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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.
| 351. |
A solid sphere falls with a terminal velocity v in air .If it is allowed to fall in vaccum,A. terminal velocity of sphere = vB. terminal velocity of sphere lt vC. terminal velocity of sphere gt vD. sphere never attains terminal velocity |
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Answer» Correct Answer - D When a solid sphere falls in vacuum, no viscous force is acting on the sphere and the sphere falls under gravity. Due to which sphere never attains terminals velocity. |
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| 352. |
In a cylindrical container water is filled up to a height of `h_(0)=1.0m`. Now a large number of small iron balls are gently dropped one by one into the container till the upper layer of the balls touches the water surface. if average density of the contents is `rho=4070 kg//m^(3)`, density of iron is `rho_(i)=7140kg//m^(3)` and density of iron is `rho_(0)=1000kg//m^(3)` find the height `h` of the water level (in S.I. units) in the container with the iron balls. |
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Answer» Correct Answer - 2 `rho=("total mass")/("total volume")=(m_(1)+m_(2))/(Ah)` `rho_(0)=(m_(1))/(Ah_(0)),rho_(i)=(m_(2))/(A(h-h_(0)))` `impliesh=(rho_(0)h_(0)+(h+h_(0))rho_(i))/(rho)=((rho_(i)-rho_(0)))/((rho_(i)-rho))=2m` |
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| 353. |
A marble of mass `x` and diameter `2r` is gently released in a tall cylinder containing honey. If the marble displaces mass `y(ltx)`of the liquid, then the terminal velocity is proportional toA. `x+y`B. `x-y`C. `(x+y)/(r)`D. `(x-y)/(r)` |
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Answer» Correct Answer - D `v_(0)=(2)/(9)r^(2)((rho-rho^(1))g)/(eta),x=(4)/(3)pir^(3)rho(or)` `rhoalpha=(x)/(r^(3)),rho^(1)alpha(y)/(r^(3)),v_(0)alpha(x-y)/(r)` |
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| 354. |
(a). It is known that density `rho` of air decreases with height y as `rho=rho_(0)e^(-y//y(_o))` ltb rgt where `p_(o)=1.25kgm^(-3)` is the density at sea level, and `y_(o)` is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant (b). A large He balloon of volume `1425m^(3)` is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise? [take `y_(o)=8000m and rho_(He)=0.18kgm^(-3)`] |
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Answer» (a). We know that rate of decrease of density `p` of air is directly proportional to the height y. it is given as `drho//dy=-rho//y_(0)` where y is constant of proportionally and -ve sign signifies that density is decreasing with increase in height. On integration, we get `underset(rho_(0))overset(rho)int(drho)/(rho)=-int_(0)^(y)(1)/(y_(0))dy` `implies[logrho]_(rho_(0))^(rho)=-[(y)/(y_(0))]_(0)^(y),` where `rho_(0)=` density of air at sea level i.e., `y=0` or `log_(2)(rho)/(rho_(0))=-(y)/(y_(0))` or `rho=rho_(0)e^(-(y)/(y_(0)))` here dimensions and units of constant `y_(0)` are same as of y. (b). Here volume of He balloon, `V=1425m^(3)`, mass of payload, `m=400kg` `y_(0)=8000m,` density of He `rho_(He)=0.18kgm^(-3)` Mean density of balloon `rho=("Total mass of balloon")/("Volume")=(m+V*rho_(He))/(V)Pa` `=(400+1425xx0.18)/(1425)=0.4608=0.46kgm^(-3)` As density of air at sea level `rho_(0)=1.25kgm^(-3)` the balloon will rise up to a height y where density of air `=`density of balloon `rho=0.46kgm^(-3)` As `rho=rho_(0)e^(-(y)/(y_(0)))` or `(rho_(0))/(rho)=e^((y_(0))/(y))` `thereforelog_(e)((rho_(0))/(rho))=(y_(0))/(y)` or `y=(y_(0))/(log_(e)((rho_(0))/(rho)))=(8000)/(log_(e)((1.25)/(0.46)))` `=8002m` or `8.0km` |
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| 355. |
A U tube contains water and methylated spirts separated by mercury columns in the two arms are in level with 10.0cm of water in one arm and 12.5 cm of spirit in the other. What is the relative density of spirit? |
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Answer» for water column in one arm of U tube, `h_(1)=10.0cm, rho("density")=1g cm^(-3)` for spirit column in other arm of U tube `h_(2)=12.5cm,rho=?` As the mercury column in the two arms of U tube are in level, therefore pressure exerted by each is equal. Hence, `h_(1)rho_(1)g=h_(2)g_(2)g` or `rho_(2)=h_(1)rho_(1)//h_(2)=10xx1//12.5=0.8gcm^(-3)` therefore, relative density of spirit `=rho_(2)//rho_(1)=0.8//1=0.8` |
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| 356. |
A U tube contains water and methylated spirts separated by mercury columns in the two arms are in level with 10.0cm of water in one arm and 12.5 cm of spirit in the other. What is the relative density of spirit?A. `5//4`B. `1//2`C. `2//5`D. `4//5` |
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Answer» Correct Answer - D `rho_(w)gh_(w)=rho_(m)gh_(m)` |
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| 357. |
A thin wire ring of 3 cm radius float on the surface of liquid. The pull required to raise the ring before the film breaks is `30.14xx10^(3)N` more than its weight. The surface tension of the liquid (in `Nm^(-1))` isA. `80 xx 10^(-3)`B. `87 xx 10^(-3)`C. `90 xx 10^(-3)`D. `98 xx 10^(-3)` |
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Answer» Correct Answer - A `T=(F)/(4pir)` |
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| 358. |
Work of `6.0xx10^(-4)`N joule is required to the done in increasing the size of soap film form `10cmxx6cm` to `10cmxx11cm`. The surface tension of the film is (in `N//m)`A. `5xx10^(-2)`B. `6xx10^(-2)`C. `1.5xx10^(-2)`D. `1.2xx10^(-2)` |
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Answer» Correct Answer - B `W=TDeltaA` |
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| 359. |
A wire is bent in the form of a `U`-shape and a slider of negligible mass is connecting the two vertical sides of the U-shape. This arrangement is dipped in a soap solution and lifted a thin soap film is formed in t he frame it supports a weight of `2.0xx10^(-2) N` if the length of the slider is 40 cm what is the surface tension of the film?A. `25Nm^(-1)`B. `2.5Nm^(-1)`C. `2.5xx10^(-2)Nm^(-1)`D. `2.5xx10^(-3)Nm^(-1)` |
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Answer» Correct Answer - C `T=(F)/(2l)` |
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| 360. |
In Fig(i) a thin film supports a small weight `3.5xx10^(-2)N` The weight supported by a film of the same liquid at the same temperature in fig.(ii) isA. `3.5xx10^(-2)N`B. `3.5xx10^(-3)N`C. `3.5xx10^(-1)N`D. `3.5xx10^(-4)N` |
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Answer» Correct Answer - A Force due to S.T`=` weight `(2)/(T)=` weight `implies2xx30xx10^(-3)xxT=3.5xx10^(-3)` Same liquid, same temperature same length of the film supports same weight. |
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| 361. |
A ring of inner and outer radii 8 and 9 cm is pulled out of water surface with a force of [S.T of water `(T)=70"dyne"//cm]`A. `26xx10^(-2)N`B. `12.6xx10^(-2)N`C. `7.48xx10^(-2)`D. `3.08xx10^(-2)` |
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Answer» Correct Answer - C `F=2pi(r_(1)+r_(2))T` |
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| 362. |
A bird of mass 1.23 kg is able to haver by imparting a downward velocity of `10 m//s` uniformly to air of density `rho kg//m^(3)` over an effective area `0.1m^(2)` the acceleration due to gravity is `10m//s^(2)` then the magnitude of `rho` in `kg//m^(3)`A. `0.34`B. `0.89`C. `1.23`D. `4.8` |
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Answer» Correct Answer - C Weight of bird `=` force excted by the bird by moving its wings `rhoAu^(2)=mg (F=Au^(2)rho)` |
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| 363. |
In magnus effect, lift occurs towards the direction whereA. Relative velocity of the fluid is geaterB. Relative velocity of the fluid is smallerC. Pressure is greaterD. Kinetic energy of the fluid is smaaler |
| Answer» Correct Answer - A | |
| 364. |
In the figure shown below, a fluid of density `2 xx 10^(3)` kg `m^(-3)` is flowing in a horizontal pipe. The speed of water at point A is 4 cm `s^(-1)`, what is its speed at point B? A. 3 cm `s^(-1)`B. 22.72 cm `s^(-1)`C. 6 cm `s^(-1)`D. 60 cm `s^(-1)` |
| Answer» Correct Answer - B | |
| 365. |
A tube of length L and radius R is joinced to another tube of length `(L)/(3)` and radius `(R)/(2)`. A fluid is flowing through this joint tube. If the pressure difference across the first tube is `P` then pressure difference across the second tube isA. `(16P)/(3)`B. `(4P)/(3)`C. `P`D. `(3P)/(16)` |
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Answer» Correct Answer - 1 `(P_(1))/(P_(2))=(R_(1))/(R_(2))=(((8nl_(1))/(pir_(1)^(4))))/(((8nl_(2))/(pir_(2)^(4))))` |
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| 366. |
Water stands at a height of 100 cm in a vessel whose side walls are vertical. A B and C are holes at height 80 cm, 50 cm, and 20 cm respectively from the bottom of the vessel. The correct system of flowing out is:A. B. C. D. |
| Answer» `R=uxxsqrt((2H)/(g))=sqrt(2gh)xxsqrt((2H)/(g))` | |
| 367. |
The work done to get n smaller identical drops to form a big spherical drop of water is proportional toA. `(1)/(n^((2)/(3))-1)`B. `(1)/(n^((2)/(3))-1)`C. `n^((1)/(3))-1`D. `n^((4)/(3))-1` |
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Answer» Correct Answer - 3 `W=4piR^(2)T(n^((1)/(3))-1)` |
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| 368. |
Twenty seven identical rain drops colaese to form a big rain drop. If this rain drop travels with terminal velocity of 18 `ms^(-1)`, what is the terminal velocity of each small rain drop?A. 4 `ms^(-1)`B. 2 `ms^(-1)`C. 6 `ms^(-1)`D. 10 `ms^(-1)` |
| Answer» Correct Answer - B | |
| 369. |
8000 identical water drops combine together to form a big drop. Then the ratio of the initial surface enrgy of all the initial surface energy of all the drops together isA. `1:10`B. `1:15`C. `1:20`D. `1:25` |
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Answer» Correct Answer - C `(E_(1))/(E_(2))=(1)/(n^(1//3))` |
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| 370. |
A capillary tube of radius 0.25 mm is dipped vertically in a liquid of density `800kgm^(-3)` and of surface tension `3xx10^(-2)Nm^(-2)`. The angle of contact of liquid -glass is given by `costheta=0.3` If `g=10ms^(-2)` the rise of liquid in the capillary tube is.. CmA. 9B. 0.9C. `9xx10^(-3)`D. `0.09` |
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Answer» Correct Answer - B `T=(rhdg)/(2costheta)` |
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| 371. |
Three capillary tubes of same radius 1 cm but of length 1 m 2 m and 3 m are fitted horizontally to the bottom of a long vessel containing a liquid at constant pressure and flowing through these. Whatis the length of a single tube which can replace the three capillaries. |
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Answer» `V_(1)=(piPr^(4))/(8etal_(1)),V_(2)=(piPr^(4))/(8etal_(2)),V_(3)=(piPr^(4))/(8etal_(3))` and `V=(piPr^(4))/(8etal)` Now `V=V_(1)+V_(2)+V_(3)implies(1)/(l)=(1)/(l_(1))+(1)/(l_(2))+(1)/(l_(3))` Substituting the values we get `l=(l_(1)l_(2)l_(3))/(l_(1)l_(2)+l_(1)l_(3)+l_(2)l_(3))=(1xx2xx3)/(1xx2+1xx3+2xx3)=(6)/(11)m` |
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| 372. |
Two capillary tubes of same radius `r` but of lengths `l_(1)` and `l_(2)` are fitted in parallel to the bottom of a vessel. The pressure to the bottom of a vessel. The pressure head is P. What should be the length of a singl tube of same radius that can replace the two tubes so that the rate of flow is same as before?A. `l_(1)+l_(2)`B. `(1)/(l_(1))+(1)/(l_(2))`C. `(l_(1)l_(2))/(l_(1)+l_(2))`D. `(1)/(l_(1)+l_(2))` |
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Answer» Correct Answer - C Fluid resistance `R=(8etaL)/(pir^(4))` in parallel `(1)/(R)=(1)/(R_(1))+(1)/(R_(2))` or `(pir^(4))/(8etal_(eq))=(pir^(4))/(8etal_(1))+(pir^(4))/(8etal_(2))` or `l_(eq)=(l_(1)l_(2))/(l_(1)+l_(2))` |
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| 373. |
When two capillary tubes A and B are immersed in water , the heights of water columns are found to be in the ratio `2:3` the ratio of the radii of tubes A and B isA. `2:3`B. `4:9`C. `9:4`D. `3:2` |
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Answer» Correct Answer - D `T=(rhdg)/(2costheta),rprop(1)/(h)` |
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| 374. |
Three horizontal capillary tubes of same radii and length `L_(1),L_(2)` and `L_(3)` are fitted side by side a little above the bottom, to the wall of a tank that is filled with water. The length of a single capillary tube of same radius that can replace the three tubes such that thwe rate of flow of water through the single tube equals the combined rate of flow through the three tubes isA. `(L_(1)L_(2)L_(3))/(L_(1)+L_(2)+L_(3))`B. `(L_(1)L_(2)L_(3))/(L_(1)L_(2)+L_(2)L_(3)+L_(3)L_(1))`C. `(L_(1)+L_(2)+L_(3))/(L_(1)L_(2)L_(3))`D. `(L_(1)L_(2)+L_(2)L_(3)+L_(3)L_(1))/(L_(1)L_(2)L_(3))` |
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Answer» Correct Answer - B `Q=Q_(1)+Q_(2)+Q_(3),Qalpha(1)/(l)` |
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| 375. |
Two capillary tubes of same length but radii `r_(1) r_(2)` are arranged horizontally side by side to the bottom of a large vessel containing water. The radius of single tube of same length that can replaced them so that the rate of volume flow through it is equal to the total rate of volume flow through the two tubes isA. `r_(1)+r_(2)`B. `(r_(1)+r_(2))^((1)/(4))`C. `(r_(1)+r_(2))^(4)`D. `(r_(1)^(4)+r_(2)^(4))^((1)/(4))` |
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Answer» Correct Answer - 4 `(V)/(t)=(V_(1))/(t)+(V_(2))/(t),(piPr^(4))/(8etal)=(piPr_(1)^(4))/(8etal)+(piPr_(2)^(4))/(8etal)` |
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| 376. |
Two liquids are allowed to flow through two capillary tubes of length in the ratio `1:2` and radii in the ratio `2:3` under the same pressure difference. If the volume rates of flow of the liquids are in the ratio `8:9` the ratio fo their coefficients of viscosity isA. `1:3`B. `3:1`C. `4:9`D. `9:4` |
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Answer» Correct Answer - C `Q=(piPr^(4))/(8etal),etaalpha(r^(4))/(Ql)` |
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| 377. |
What is the pressure on a swimmer 10m below the surface of lake? `g=10ms^(-2)`, atmospheric pressure = `1.01 xx 10^(5)Pa` |
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Answer» Here `h=10m` and `rho=1000kg` `m^(-3)`. Take `g=10ms^(-2)` `P=P_(a)+rhogh` `=1.01xx10^(5)Pa+1000kgm^(-3)xx10ms^(-2)xx10m` `=2.01xx10^(5)Pa` `~~2atm` This is a 100% increase in pressure from surface level. At a depth of 1 km, the increase in pressure is 100 atmm! submarines are designed to withstand such enormous pressures. |
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| 378. |
Eight equal drops of water each of radius `r=2mm` are falling through air with a terminal velocity of `16(cm)/(s)`. The eight drops combine to form a big drop. Calculate the terminal velocity of big drop.A. `16(cm)/(s)`B. `32(cm)/(s)`C. `64(cm)/(s)`D. none of these |
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Answer» Correct Answer - C `V_("bigger")=n^((2)/(3))V_("small")` |
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| 379. |
Two mercury drops (each of radius r) merge to form a bigger drop. The surface energy of the bigger drop, if `T` is the surface tension isA. `2^(5//3)pir^(2)`B. `4pir^(2)T`C. `2pir^(2)T`D. `2^(8//3)pir^(2)` |
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Answer» Correct Answer - D Let `R` be the radius of the bigger drop, then volume of bigger drop `=2xx` volume of small drop `(4)/(3)piR^(3)=2xx(4)/(3)pir^(3),R=2^(1//3)`r Surface energy of bigger drop `E=4piR^(2)T=4xx2^(2//3)pir^(2)T=2^(8//3)pir^(2)T` |
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| 380. |
The terminal velocity of a copper ball of radius `2mm` falling through a tank of oil at `20^(@)C` is `6.5cm//s`. Find the viscosity of the oil at `20^(@)C`. Density of oil is `1.5xx10^(3)Kg//m^(3)`, density of copper is `8.9xx10^(3)Kg//m^(3)`. |
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Answer» We have `v_(t)=6.5xx10^(-2)ms^(-1),a=2xx10^(-3)m`. `g=9.8ms^(-2),rho=8.9xx10^(3)kgm^(-3)` `sigma=1.5xx10^(3)kgm^(-3)`. From Eq. `eta=(2)/(9)xx((2xx10^(-3))^(2)m^(2)xx9.8ms^(-2))/(6.5xx10^(-2)ms^(-1))xx7.4xx10^(3)kgm^(-3)` `=9.9xx10^(-1)kgm^(-1)s^(-1)` |
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| 381. |
Neglecting the density of air, the terminal velocity obtained by a raindrop of radius `0.3 mm` falling through the air of viscosity `1.8 xx10^(-5) N//m^(2)` will beA. `10.9(m)/(s)`B. `8.3(m)/(s)`C. `9.2(m)/(s)`D. `7.6(m)/(s)` |
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Answer» Correct Answer - B `v_(t)=(2r^(2)rhog)/(9eta)` |
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| 382. |
A drop of water of radius 0.0015 mm is falling in air .If the cofficient of viscosity of air is `2.0 xx 10^(-5) kg m^(-1)s^(-1)` ,the terminal velocity of the drop will be (The density of water = `10^(3) kg m^(-3)` and g = `10 m s^(-2)` )A. `1.0 xx 10^(-4) m s^(-1)`B. `2.0 xx 10^(-4) m s^(-1)`C. `2.5 xx 10^(-4) m s^(-1)`D. `5.0 xx 10^(-4) m s^(-1)` |
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Answer» Correct Answer - C Here, r=0.0015 mm `=0.0015xx10^(-3) m` `eta=2.0xx10^(-5) kg m^(-1) s^(-1)` `rho=1.0xx10^(3) kg m^(-3)` `g=10 m s^(-2)` Neglecting the density of air, the terminal velocity of the water drop is `v_(T)=(2)/(9)(r^(2)rhog)/(eta)=(2xx(0.0015xx10^(-3))^(2)xx1.0xx10^(3)xx10)/(9xx2.0xx10^(-5))` `=2.5xx10^(-4) m s^(-1)` |
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| 383. |
A small ball of density `rho` is immersed in a liquid of density `sigma(gtrho)` to a depth `h` and released. The height above the surface of water up to which the ball will jump is |
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Answer» Volume of ball `V=(m)/(rho)` Acceleration of ball moving in upward direction inside the liquid. `a=(F_("net"))/(m)=("upthrust-weight")/(m)=(V_("total")rho_(1)g-mg)/(m)` `a=(((m)/(rho))(3rho)(g)-mg)/(m)=2g` (upwards) `:.` velocity of ball while crossing the surface `v=sqrt(2ah)=sqrt(4gh)` `:.` the ball will jump to a height `H=(v^(2))/(2g)=(4gh)/(2g)=2h` |
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| 384. |
A ball is made of a material of density `rho` where `rho_(oil)ltrholtrho_(water)` with `rho_(oil)` and `rho_(water)` representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?A. B. C. D. |
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Answer» Correct Answer - B `rho_("oil")ltrholtrho_("water")` Oil is the least dense of them so it should settle at the top with water at the base. Now the ball is denser than oil but less denser than water. So, it will sink through oil but will not sink in water. So it will staty at the oil water interface. |
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| 385. |
if a ball of steel (density `rho=7.8 g//cm^(3)`) attains a terminal velocity of `10 cm//s` when falling in a tank of water (coefficient of viscosity, `eta_(water) =8.5xx10^(-4)`Ps s), then its terminal velocity in glycerine `(rho =1.2 g//cm^(2), eta =13.2 Pa s)` would be nearlyA. `1.6xx10^(-5)cms^(-1)`B. `6.25xx10^(-4)cms^(-1)`C. `6.45xx10^(-4)cms^(-1)`D. `1.5xx10^(-5)cms^(-1)` |
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Answer» Correct Answer - B `vprop(rho-rho_(0))/(eta),(v_(2))/(v_(1))=(rho-rho_(2))/(rho-rho_(1))xx(eta_(1))/(eta_(2))=6.25xx10^(-4)cms^(-1)` |
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| 386. |
Work done in increasing the size of a soap bubble from a radius of 3cm to 5cm is nearly (Surface tension of soap solution `=0.03Nm^-1`)A. `0.2pimJ`B. `2pimJ`C. `0.4pimJ`D. `4pimJ` |
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Answer» Correct Answer - C Work done `=` change in surface energy `impliesW=2Txx4pi(R_(2)^(2)-R_(1)^(2))` `=2xx0.03xx4pi[(5)^(2)-(3)^(2)]xx10^(-4)J=0.4pimJ` |
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| 387. |
The area of hole (1) is `A_(1)` and of hole (2) is `A_(2)`. The velocity of efflux and range of liquid is `v_(1), v_(2) and x_(1), x_(2)` for holes (1) and (2) respectively. The time taken by liquid, jet to reach the ground from holes (1) and (2) are `t_(1) and t_(2)` respectively. Cross section of tank is very very large compared to size of holes. `{:(,"Column-I",,"Column-II"),((A),v_(1):v_(2),(p),2 : 1),((B),x_(1) : x_(2),(q),1 : 2),((C),t_(1) : t_(2),(r),1 : 1),((D),"if"A_(1) : A_(2) = 2 : 1",then"v_(1) : v_(2) =,(s),4 : 1):}` |
| Answer» Correct Answer - A(q), B(r), C(p), D(q) | |
| 388. |
A garden pipe has an internal diameter of 4 cm. It is connected to a lawn sprinkler that consists of 24 holes at the other end. Each hole is of radius 0.06 cm. If the water in the pipe has a speed of 80 cm `s^(-1)`, find its speed through sprinkler holes. |
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Answer» Given that speed of water through pipe, `v_(1) = 80 cm s^(-1)` Area of cross-section of pipe, `A_(1) = (pi xx (4)^(2))/(4)` `= 4 pi cm^(2)` Let the speed of water through hole = `v_(2)` Area of cross- section available to water through sprinkler `A_(2) = 24 xx` (Area of one hole) = `24 xx (pi xx (0.06)^(2))` `= 24 pi xx 36 xx 10^(-4) cm^(2)` From the equation of continuity `A_(1)v_(1) = A_(2) v_(2)` `rArr 80 xx 4 pi = (24 pi xx 36 xx 10^(-4)) xx (v_(2))` `rArr (80 xx 4 pi)/(24 pi xx 36 xx 10^(-4)) = v_(2)` = `0.3703 xx 10^(4) cm s^(-1) = v_(2)` `rArr v_(2) = 37 m s^(-1)` |
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| 389. |
At What velocity does water emerge from an orifice in a tank in which gauge pressure is `3 xx 10^(5) Nm^(-2)` before the flow starts ? (Take the density of water `=1000 kg m^(-3)`.)A. `24.5 m s^(-1)`B. `14.5 m s^(-1)`C. `34.5 m s^(-1)`D. `44.5 m s^(-1)` |
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Answer» Correct Answer - A Here, `P=3xx10^(5) Nm^(-2), rho=1000 kgm^(-3), g=9.8 ms^(-2)` As `P=hrhog` `therefore h=(P)/(rhog)=(3xx10^(5))/(1000xx9.8)m` Velocity of efflux, `v=sqrt(2gh)=sqrt((2xx9.8xx3xx10^(5))/(1000xx9.8))` `=sqrt(600)=24.495 ms^(-1)~~24.5 ms^(-1)` |
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| 390. |
A tank containing water has an orifice 2 m below the surface of water. If there is no wastage of energy. Find the speed of discharge. |
| Answer» Correct Answer - 6.32 m `s^(-1)` | |
| 391. |
A drop of water breaks into two droplets of equal size. In this process which of the following statements is correct? (1). The sum of temperature of the two droplets together is equal to the original temperature of the drop. (2).the sum of masses of the two droplets is equal to the original mass of the drop. (3). the sum of the radii of the two droplets is equal to the radius of the original drop. (4). the sum of the surface areas of the two droplets is equal to the surface area of the original drop.A. 1 is correctB. 2 is correctC. 3 is correctD. 4 is correct |
| Answer» Correct Answer - B | |
| 392. |
The potential energy of molecule on the surface of a liquid as compared to in side the liquid isA. zeroB. smallerC. the sameD. greater |
| Answer» Correct Answer - D | |
| 393. |
For a surface molecule,A. the net force on it is non zeroB. the net force on it zeroC. ther is net downward forceD. there is net upward force |
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Answer» Correct Answer - C A surface molecule experiences a net downward force. |
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| 394. |
A light wire AB of length 10 cm can slide on a vertical frame as shown in figure. There is a film of soap solution trapped between the frame and the wire. Find the load W that should be suspended from the wire to keep it in equilibrium. Neglect friction. Surface tension of soat solution `=25 dyn cm^-1`. Take g=10 ms^-2` A. 0.2 gB. 0.3 gC. 0.4 gD. 0.5 g |
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Answer» Correct Answer - D `25xx10^(-3)xx2xx0.1=mxx10` or `m=(5xx10^(-3))/(10)kg=0.5g` |
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