. `(rho gH)/(2 cos theta)`D. `(rho g H cos theta)/(2)` |
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Answer» Correct Answer - B (b) Gauge pressure ` - h rho g` `h = (H)/(2)` :. Gauge pressure is `(H rho g)/(2)` |
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| 295. |
If pressure at half the depth of a lake is equal to 2//3 pressure at the bottom of the lake then what is the depth of the lake ?A. 10 mB. 20 mC. 60 mD. 30 m |
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Answer» Correct Answer - B (b) Pressure at bottom of the lake ` = P_0 + h rho g` and pressure at half the depth of a lake ` = P_0 + (h)/(2) rho g` According to given condition `P_0 + (1)/(2) h rho g = (2)/(3) (P_0 + h rho g) rArr (1)/(3) P_0 = (1)/(6)h rho g` `rArr h = (2P_0)/(rhog) = (2xx 10^5)/(10^3 xx 10) = 20 m.` |
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| 296. |
Assertion : A ball is released from the bottom of a tank filled with a liquid. It moves upwards . In moving upwards upthrust will decrease. Reason : Density of ball is less than the density of liquid.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 - A | |
| 297. |
The side wall of a wide vertical vessel of height `h=75 cm` a narrow slit (vertical) running all the way down to the bottom of the vessel. The length of the slit is `l=50 cm` and the width is `b=1 mm`. With the slit closed, water is filled to the top. Find the resultant reaction force of water coming out as the slit is opened. brgt |
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Answer» Let us consider an infinitensimal portion of length `dx` of the slit at a depth `x` below water level. Reaction force due to this portion is given by `dF=Av^(2)rho=(bdx)2gxrho` Total reaction force is `F=intdf=int_(h-t)^(h)2grhob(xdx)` `=(2grhob)/2[h^(2)-(l-h)^(2)]=grhobl(2h-l)` |
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| 298. |
A glass plate of length `10 cm`, breadth `1.54 cm` and thickness `0.20 cm` weigh `8.2 g` in air. It is held vertically with the long side horizontal and the lower half under water. Find the apparent weight of the plate. Surface tension of water `=7.3xx10^(-2)N//m` and `=9.8ms^(-12)` |
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Answer» Correct Answer - A Volume of the portion of the plate immersed in water is `10xx(1)/(2)(1.54)xx0.2=1.54cm^(3)` Therefore, if the density of water is taken as 1, Then uptrust =weight of the water displaced `=1.54xx1xx981509.2 dynes` Now, the total length of the plate in contact with the water surface is `2(10+0.2)=20.4 cm` `:.` Downward pull upon the plate due to surface tension `=20.4xx73=1489.2 dynes` `:.` Resultant upthrust `=1509.2-1489.2` `=20.0 dynes=(20)/(980)` `=0.0204 gm-wt` `:.` Apparent weight of the plate in water =weight of the plate in air-resultant uothrust `=8.2-0.0204=8.1796 gm`. |
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| 299. |
A solid sphere a cone and a cylinder are floating in water. All have same mass, density and radius. Let `f_(1)f_(2)`,and `f_(3)` are the fraction of their volumes inside the water and `h_(1),h_(2)` and `h_(3)` are depths inside water. Then A. `f_(1)=f_(2)=f_(3)`B. `f_(3)gtf_(2)gtf_(1)`C. `h_(3) lth_(1)`D. `h_(3)lth_(2)` |
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Answer» Correct Answer - A::C::D Fraction of volume immersed in given by `f=(rho_s)/(rho_l)` `rho_(s)` and `rho_(l)` are same, Hence: `f_(1)=f_(2)=f_(3)` Base area in third case is uniform. Hence `h_(3)` is minimum. |
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| 300. |
A small sphere of mass m is dropped from a height After it has fallen 100 m it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first 100 m of fall is-A. Greater than the work done by air friction in the second 100 mB. Less than the work done by air friction in the second 100 mC. Equal to 100 mgD. Greater than 100 mg |
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Answer» Correct Answer - B |
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