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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.
| 201. |
The nature of r-h graph (`r` is radius of capillary tube and `h` is capillary rise) isA. straight lineB. parabolaC. ellipseD. recangular hyperbola |
| Answer» Correct Answer - D | |
| 202. |
In a capillary rise, find the heat developed taking all standard notations as described in the foregoing section.A. `Q=(2piTcos^(2)theta)/(rhog)`B. `Q=(2pir^(2)Tcos^(2)theta)/(rhog)`C. `Q=(2piT^(2)sin^(2)theta)/(rhog)`D. `Q=(2piT^(2)cos^(2)theta)/(rhog)` |
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Answer» Correct Answer - D As the liquid rises, positive work is done by surface tension in pulling the liquid up by a distance h which is given as `W_(ST)=F_(y)h` where `F_(y)=(Tcostheta)2pir` `W_(gr)=-DeltaU=mgy_(CM)`, where `Y_(CM)=(h)/(2)` and `m=(pir^(2)h)rho` Then, `DeltaU=(pir^(2)rhogh^(2))/(2)` where `h=(2Tcostheta)/(rhogr)` This gives applying first law of thermodynamics, the heat dissipated Q can be gives as `W_(ST)=Q+DeltaU` From the above equations, we have `Q=(2piT^(2)cos^(2)theta)/(rhog)` |
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| 203. |
Viscosity is the property by virtue of which a liquid.A. occupies minimum surface areaB. offers resistance for the relative motion between its layers.C. becomes spherical in shape.D. tends to gain its deformed position. |
| Answer» Correct Answer - B | |
| 204. |
Statement -1: A smooth block of mass `2kg` and specific gravity `2.5` is attached with a spring of force constant `k = 100 N//m` and is half dipped in water. If the extension in the spring is 1cm, the force exteted by the bottom of tank on the block is `19N`. Statement-2 : In the arrangement shown, the buoyant force acting on the block is equal to weight of liquid displaced. A. Statement I is true, statement II is true and statement II is correct explanation for statement IB. Statement I is true, statement II is true and statement II is not the correct explanation for statement IC. 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 `f_(B)+kx=mgimpliesf_(B)=20-1=19N` |
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| 205. |
The excess pressure inside a small air bubble of radius 0.05 mm in water of surface tension 70 dyne `cm^(-1)` (in pascal)A. 28.2B. `2.8xx10^(2)`C. `2800`D. `280` |
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Answer» Correct Answer - 3 `DeltaP=(2T)/(r)` |
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| 206. |
The force required to take away flat plate of radius `4cm` from the surface of water is (surface tension os water `=70"dyne"//cm)`A. `1221.2 "dyne"`B. `1589.2"dyne"`C. `1645.3"dyne"`D. `1758.4"dyne"` |
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Answer» Correct Answer - D `F=2pirxxT` |
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| 207. |
Fig, shown a thin film supporting a small weight = `4.5xx10^(-2)N`. What is the weight supported by a film of the same liquid at the same temperature in fig. explain your answer physically. . |
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Answer» (a). Here, length of the film supporting the weight `=40cm=0.4m` total weight supported (or force)= `4.5xx10^(-2)N` Film has two free surfces, surfaces tension `S=4.5xx10^(-2)//2xx0.4=5.625xx10^(-2)Nm^(-1)` Sice the liquid is same for all the cases (a), (b) and (c) and temperature is also same, therefore surface tension for cases (b) and (c) will also be the same `=5.625xx10^(-2)`. in figure 7(b), 38(b) and (c), the length of the film supporting the weight is also the saihe as that of (a), hence the total weight supported in each case is `4.5xx10^(-2)N`. |
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| 208. |
The work done in increasing the radius of a soap bubble from 4 cm to 5 cm is Joule (given surface tension of soap water to be `25xx10^(-3)N//m)`A. `0.5657xx10^(-3)`B. `5.657xx10^(-3)`C. `56.5xx10^(-3)`D. `565xx10^(-3)` |
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Answer» Correct Answer - A `W=8pi(r_(2)^(2)-r_(1)^(2))T` |
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| 209. |
Which of the following substances has the greatest viscosity?A. MercuryB. WaterC. KeroseneD. Glycerin |
| Answer» Correct Answer - D | |
| 210. |
Viscosity is most closely related toA. densityB. velocityC. frictionD. energy |
| Answer» Correct Answer - C | |
| 211. |
Machine parts are jammed in winter due toA. increase in viscosity of libricantB. decrease in viscosity of libricantC. increase in surface tension of lubricantD. decrease in surface tension of lubricant |
| Answer» Correct Answer - A | |
| 212. |
When a solid ball of volume V is falling through a viscous liquid, a viscous force F acts of it. If another ball of volume 2 V of the same meterial is falling through the same liquid then the viscous force experienced by it will be (when both fall with terminal velocities).A. `F`B. `(F)/(2)`C. `2F`D. `(F)/(4)` |
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Answer» Correct Answer - C `F=6pietarv(becausevpropr^(2),Fpropr^(3),FpropV)` `implies(F_(1))/(F_(2))=(V_(1))/(V_(2))` |
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| 213. |
A metalic wire of diameter `d` is lying horizontally o the surface of water. The maximum length of wire so that is may not sink will beA. `sqrt((2T)/(pidg))`B. `sqrt((2Tg)/(pid))`C. `sqrt((2pid)/(Tg))`D. any length |
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Answer» Correct Answer - D `mg=2Tl rArr pir^(2) l rho g = 2Tl rArr pir^(2) rho g = 2T` the expression is independent of length |
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| 214. |
A woman of mass 50 kg stands on a wooden block placed over a tank of water. The wooden block is such that the woman is entirely above water. If relative density of wood is 0.85, the volume of the wooden block is:A. `0.5xx10^(-1)m^(3)`B. `0.585xx10^(-1)m^(3)`C. `0.33m^(3)`D. `0.54xx10^(-1)m^(3)` |
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Answer» Correct Answer - 3 `W_("woman")+W_("wood")=F_(B),F_(B)=` buoyant force |
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| 215. |
A certain block weighs 15 N in air. But is weighs only 12 N when completely immersed in water. When immersed completely in another liquid, it weighs 13 N. Calculate the relative density of (i) the block and (ii) the liquid.A. `5,(2)/(3)`B. `(2)/(3),5`C. `(4)/(5),5`D. `5,(4)/(5)` |
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Answer» Correct Answer - 1 RD of body `R.D.=(W_("air"))/(W_("air")-W_("water"))` RD of liquid `=(W_("air")-W_("liquid"))/(W_("air")-W_("water"))` |
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| 216. |
A body is just floating on the surface of a liquid. The density of the body is the same as that of the liquid. The body is slightly pushed down. What will happen to the body?A. come back slowly to its earlier positionB. remain submerged where it is leftC. sink in liquidD. come out vigoursly |
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Answer» Correct Answer - C When a body is just floating in a liquid whose density is equal to the density of body is pushed down slighty then downward force on the body increases due to atmospheric pressure and due to water column above the body. As a result of which the body sinks in the liquid. |
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| 217. |
In the syphon as shown which of the option is not correct, if `h_(2)gth_(1)` and `h_(3)lth_(1)`?A. `p_(E)ltp_(D)`B. `p_(E)gtp_(C)`C. `p_(B)=p_(C)`D. `p_(E)ltp_(B)` |
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Answer» Correct Answer - D If `p_(0)=` atmospheric pressure `P_(A)=P_(0)=P_(B)+rhoh_(1)g,P_(D)=P_(0)=P_(E)+rhoh_(3)g` Since, `h_(3)gth_(1), so, P_(E)gtP_(B)` |
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| 218. |
A column of water `40 cm` high supports a `30 cm` column of an unknown liquid . What is the density of the liquid ? |
| Answer» `1.33 xx 10^(3)"kg m"^(-3)` | |
| 219. |
Atmospheric pressure is nearly `100 kPa.` How large the force does the air in a room exert on the inside of a window pan that is `40 cm xx 80 cm`? |
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Answer» Atmospheric pressure = 100 kPa `= 10^(5) Pa` Area, A = 40 cm `xx` 80 cm = 40 `xx 80 xx 10^(-4) m^(2)` F = P `xx` A `= 10^(5) xx 3200 xx 10^(-4)` = 32 kN. |
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| 220. |
Find the pressure exerted at the tip of a board pin, of area 0.1 `mm^(2)`, if it is pressed against the board with a force 15 N. |
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Answer» Normal force applied on the pin, F = 15 N Area of the tip of the board pin, A = 0.1 `mm^(2)` = `0.1 xx 10^(-6) m^(2)` Pressure, `P = (F)/(A) = (15)/(0.1 xx 10^(-6)) N m^(-2)` `= 15 xx 10^(7) N m^(2)` |
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| 221. |
The force does water exert on the base of a house tank of base area `1.5m^(2)` when it is filled with water up to a height of 1 m is `(g=10(m)/(s^(-2)))`A. 1200 kgwtB. 1500 kgwtC. 1700 kgwtD. 2000 kgwt |
| Answer» Correct Answer - `F=PA,P=hrhog` | |
| 222. |
What is force on the base of a tank of base area `1.5m^(2)` when it is filled with water upto a height of 1m `(rho_("water")=10^(3)kg//m^(3),P_(0)=10^(5)Pa` and `g=10m//s^(2))` |
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Answer» Absolute pressure at the bottom of the container is `P=P_(0)+hrhog=10^(5)+1xx10^(3)xx10=1.1xx10^(5)Pa` Then force on the base is `F_("bese")=PA=(1.1xx10^(5))(1.5)=1.65xx10^(5)N` |
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| 223. |
At the mount of the tap area of cross-section is `2.0cm^(2)` and the speed of water is `3m//s`. The area of cross-section of the water column 80 cm below the tap is (use `g=10m//s^(2))`A. `0.6cm^(2)`B. `1.2cm^(2)`C. `1.5cm^(2)`D. `2.0cm^(2)` |
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Answer» Correct Answer - B `v_(2)^(2)=v_(1)^(2)+2gh,A_(1)v_(1)=A_(2)v_(2)` |
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| 224. |
A conical cup of height 1m, rests, open end down on a flat surface. What will be the upward lifting force on the cup when the water in its is as shown? |
| Answer» Correct Answer - 303 N | |
| 225. |
A cylindrical tank `1 m` in radius rests on a platform `5 m` high. Initially the tank is filled with water to a height of `5 m`. A plug whose area is `10 ^-4 m^2`, is removed from an orifice on the side of the tank at the bottom. Calculate the following : (a) Initial speed with which the water flows from the orifice. (b) Initial speed with which the water strikes the ground, ( c) Time taken to empty the tank to half its original value.A. 10B. 5C. `5sqrt(2)`D. `10sqrt(2)` |
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Answer» Correct Answer - D `v_(x)=sqrt(2gh_(1)),v_(y)=sqrt(2gh_(2)),v=sqrt(v_(x)^(2)+v_(y)^(2))` |
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| 226. |
The equation of continuity leads toA. law of conservation of moments of liquid flowB. law of conservation of energyC. law of equipartition of energyD. law of conservation of mass. |
| Answer» Correct Answer - D | |
| 227. |
The excess pressure in soap bubble is `10(N)/(m^(2))` if eight soap bubble are combined to form a big soap bubble excess pressure in big bubble is (in `(N)/(m^(2)))`A. 5B. 10C. 20D. 2.5 |
| Answer» `(4)/(3)piR^(3)=n(4)/(3)pir^(3),R=2r,p=(4T)/(R)implies(P_("big"))/(P_("small"))=(r)/(R)` | |
| 228. |
Two air bubbles of radii 0.002 m and 0.004 m of same liquid come together to form a single bubble under isothermal condition. Find the radius of the buble formed. Given surface tension of liquid is `0.072 Nm^(-1)`A. 6 mm with concave surface towards smaller bubble.B. mm with concave surface towards bigger bubble.C. 4 mm with concave surface towards smaller bubble.D. 4 mm with concave surface towards bigger bubble. |
| Answer» `r=(r_(1)r_(2))/(r_(2)-r_(1))` | |
| 229. |
When a cylindrical tube is dipped vertically into a liquid the angle of contact is `140^(@)` . When the tube is dipped with an inclination of `40^(@)` the angle of contact isA. `100^(@)`B. `140^(@)`C. `180^(@)`D. `60^(@)` |
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Answer» Correct Answer - B Angle of contact is independent of tilting angle. |
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| 230. |
A car moving on a road when overtaken by a busA. is pulled towards the busB. is pushed away from the busC. is not affected by the busD. information is insufficient. |
| Answer» Correct Answer - A | |
| 231. |
A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength `(81pi)/(7)xx10^5Vm^-1`. When the field is switched off, the drop is observed to fall with terminal velocity `2xx10^-3ms^-1`. Given `g=9.8ms^-2`, viscoisty of the air `=1.8xx10^-5Nsm^-2` and the denisty of oil `=900kg m^-3`, the magnitude of q isA. `1.6 xx 10^(-19)` CB. `3.2 xx 10^(-19)` CC. `4.8 xx 10^(-19)` CD. `8.0 xx 10^(-19)` C |
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Answer» Correct Answer - D Here, `E=(81pi)/(7)xx10^(5) V m^(-1),v=2xx10^(-3) m s^(-1)`, `eta=1.8xx10^(-5) N s m^(-2), rho=900 kg m^(-3)` When drop is balanced in still air under the effect of electric field, then `qE=(4)/(3)pir^(3)rhog` or `q=(4)/(3E)pir^(3)rhog` When the electric field is switched off, let the drop falls with terminal velocity `v`, then `v=(2r(rho-sigma)g)/(9eta)` or `r=[(9veta)/(2(rho-sigma)g)]^(1//2)` `therefore q=(1)/(E)xx(4)/(3)pirhog[(9etav)/(2(rho-sigma)g)]^(3//2)` `=(7)/(81pixx10^(5))xx(4)/(3)xxpixx900xx9.8xx[(9xx1.8xx10^(-5)xx2xx10^(-3))/(2xx900xx9.8)]^(3//2)` On solving we get, `q=8xx10^(-19)C` |
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| 232. |
In car lift compressed air exerts a force `F_1` on a small piston having a radius of 5 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car to be lifted is 1350 kg, what is `F_1 ?` What is the pressure necessary to ac complish this task ?A. `14.7xx10^(3)N`B. `1.47xx10^(3)N`C. `2.47xx10^(3)N`D. `24.7xx10^(3)`N |
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Answer» Correct Answer - B `(F_(1))/(A_(1))=(F_(2))/(A_(2))` |
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| 233. |
A solid sphere moves at a terminal velocity of `20ms^-1` in air at a place where `g=9.8ms^-2`. The sphere is taken in a gravity free hall having air at the same pressure and pushed down at a speed of `20 ms^-1`A. its initial acceleration will be `9.8ms^(-2)` downward.B. its initial acceleration will be `9.8ms^(-2)` upwardC. The magnitude of acceleration will decrease as the time passes.D. it will eventually stop. |
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Answer» Correct Answer - B::C::D `v_(t)=(2r^(2)rhog)/(9eta)` |
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| 234. |
A glass rod of radius `r_(1)` is inserted symmetrically into a vertical capillary tube of radius `r_(2)` such that their lower ends are at the same level. The arrangement is now dipped in water. The height to which water will rise into the tube will be (`sigma =` surface tension of water, `rho = ` density of water)A. `(2sigma)/((r_(2)-r_(1))rhog)`B. `(2sigma)/((r_(2)-r_(1))rhog)`C. `(2sigma)/((r_(2)-r_(1))rhog)`D. `(2sigma)/((r_(2)^(2)+r_(1)^(2))rhog)` |
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Answer» Correct Answer - A Total upward force dur to surface tension `=sigma(2pir_(1)+2pir_(2))` This supports the weight of the liquid column of height h. Weight of liquid column liquid column of height h. Weight of liquid column `=h=[pir_(2)^(2)+2pir_(2))]rhog` Equating, We get `hpi(r_(2)+r_(1))(r_(2)-r_(1))rhog=2pisigma(r_(1)+r_(2))` or `h(r_(2)-r_(1))rhog=2sigma` or `h=(2sigma)/((r_(2)-r_(1))rhog)` |
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| 235. |
When a capillary tube is dipped in a liquid, the liquid rises to a height h in the tube. The free liquid surface inside the tube is hemispherical in shape. The tube is now pushed down so that the height of the tube outside the liquid is less than `h`. ThenA. the liquid will come out of the tube like in a small fountainB. The liquid will oze out of the tube slowlyC. the liquid will fill the tube but not come out of its upper endD. the free liquid surface inside the tube will not hemispherical. |
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Answer» Correct Answer - C::D The angle of contact at the free liquid surface inside the capillary tube will change such that the certical component of the surface tension forces just balances the weight of the liquid column. |
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| 236. |
If `n` drops of a liquid, form a single drop, thenA. some energy will be released in the processB. some energy will be released in the processC. the energy released or absorbed will be `E(n-n^((2)/(3)))`D. the enrgy released or absorbed will be `nE(2^((2)/(3))-1)` |
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Answer» Correct Answer - A::C Let S `=` surface energy per unit drop `r=` radius of each small drop `R` radius of a single drop `nxx(4)/(3)pir^(3)=(4)/(3)piR^(3)` or `R=rn^((1)/(3))` Initial surface energy `E_(i)=nxx4pir^(2)xxS=nE` Final surface energy `E_(f)=4piR^(2)S=4piR^(2)n^((2)/(3))S=n^((2)/(3))E` Therefore, energy released `E_(i)-E_(f)=(n-n^((2)/(3)))` |
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| 237. |
Two vertical parallel glass plates are partically submerged in water. The distance between the plates is `d=0.10mm`, and their width is `l=12cm` assuming that the water between the plates does not reach the upper edges of the plates and the wetting is complete. Find the force of their mutual attractionA. 13 NB. 26 NC. 39 ND. 6.5 N |
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Answer» Correct Answer - A `(drhog)A=Tl,A=(Tl)/(drhog)` `F=(2T)/(d)A=(2T)/(d)xx(Tl)/(drhog)` or `F=(2T^(2)l)/(d^(2)rhog)` |
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| 238. |
Some iron beads are embedded in wax ball which is just floating in water. The volume of ball is 18 `cm^(3)` and relative density of waxis 0.9 .Then mass of the iron trapped in the ball isA. 1.8 gB. 2.7 gC. 16.8 gD. 8.1 g |
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Answer» Correct Answer - A Let the mass of iron trapped in the ball be m. According to law of floatation , Weight = upthrust (m + mass of wax)g `=18 xx 1xx g` Ignoring the volume of iron since it is in the form of small beads `therefore m+18 xx0.9-18xx1` or `m=1.8 g` |
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| 239. |
There are two holes `O_(1)` and `O_(2)` in a tank of height H. The water emerging from `O_(1)` and `O_(2)` strikes the ground at the same points as shows in fig. Then:A. `H=h_(1)+h_(2)`B. `H=h_(2)-h_(1)`C. `H=h_(1)h_(2)`D. `H=(h_(2))/(h_(1))` |
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Answer» Correct Answer - A `R=V_("efflux")xxTime,R_(1)=R_(2)` |
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| 240. |
As Fig. shows, `S_(1)` and `S_(2)`, are spring balances. Block `A` is hanging from spring balance `S_(1)` and immersed in liquid `L` which is contained in beaker `B`. The mass of beaker `B` is `1 kg` and mass of liquid `L` is `1.5 kg`. Balances `S_(1)` and `S_(2)` reads `2.5 kg` and `7.5 kg`, respectively. What will be the readings of `S_(1)` and `S_(2)` when block `A` is pulled up out of the liquid. Find the reading of `S_(1)` and `S_(2)`? A. 7.5 kgB. 2 kgC. 3.5 kgD. 2.1 kg |
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Answer» Correct Answer - A Balance D reads weight minus buoyancy force whereas E reads weight and reaction forces. |
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| 241. |
The top view of closed compartment containing liquid is moving with an acceleration along x-axis as shown. Find the incorrect statementA. The pressure at A and O is sameB. The pressure at O and `O_(1)` is sameC. The pressure at B and C is sameD. The pressure at D and E is same |
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Answer» Correct Answer - C `(dP)/(dx)=-rhox` acceleration The pressure decreases in positive x-direction The pressure is lower in fron side. The pressure at B and C can not be same. |
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| 242. |
A sphere of solid material of specific gravity `8` has a concentric spherical cavity and just sinks in water. The ratio of radius of cavity to that of outer radius of the sphere must beA. `(7^(1//3))/(2)`B. `(5^(1//3))/(2)`C. `(9^(1//3))/(2)`D. `(3^(1//3))/(2)` |
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Answer» Correct Answer - A Let `rho` the density of the material `rho_(0)` be the density of water. When the sphere has just started sinking the weight of the sphere`=` weight of water displaced (approx) `implies(4)/(3)pi(R^(3)-r^(3))rhog=(4)/(3)piR^(3)rho_(0)g` `implies(R^(3)-r^(3))rho=R^(3)rho_(0)implies((R^(3)-r^(3)))/(R^(3))=(rho_(0))/(rho)` `implies(r)/(R)=((7)^(1//3))/(2)` |
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| 243. |
A soap bubble is blown to a radius of 3 cm. if it is to be further blown to a radius of 4 cm what is the work done? (surface tension of soap solution `=3.06xx10^(-2)Nm^(-1))` |
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Answer» Initial radius of soap bubble `R_(1)=3cm=3xx10^(-1)m` Final radius of soap bubble `R_(2)=4cm=4xx10^(-2)m` Wirk done in blowing soap buble from radius `R_(1)` is `R_(2)` is `impliesW=8pi(R_(2)^(2)-R_(1)^(2))T` `=8xx(22)/(7)xx3.06xx10^(-2)(16-9)xx10^(-4)` `=176xx3.06xx10^(-6)J=539.6xx10^(-6)J` |
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| 244. |
Two water pipes of diameters 4 cm and 8 cm are connected with main supply line. The velocity of flow of water in the pipe of 8 cm diameter is hwo many times to that of 4 cm diameter pipe?A. 4B. `(1)/(4)`C. `2`D. `(1)/(2)` |
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Answer» Correct Answer - 2 `A_(1)V_(1)=A_(2)V_(2)` |
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| 245. |
Two horizontal pipes of different diameters are connected together by a value. Their diameters are 3 cm and 9 cm. Water flows at speed 6 m `s^(-1)` in the first pipe and the pressure in it is `2 xx 10^(5) N m^(-2)`. Find the water pressure and speed in the second pipe. |
| Answer» 0.66 m `s^(-1)`, 2.18 `xx 10^(5) Nm^(-2)` | |
| 246. |
Calculate the minimum pressure difference required to force the blood from the heart to the top of the head (vertical distance about 50 cm). Neglect friction. |
| Answer» `5.2 xx 10^(3) N//m^(2)` | |
| 247. |
Water flows through a horizontal pipe of varying area of cross section at the rate 15 cubic metre per minute. Find the radius of pipe where water velocity is 3 `ms^(-1)` |
| Answer» Correct Answer - 16.3 cm | |
| 248. |
An incompressible liquid flows through a horizontal tube LMN as shown in the figure. Then the velocity `V` of the liquid throught he tube N is:A. `1ms^(-1)`B. `2ms^(-1)`C. `4.5ms^(-1)`D. `6ms^(-1)` |
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Answer» Correct Answer - D `A_(1)v_(1)+A_(2)v_(2)=A_(3)v_(3)` |
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| 249. |
Water flows through a non-uniform tube of area of cross section A, B and C whose values are 25, 15 and `35cm^(2)` resprectively. The ratio of the velocities of water at the sections A,B and C isA. `5:3:7`B. `7:3:5`C. `21:35:15`D. `1:1:1` |
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Answer» Correct Answer - C `Av=const,A_(1)v_(1)=A_(2)v_(2)` |
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| 250. |
The velocity of the wind over the surface of the wing of an aeroplane is `80ms^(-1)` and under the wing `60ms^(-1)`. If the area of the wing is `4m^(2)`, the dynamic lift experienced by the wing is [density of air `=1.3kg.m^(-3)`]A. 3640 NB. 7280 NC. 14560 ND. 72800 N |
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Answer» Correct Answer - B `F=(d)/(2)(V_(2)^(2)-V_(1)^(2))A` |
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