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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.
| 451. |
A square plate of 0.1 m side moves parallel to a second plate with a velocity of `0.1 m//s`, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 poise , distance between the plates in m isA. `0.1`B. `0.05`C. `0.005`D. `0.0005` |
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Answer» Correct Answer - D `A=(0.1)^(2) = 0.01m^(2),.^ (eta=0.01 "Poise" = 0.001 "decapoise") (M.K.S. "unit")`, `dupsilon = 0.1m//s` and `F = 0.002N` `F=etaA(dv)/(dx) " "therefore dx = (etaAdv)/(F) = (0.001xx0.01xx0.01)/(0.002) = 0.0005m`. |
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| 452. |
The diagram shows a cup of tea seen from above. The tea has been stirred and is now rotating without turbulence. A graph showing the speed `v` with which the liquid is crossing points at a distance `X` from `O` along a radius `OX` would took likeA. B. C. D. |
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Answer» Correct Answer - D When we move from the centre to the circumference the velocity of liquid goes on decreasing and finally becomes zero |
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| 453. |
A capillary tube of radius R is immersed in water and water rises in it to a height H . Mass of water in the capillary tube is . M If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now beA. `M//2`B. MC. 2MD. 4M |
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Answer» Correct Answer - C `h=(2Tcostheta)/(rrhog)" or "hprop1/r` =`m=(pir^(2)h)rho" or "mpropr^(2)h" or "mpropr` |
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| 454. |
A capillary tube of radius R is immersed in water and water rises in it to a height H . Mass of water in the capillary tube is . M If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now beA. `M`B. `2M`C. `M//2`D. `4M` |
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Answer» Correct Answer - B Mass of the liquid in capillary tube `M = Vrho = (pir^(2)h)rho " "{:(therefore M prop r^(2) h prop r),(["As " h prop (1)/(r )]):}` So if radius of the tube is doubled, mass of water will becomes `2M`, which will rise in capillary tube. |
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| 455. |
Water rises to a height of 30 mm in a capillary tube. If the radius of the capillary tube is made `3//4` of its previous value. The height to which the water will rise in the tube isA. 30 mmB. 20 mmC. 40 mmD. 10 mm |
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Answer» Correct Answer - C `h_(1)r_(1)=h_(2)r_(2)rArrh_(2)=30xx4//3=40 mm` |
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| 456. |
On dipping one end of a capiilary in liquid and inclining the capillary at an angles `30^(@)` and `60^(@)` with the vertical, the lengths of liquid columns in it are found to be `l_(1)` and `l_(2)` respectively. The ratio of `l_(1)` and `l_(2)` isA. `1:sqrt(3)`B. `1:sqrt(2)`C. `sqrt(2):1`D. `sqrt(3):1` |
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Answer» Correct Answer - A `l_(1) = (h)/(cos alpha_(1))` and `l_(2) = (h)/(cos alpha_(2)) therefore (l_(1))/(l_(2)) = (cos alpha_(2))/(cos alpha_(1)) = (cos 60^(@))/(cos 30^(@)) = (1//2)/(sqrt(3)//2) = 1:sqrt(3)` |
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| 457. |
In the figure shown find the value of `h_(2)` for maximum range `R` if (a) `h_(1)=4H` (b)`h_(1)=8H` Also find the value of this maximum range in both cases. |
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Answer» Correct Answer - A::B::C (a) `h_(1)+6H=4H+6H=10H=h`(say) Maximum range will be obtained from `h_(2)=(h)/(2)=5H`. and this maximum range will be, `R_(max)=h=10H`. (b) In this case, `h=h_(1)+(6H)=(8H)+(6H)=14H` `(h)/(2)` or `7H` point lies on the table. so, maximum range will be obtain from the bottommost point of the liquid container or `h_(2)=6H` and this maximum range will be `R=2 sqrt(h_(2)(h-h_(2))` `=2 sqrt(6Hxx8H)=8 sqrt(3) H`. |
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| 458. |
`A` drop of water of volume `V` is pressed between the two glass plates so as to spread to an area. `A`. If `T` is the surface tension, the normal force required to separate the glass plates isA. `(TA^(2))/(V)`B. `(2TA^(2))/(V)`C. `(4TA^(2))/(V)`D. `(TA^(2))/(2V)` |
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Answer» Correct Answer - B Force required to separate the glass plates `F=(2AT)/(t) xx (A)/(A) = (2TA^(2))/((Axxt)) = (2TA^(2))/(V)` |
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| 459. |
A drop of water of volume `0.05 cm^(3)` is pressed between two glass plates, as a consequence of which, it spreads and occupies an are of `40 cm^(2)`. If the surface tension of water is `70 "dyne"//cm`, find the normal force required to separate out the two glass plates is newton. |
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Answer» Correct Answer - D We have discussed above, `F=(2AT)/(d) = (2A^(2)T)/(Ad)` But, `Ad =` volume `:. F = (2A^(2)T)/(V)` Substituting the values we get, `F =(2xx(40xx10^(-4))^(2)xx(70xx10^(-3)))/(0.05 xx 10^(-6))` `= 45 N` |
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| 460. |
for the arrangement shown in figure, initially the balance `A`and `B` reads `F_(1)` and `F_(2)` respectively and `F_(1)gtF_(2)`. Finally when the block is immersed in the liquid then the readings of balance `A` and `B` are `f_(1)` and `f_(2)` respectively. Indentify the statement which is not always (where, F is some force ) correct statement. .A. `f_1)gtf_(2)`B. `F_(1)+FgtF_(2)+F`C. `f_(1)+f_(2)=F_(1)+F_(2)`D. None of these |
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Answer» Correct Answer - A `F_(1)` will decrease and `F_(2)` will increase. So, `f_(1)` may or may not be grater than `f_(2)`. total weight to system in both conditions will remain same. Hence, `f_(1)+f_(2)=F_(1)+F_(2)`. |
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| 461. |
Three identical blocks each of mass `m = 1 kg` and volume `3 xx 10^(-4) m^(3)` are suspended by massless strings from a support as shown. Underneath are three identical containers containing the same amount of water that are placed over the scales. In Fig. (a), the block is completely out of the water, in Fig. (b), the block is completely submerged but not touching the beaker and in Fig. (c), the block rests on the bottom of the beaker. The scale in Fig (a) reads `14 N`. A. The tension in the string in (b) is `10 N`.B. The tension in the string in (b) is `7 N`.C. The reading of the scale in (b) is `17 N`D. The reading of the scale in (c) is `24 N`. |
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Answer» Correct Answer - B::C::D Density of the block `=1/(3xx10^(-4)) kg//m^(3)=10/3xx10^(3)kg//m^(3)` Buoyant force `B=3xx10-4xx103xxg=3N` `:. TB=10=3=7N` Reading of `B=14+3=17N` Reading of `C=[14+(10-3)+3]N=24N` |
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| 462. |
A cylindrical vessel of `90 cm` height is kept filled up to the brim. It has four holes `1,2, 3` and `4` which are, respectively, at heights of `20 cm, 30 cm, 40 cm` and `50 cm` from the horizontal floor `PQ`. The water falling at the maximum horizontal distance from the vessel comes from A. hole number `4`B. hole number `3`C. hole number `2`D. hole number `1` |
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Answer» Correct Answer - A::B The maximum horizontal distance from the vessel comes from hole numbers `3` and `4`. Now `v=sqrt(2gh)` where `h` is the height of the hole from the top. Horizontal distance `x=vt=sqrt(2gh)sqrt((2(H-h))/g)=2sqrt(h(H-h))` |
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| 463. |
A cylindrical vessel of `90 cm` height is kept filled up to the brim. It has four holes `1,2, 3` and `4` which are, respectively, at heights of `20 cm, 30 cm, 40 cm` and `50 cm` from the horizontal floor `PQ`. The water falling at the maximum horizontal distance from the vessel comes from A. Hole number 4B. Hole number 3C. Hole number 2D. Hole number 1 |
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Answer» Correct Answer - B |
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| 464. |
A cylindrical vessel of `90 cm` height is kept filled up to the brim. It has four holes `1,2, 3` and `4` which are, respectively, at heights of `20 cm, 30 cm, 40 cm` and `50 cm` from the horizontal floor `PQ`. The water falling at the maximum horizontal distance from the vessel comes from A. hole number `4`B. hole number `3`C. hole number`2`D. hole number `1` |
| Answer» Correct Answer - B | |
| 465. |
A cylinderical vessel filled with water is released on a fixed inclined surface of angle `theta` as shown in figure. The friction coefficient of surface with vessel in `mu(lt tan theta)`. Then the constant with the incline will be (Neglect the viscosity of liquid) A. `tan^(-1) mu`B. `theta-tan^(-1) mu`C. `theta+tan^(-1)mu`D. `cot^(-1)mu` |
| Answer» Correct Answer - D | |
| 466. |
A copper piece of mass 10 g is suspended by a vertical spring. The sprig elongates 1 cm over its natural length to keep the piece in equilibrium. A beaker containing water is now placed below the piece so as to immerse the piece completely in water. Find the elongation of the spring. Density of copper =9000 kgm^-3. `Take g=10ms^-2`A. 0.45 cmB. 0.89 cmC. 1.02 cmD. 1.86 cm |
| Answer» Correct Answer - B | |
| 467. |
A metal piece of mass `10 g` is suspended by a vertical spring. The spring elongates `10 cm` over its natural length to keep the piece in equilibrium. A beaker containing water is now placed below the piece so as to immerse the piece completely in water. Find the elongation of the spring. Density of metal `= 9000 kg//m^(3)`. Take `g = 10 m//s^(2)`. |
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Answer» Let the spring constant be `k`. When the piece is hanging in air, the equilibrium condition gives `k(10 cm) = (0.01 kg) (10 m//s^(2))` or `k (10 cm) = 0.1 N`. ………..(i) The volume of the metal piece `= (0.01kg)/(9000kg//m^(3)) = (1)/(9) xx 10^(-5)m^(3)`. This is also the volume of water displaced when the piece is immersed in water. The force of buoyancy `=` weight of the lilquid displace `=(1)/(9) xx 10^(-5)m^(3) xx (1000 kg//m^(3)) xx (10 m//s^(2))` `= 0.011 N`. If the elongation of the spring is `x` when the piece is immersed in water, the equilibrium condition of the piece gives, `kx = 0.1 N - 0.011 N = 0.089 N`. ..........(ii) By `(i)` and `(ii), x = (0.089)/(10) cm = 0.0089 cm`. |
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| 468. |
A tank is filled up to a height `2H` with a liquid and is placed on a platform of height `H` from the ground, the distance x from the ground where a small hole is punched to get the maximum range R is .A. `H`B. `1.25H`C. `1.5H`D. `2H` |
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Answer» Correct Answer - C `h_(Top)=(2H+H)/(2)=1.5H`. Since, this point lies in the tank. So hole should be made at this point. |
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| 469. |
A tank is filled up to a height `2H` with a liquid and is placed on a platform of height `H` from the ground, the distance x from the ground where a small hole is punched to get the maximum range R is .A. `H`B. `1.25H`C. `1.5H`D. `2H` |
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Answer» Correct Answer - C As `x=1/2"gt"^(2)implies t=sqrt((2x)/r)` velocity of efflux `v=sqrt(2g(H-x))` Hence `R=vt=2sqrt(x(3H-x))` For range to be maximum `(dR)/(dx)=0` which gives `x=3/2H` Alternatively : Maximum range is also equal to average of two heights `x=(H+2H)/2=(3H)/2` |
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| 470. |
A bent tube is lowered into a water stream as shown in figure. The velocity of the stream is `V_(1)`. The closed upper end is located at a height `h_(0) ` and has a small orifice. To what height `h` will the liquid get spurt? brgt |
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Answer» `1/2rhov_(1)^(2)+P_(0)+0=1/2rhov_(2)^(2)+P_(0)+rhogh_(0)impliesV_(2)^(2)=V_(1)^(2)=2gh_(0)` If the liquid reaches a height `h` then `V_(2)^(2)=2gh`, `implies2gh, V_(1)^(2)-2gh_(0)implies h=(V_(1)^(2))/(2g)=h_(0)` |
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| 471. |
Water is flowing through a horizontal pipe of non-uniform cross- section. At the extreme narrow portion of the pipe, the water will haveA. Maximum speed and least pressureB. Maximum pressure and least speedC. Both pressure and speed maximumD. Both pressure special least |
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Answer» Correct Answer - A |
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| 472. |
A pipe of non uniform cross section has two distinct section ` 1` and `2 ` with areas `2cm^(2)` and `4cm^(2)` respectively. If the velocity of flowing liquid at section `1` is `5cm//s`, determine the velocity at section `2`. |
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Answer» From the equation of continuity `a_(1)v_(1)=a_(2)v_(2)` Given `a_(1)=2cm^(2),a_(2)=4cm^(2),v_(1)=5cm//s, v_(2)=?` `:. 2(5)=4(v_(2))impliesv_(2)=2.5cm//s` |
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| 473. |
Two soap bubbles in vacuum of radius `3cm` and` 4cm` coalesce to form a single bubble, in same temperature. What will be the radius of the new bubbleA. `7cm`B. `(12)/(7) cm`C. `12cm`D. `5cm` |
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Answer» Correct Answer - D `n=n_(1)+n_(2)` ` (pV)/(RT)=(p_(1)V_(1))/(RT)+(p_(2)V_(2))/(RT)` or, `pV=p_(1)v_(1)+p_(2)v_(2)` `:. ((4t)/(r ))(4/3 pir^(3))=((4t)/(r_(1) ))(4/3 pir_(1)^(3))` `+((4t)/(r_(2)))(4/3 pir_(2)^(3))` `:. r=sqrt(r_(1)^(2)+r_(2)^2)` `=5cm`. |
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| 474. |
A liquid is flowing through a pipe of non-uniform cross-section. At a point where area of cross-section of the pipe is less, match the following columns. |
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Answer» Correct Answer - A::B::C From, `A_(1)v_(1)=A_(2)v_(2) " or " v prop (1)/(4)` At a point where area of cross-section is less, volume of liquid flowing per second is same but speed is more. Therefore, here kinetic energy is more or the pressure will be less. |
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| 475. |
There are two point A and B inside a liquid as shown in figure. Now the vessel starts moving upwards with an acceleration a. Match the following. `|{:(,"Table-1",,"Table-2"),((A),"Pressure at B will",(P),"increase"),((B),"Pressure difference",(Q),"decrease"),(,"between A and B will",,),((C),"Upthrust on an object",(R),"remain same"),(,"inside the verssel will",,):}|` |
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Answer» Correct Answer - A::B::C At depth h p `p=p_(0)+p(g+a)h rArr Deltap=p(g+a)Deltah` Upthrust will also increase because upthrust `F=V_(i)(g+a)rho` Here, `V_(i)` is the immersed volume. |
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| 476. |
A body floats in a liquid contained in a vessel. The vessel falls vertically with an acceleration ( a (`lt` g). If `V_i` and `V_f` be the initial and final volume of the body immersed in the liquid then :A. `V_i gt V_f`B. `V_i lt V_f`C. `V_i = V_f`D. Data Insufficient |
| Answer» Correct Answer - D | |
| 477. |
A circular cylinder of height `h_(0)=10 cm ` and radius `r_(0)`=2 cm is opened at the top and filled with liquid. It is rotated about its vertical axis. Determine the speed of rotation so that half the area of the bottom gets exposed (`g=10 m//s^(2)`):A. 25 rad/sB. 50 rad/sC. 100 rad/sD. 200 rad/s |
| Answer» Correct Answer - D | |
| 478. |
A cylindrical wooden float whose base area S and the height H drift on the water surface. Density of wood d and density of water is `rho`. What minimum work must be performed to take the float out of the water ?A. `(S^(2)gd)/(2rho)`B. `(Sgd^(2)H^(2))/(rho)`C. `(Sgd^(2)H^(2))/(2rho)`D. `(2S^(3)gd^(2))/(rhoH^(2))` |
| Answer» Correct Answer - C | |
| 479. |
Statement-1: A block is immersed in a liquid inside a beaker which is falling freely buoyant force acting on block is zero. Statement-2: In case of freely falling liquid there is no pressure difference between any two points. A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A According to equation of continuity av = constant. |
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| 480. |
If a liquid does not wet glass, its angle of contact isA. zeroB. acuteC. right angleD. |
| Answer» (c ) For the liquids wich do not wet the glass the liquid meniscs is convex upward ,so angle of contact is obtuse. | |
| 481. |
Two different liquids are flowing in two tubes of equal radius. The ratio of coefficients of viscosity of liquids is 52:49 and the ratio of their densities is `13:1`, then the ratio of their critical velocities will beA. `4 : 49`B. `49 : 4`C. `2 : 7`D. `7 : 2` |
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Answer» Correct Answer - A |
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| 482. |
Viscosity of liquidsA. increases with increase in temperatureB. is independent of temperatureC. decrease with decreases in temperatureD. decreases with increase in temperature. |
| Answer» Correct Answer - D | |
| 483. |
At critical temperature, the surface tension of a liquidA. is zeroB. is infinityC. is same as that at any other temperatureD. cannot be determined |
| Answer» Correct Answer - A | |
| 484. |
When the temperature increased the angle of contact of a liquidA. increases angle of contactB. decreasesC. remains the sameD. first increases and then decreases |
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Answer» Correct Answer - B Cohesive force decreases. So angle of contact decreases. |
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| 485. |
Air is blown through a hole on a closed pipe containing liquid. Then the pressure willA. Increase on sidesB. Increase downwardsC. Increase in all directionsD. Never increases |
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Answer» Correct Answer - C |
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| 486. |
The lower end of a capillary tube touches a liquid whose angle of contact is `110^(@)`, the liquid .A. rises into the tubeB. falls in the tubeC. may rise or fall insideD. neither rises nor falls inside the tube |
| Answer» Correct Answer - B | |
| 487. |
A liquid will not wet the surface of a solid if the angle of contact isA. `0^(@)`B. `45^(@)`C. `60^(@)`D. `gt90^(@)` |
| Answer» Correct Answer - D | |
| 488. |
When the temperature increased the angle of contact of a liquidA. increasesB. decreasesC. remains constantD. first increases and then decreases |
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Answer» Correct Answer - B With the increase in temperature, the surface tension of liquid decrease and angle of contact also decreases. |
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| 489. |
Radius of a soap bubble is increased from R to 2 R work done in this process in terms of surface tension isA. `24piR^(2)S`B. `48piR^(2)S`C. `12piR^(2)S`D. `36piR^(2)S` |
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Answer» Correct Answer - A `W = 8piT (R_(2)^(2)-R_(1)^(2)) = 8piS[(2R)^(2)-(R )^(2)] = 24 pi R^(2)S` |
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| 490. |
The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (T=surface tension of the solution)A. `2 pi (D^(2)-d^(2))S`B. `pi(D^(2)-d^(2))S`C. `4pi (D^(2)-d^(2))S`D. `8pi(D^(2)-d^(2))S` |
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Answer» Correct Answer - A Change in surface area `=2 xx 4pi(D/2)^(2) -(d/2)^(2)=2pi(D^(2)-d^(2))` `therefore` Work done = surface x change in area `=2piS(D^(2)-d^(2))` |
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| 491. |
The work done in blowing a soap bubble of 10 cm radius is (Surface tension of the soap solution is `3/100` N/m)A. `75.36xx10^(-4)J`B. `37.68xx10^(-4)J`C. `150.72xx10^(-4)J`D. `75.3^(6J)` |
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Answer» Correct Answer - A `W = 8piR^(2)T = 8pi(10xx10^(-2))^(2) (3)/(100) = 75.36 xx 10^(-4)J` |
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| 492. |
The work done in blowing a bubble of volume `V` is `W`, then what is the work done in blowing a soap bubble of volume `2V` ?A. `W//2`B. `sqrt(2)W`C. `.^(3)sqrt(2)W`D. `.^(3)sqrt(4)W` |
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Answer» Correct Answer - D As volume of the bubble `V=(4)/(3) pi R^(3) implies R = ((3)/(4pi))^(1//3)V^(1//3) implies R^(2) = ((3)/(4pi))^(2//3) V^(2//3) implies R^(2) prop V^(2//3)` Work done in blowing a soap bubble `W = 8piR^(2)T implies W prop R^(2) prop V^(2//3)` `therefore (W_(2))/(W_(1)) = ((V_(2))/(V_(1)))^(2//3) = ((2V)/(V))^(2//3) = (2)^(2//3) = (4)^(1//3) implies W_(2) = .^(3)sqrt(4)W` |
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| 493. |
Calculate the work done by a boy In making a soap bubble of diameter `1.4 cm ` by blowing .if he surface tension of soap solution is `0.03 N//m`A. `3xx10^(-5) J`B. `3.696xx10^(-5) J`C. `2xx10^(-5) J`D. `4.2xx10^(-5) J` |
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Answer» Intial area of soap bubble =0 Soap bubble has two surfaces internal and external. Increase in surface area `=DeltaA=2xx4pir^(2)` `=2xx4xx(22)/(7)xx(0.7xx10^9-2)^(2)` `=12.32xx10^(-4)m^(2)` Work done `=W-SDeltaA` `=0.03xx12.32xx10^(-4)` `=3.696xx10^(-5) J` |
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| 494. |
What will happen if n drops of a liquid each has surface energy E, combine to form a single drop.A. No energy will be released in the precessB. Some energy will be absorbed in the processC. Energy released or absorbed will be `E(n-n^(2//3))`D. Energy released or absorbed will be `nE(n-n^(2//3-1))` |
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Answer» Let radius of smalll drop =r Radius of single drop =R Surface tension S= surface energy per unit area `nxx(4)/(3)xxpir^(3)=(4)/(3)piR^(2)` `R=n^(1//3) r` 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` Energy released `=E_(i)-E_(f)=E(n-n^(2//3))` |
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| 495. |
A drop of radius r is broken into n equal drips. Calulate the work done if surface tension of water is T.A. `4rpiR^(2)nT`B. `4rpiR^(2)T(n^(2//3-1))`C. `4rpiR^(2)T(n^(1//3-1))`D. None of the above |
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Answer» The volume of n smaller drop =volume of bigger drop `:. " " n(4)/(3)pir^(3)=(4)/(3)piR^(2)` `:. " " R=n^(1//3)r` `:. " " r=(R)/(n^(1//3))` Here, R=radius of bigger drop r=radius of smallar drop `:. W=4pi(nr^(2)-R^(2))T` `4pi[n((R)/(n^(1//2)))^(2)-R^(2)]T=4pi{(nR^(2))/(n^(2//3))-R^(2)}T` `=4piR^(2)(n^(1-2//3)-1)T=4piR^(2)T(n^(1//3)-L)` |
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| 496. |
As the temperature of water increases, its viscosityA. Remains unchangedB. DecreasesC. IncreasesD. Increases or decreases depending on the external pressure |
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Answer» Correct Answer - B |
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| 497. |
The coefficient of viscosity for hot air isA. Greater than the coefficient of viscosity for cold airB. Smaller than the coefficient of viscosity for cold airC. Same as the coefficient of viscosity for cold airD. Increases or decreases depending on the external pressure |
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Answer» Correct Answer - A |
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| 498. |
A block of iron is kept at the bottom of a bucket full of water at `2^@C`. The water exerts bouyant force on the block. If the temperature of water is increased by `1^@C` the temperature of iron block also increases by `1^@C`. The bouyant force on the block by waterA. will increaseB. will decreaseC. will not changeD. may decrease or increase depending on the values of their coefficient of expansion |
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Answer» Correct Answer - A Increasing the temperature of water from `2^(0)C` to `3^(0)C` increases its density while decreases the density of iron. Hence the bouyant force increases. |
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| 499. |
Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of `6cms^(-1)`. If they coalesce to form one big drop, what will be its terminal speed? Neglect the buoyancy due to airA. `1.5 cms^(-1)`B. `6 cms^(-1)`C. `24 cms^(-1)`D. `32 cms^(-1)` |
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Answer» Correct Answer - C Equanting the volume, we have `8((4)/(3) pir^(3))=(4)/(3) piR^(3)` or, `R=2r` `v_(T )prop (radius^(2))` Radius has become two times. Therefore, terminal velocity will become 4 times. |
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| 500. |
A certain number of spherical drops of a liquid of radius `r` coalesce to form a single drop of radius `R` and volume `V`. If `T` is the surface tension of the liquid, thenA. energy=4VT `((1)/(r )-(1)/(R ))` is releasedB. energy = 3 VT `((1)/(r )+(1)/( R))` is absorbedC. energy `=3V T ((1)/(R )-(1)/(r ))` is releasedD. energy is neither released nor absorbed |
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Answer» Correct Answer - C Here, if the surface area changes, it will change the surface energy as well. If the surface area decreases, it means that energyu is released and vice versa. Change in surface energy `Deltaa xx T…..(i)` Let we have n number of drops initially `So, DeltaA=4piR^(2)-n(4pir^(2))...(ii)` Volume is constant `So, n(4)/(3)pir^(3)=(4)/(3)piR^(3)=V...(iii)` From (ii) and (iii) we get `DeltaA=(3)/(R)(4pi)/(3)xxR^(3)-(3)/(r)(n(4pi)/(3r^(3)))=(3)/(R)xxV-(3)/(r)V` `Rightarrow DeltaA=3V ((1)/(R)-(1)/(r))="negative value"` As, `R gt r, "so" DeltaA` is negative. It means that surface areais decreased, so energy must be released. `"Energy released"=DeltaAxxT=-3VT((1)/(r)-(1)/R)` |
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