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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.
| 251. |
When water flows through a tube of radius r placed horizontally, a pressure difference p develops across the ends of the tube. If the radius fo the tube is doubled and the rate fo flow halved, the pressure differnece will beA. 80 pB. pC. `p//8`D. `p//32` |
| Answer» Correct Answer - D | |
| 252. |
A small drop of water of surface tension S is squeezed between two clean glass plates so that a thin layar of thickness d and area A is formed between them. If the angle of contact is zero, what is the force required to pull the plates apart? |
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Answer» An extremely thin layer of a liquid can be consirdered as the collection of large number of hemispherical drops. In the case of a spherical drop, the excess of pressure = ` (2S)/(r )`. But in the case of thin layer of liquid , which is a combination of hemispherical drops, the case of thin layar of liquid, which is a combination of hemispherical drops, the excess pressure is `p=S/e`, where `r= d//2`. Therefore, ` p=(S)/(d//2) = (2S)/(d)` force dure to surface tension pushing the two plates together is `F= p xx A = (2SA)/(d)`. |
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| 253. |
It becomes easier to spray the water in which some soap is dissolved. Why? |
| Answer» When some soap is dissolved in water the surface tension of water decreases. Now less energy is needed for spraying the water. That is why, it becomes easier to spray the water in which some soap is dissolved. | |
| 254. |
It is better to wash the clothes in hot soap solution. Why? |
| Answer» The soap solution has less surface tension as compared to ordinary water and its surface tension decreses further on heating. The hot soap solution can, therefore, spread over large surface area and as such it has more wetting power. It is on ac count of this property that hot soap solution can penetrate the pores in the threads of cloth where dust may be present and can clean the cloths better than the ordinary water. | |
| 255. |
The neck and bottom of a bottle are 2 cm and 20 cm in diameter respectively if the cork is pressed with a force of 1.2 kgf in the neck of the bottle, calculate the force exerted on the bottom of the bottle. |
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Answer» Correct Answer - 120 kgf `2 r_1 = 2cm or r_1 =1 cm` and `2 r_2 = 20 cm or r_2 =10 cm , ` `A_1= pi r_1^2 and A_2 = pi r_2^2 ,` `F_1 = 1.2 kg f = 1.2 xx 9.8 N ,` `F_2 = (F_1)/(A_1)xxA_2 = (1.2xx9.8)/(pi r_1^2) xx pi r_2^2 = 1.2 xx 9.8 xx r_2^2 // r_1^2` ` = 1.2 xx 9.8 xx (10)^2//(1)^2 = 1.2 xx 9.8 xx 100 N` = 120 kg.f` |
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| 256. |
The neck and the bottom of a bottal of a bottle filled with incompressible liquid are 3 cm and 8 cm in diameter respectively. If the cork is pressud with a force 1.4 kg f in the neck of the bottle, find the force exerted on the bottom of the liquid. |
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Answer» Correct Answer - 9.95 kg f Here, `2R_1 = 3 cm or R_1 = 1.5 cm = 1.5 xx 10^(-2) m` `R_2 = 8//2 = 4 cm = 4xx10^(-2)m,` `F_1 = 1.4 kg f = 1.4 xx 9.8 N` `F_2 = (F_1)/(pi R_1^2) xx pi R_2^2` `F_2 = F_1 (R_2^2)/(R_1^2)= (1.4 xx9.8)xx(4xx10^(-2))/(1.5 xx 10^(-2))^2` `= 9.95 xx 9.8 N = 9.95 kg f` |
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| 257. |
Assertion: The shape of an automobile is so designed that its front resembles the stream line pattern of the fluid through which it moves. Reason: The resistance offered by the fluid is maximum.A. both Assertion and Reason are true and the Reason in the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false. |
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Answer» Correct Answer - C The shape of the automobile is made stream lined in order to reduce air resistance. Thus, assertion is true and Reason in false. |
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| 258. |
Can two stramlines cross each other in a flowing liquid ? Explain. |
| Answer» No. If they do so it would mean that at the point of intersection of two tangents can be drawn, which will give us two directions of liquid flow at that point which is imposible because then the flow can never be streamlined. | |
| 259. |
What happens when the velocity of the liquid flowing through a horizontal tube is gradually increased? |
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Answer» When a liquid flowing through a horizontal tube with a velocity less than its critical, the flow of liquid is laminalr flow. When the velocity , the liquid flow becomes turbulent flow. |
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| 260. |
A cubical vessel of height `1` m is full of water. What is the workdone in pumping water out of the vessel? |
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Answer» Correct Answer - [4900 j] Let l be the length of each side of cube `rho` be the density of water. Mass of water in cubical vesssel, `m=l^(3) rho`. Mean heigh of water from ground, `h=l//2`. Workdone in pumping out the water is `W=mgh=l^(3)rhoxxgxxl//2=(1)/(2)l^(4)g rho` `=(1)/(2)xx1^(4)xx(9.8)xx(10^(3))=4900 j` |
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| 261. |
Assertion : Water is flowing through a horizontal tube, the static pressure and total pressure at any point are `1.20xx10^(5)` P and `1.28xx10^(5)` P . If the density of water is `1000 kg//m^(3)` , the velocity of liquid flow is 4 m/s. Reason : Work done on the liquid by difference in pressure is equal to gain in K.E. of liquidA. both Assertion and Reason are true and the Reason in the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false. |
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Answer» Correct Answer - A Difference of pressure, `p = (1.28xx10^(5) - 1.20xx10^(5)) = 0.08xx10^(5)` Pa If a is the area of cross-section of the tube and v is the velocity of liquid flowing through it, then volume liquid flowing per second through the tube = a v. Mass of liquid flowing per second through the tube = a v `rho` Work done on the liquid by excess of pressure `=Pxx a v` Gain in K.E. `=(1)/(2) (a v rho)v^(2)` Hence, `p a v =(1)/(2) a v rhoxx v^(2)` or `v= sqrt((2p)/(rho)) = sqrt((2xx0.08xx10^(5))/(1000)) =4m//s` Hence, both Assertion and Reason are correct and Reason is the correct explanation of Assertion. |
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| 262. |
The workdone in increasing the size of a soap film from `10 cm xx 6 cm` to `10 cm xx 11 cm` is `3 xx 10^(-4)J.` The surface tension of the film isA. `1.5 xx 10^(-2) N m^(-1)`B. `3.0 xx 10^(-2) N m^(-1)`C. `6.0 xx 10^(-2) N m^(-1)`D. `11.0xx10^(-2) Nm^(-1)` |
| Answer» Correct Answer - B | |
| 263. |
Tow soap bubbles A and B are formed at the two open ends of a tube. The bubble A is smaller then bubble B. If the valve on the tube connecting the two bubbles is opened and air can flow freely between the bubbles, thenA. there is no change in the size of the bubblesB. the two bubbles will becomes of equal sizeC. A will becomes smaller and B will become largerD. B will becomes smaller and A will becomes larger |
| Answer» Correct Answer - C | |
| 264. |
Statement -1 : A large soap bubble expands while a small bubble shrink, when they are connecteed to each other by a capillary tube. Statement -2 : The excess pressure (due to surface tension) inside a spherical bubble increasesee, as its volume decreases.A. Statement -1 is true , Statement -2 is true , Statement -2 is a correct explanation of Statement -1.B. Statement -1 is true , Statement -2 is true , Statement -2 is not a correct explanation of 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 Here, both statement - 1 and statement - 2 are correct and statement - 2 is the correct explanation of statement - 1. |
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| 265. |
In winter, if we touch steel chair , we feel cold but so when touched wooden chair , though both are at the same temperature. Why? |
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Answer» We know that steel is nuch better conductor of heat than wood. When we touch the steel chair, heat flows out from hand to the steel chair which in turn is conducted to other part of chair. So, more heat flows from our body to steel chair. Due to it, we feel cold. When we touch the wooden chair, the heat flows from our body to the touched part of wooden chair but heat is not conducted to other part of chair. Hence , less amount of heat flows from our body to wooden chair. that is why, wooden chair does not appear cold. |
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| 266. |
A `16cm^(3)` volume of water flows per second through a capillary tube of radius r cm and length l cm , when connected to a pressure head of h cm of water. If a tube of same length and radius `r//2` is connected to the same pressure head, find the mass of water flowing per minutes through the tube. |
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Answer» Here, `V_(1) = 16 cm^(3)//sec , p_(1) = h rho g , r,r_(1) =r, l_(1)=l` `V_(2)=?, l_(2) =l , r_(2)=r//2 , p_(2) = h rho g , so, p_(1)=p_(2)` Now , `V_(1) = (pi p_(1)r_(1)^(4))/(8 pi l_(1)) and V_(2) = (pi p_(2)r_(2)^(4))/(8 pi l_(2))` `:. (V_2)/(V_1) = (p_(2))/(P-1) xx (r_(2)^(4))/(r_(1)^(4)) xx (l_1)/(l_2) = ((r//2)^(4))/(r^4) xx l/l = (1/2)^(4) = 1/16` `V_(2) = (V_1)/(16) = 16/16 = 1 cm^(3)//s`, Volume of water flowing per minute = ` 1xx 60 = 60cm^(3)//mi n` `:.` Mass of water flowing per minute = `60 xx 1 = 60 gram`. |
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| 267. |
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 l and h increasesD. both l and h decrease |
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Answer» Correct Answer - D When the coin falls, the block moves upwards, hence l decreases , h will also decrease because when the coin is the water it will displace equal volume of water of water, whereas when it is on the block an equal weight of water is displaced. |
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| 268. |
A wooden cylinder floats in a vessel containing water with it axis horizontal. How will level of water in the vessel change if the cylinder floats in a vessel with its axis vertical? |
| Answer» The level of water will remain unchanged in both the cases because in each case the floating cylinder will displace equal volume of water. | |
| 269. |
Find the pressure at the tip of a drawing pin on area `0.2 mm` square if it is pushed against a board with a force of 5 kg wt. (use `g=10m//s^(2))` |
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Answer» Here, `A = 0.2 mm sq` `=0.2 xx 0.2 sq mm = 0.04 xx 10^(-6)m^(2)` `F=5 kg wt = 5 xx 10 N` Pressure ,`P=F/A = (5 xx 10)/(0.04 xx 10^(-6))=1.25 xx 10^(9) Pa`. |
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| 270. |
A glass plante of negligible mass and thickness is held against the end of a tube and pushed 10 cm under the surface of water. When released, the plate does not fall off. What depth of kerosene oil of relative density 0.8 must be poured into the tube so that the plate just falls off? |
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Answer» Let `A cm^(2)` be the area of the base of the plate and h cm be height of kerosene column at which the plate just falls off. Then downward thrust due to h cm of kerosene = upward thrust due to 10 cm of water column. i.e., ` h xx 0.8 xx g xx A = 10 xx 1 xx g xx A` or ` h=10/0.8 = 12.5 cm`. |
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| 271. |
Two pieces of cork, one small and another big are pushed below the surface of water. Which has greater tendency to rise swifty? |
| Answer» The upthrust on a piece of cork is equal to the weight of water displaced by a cork. As bigger piece of cork will displace more weight of water than small piece of cork, hence upthrust is greater on the bigger piece of cork. As a result of it, the bigger piece of cork has greater tendency to rise swiftly. | |
| 272. |
Do water and ice have the same specific heats? |
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Answer» No, their specific heats are different. For water, `s = 1 cal.g^(-1) .^(@)C^(-1)` and for ice, `s = 0.5 cal.g^(-1) .^(@)C^(-1)`. |
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| 273. |
What is the most likely value for `C_(T)` (molar heat capacity at constant temperature) ? |
| Answer» `C_(T) = (Delta Q)/(n Delta T) ` , when T is constant , `Delta T = 0` so `C_(T) = oo` (i.e., infinity). | |
| 274. |
The perciprocal of viscosity is called…….. |
| Answer» Correct Answer - fluidity | |
| 275. |
At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) what is the auge pressure? (c ) Find the force acting on the windown of area `20 cm xx 20cm` of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. The density of sea water is `1.03 xx 10^(3) kg m^(-3)`, `g=10ms^(-2)`. Atmospheric pressure = `1.01 xx 10^(5) Pa`. |
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Answer» Here `P_(a) = 1.01 xx 10^(5)Pa`, `rho = 1.03 xx 10^(4)=3 kg m^(-3), h=1000m` `A= 0.20 xx 0.20 =0.04 m^(2)` (a) Absolute pressure `P=P_(a) + rho g h` `= 1.01 xx 10^(5) + 1.03 xx 10^(3) xx 10 xx 1000` `=104.01 xx 10^(5) Pa` (b) Gauge pressure, `P=P-P_(a) = rho g h` `=(1.03 xx 10^(3)) xx 10 xx 1000 = 103 xx 10^(5)Pa` (c ) The pressure outside the submarine is `P=P_(a)+rho gh` and the pressure inside the submarine is `P_(a)`. Pressure difference on the window of the submarine =`P-P+(a) = rho gh` Force on the window = pressure difference `xx` area of window `= rho gh xx A = 1.3 xx 10^(5) xx (0.04)` `=4.12 xx 10^(5) N`. |
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| 276. |
At a depth of 500m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? Density of sea water is `1.03 xx 10^(3) kg//m^(3), g=10ms^(-2)`. Atmospheric pressure = ` 1.01 xx 10^(5)Pa`. |
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Answer» Here,`h=500m, rho = 1.03 xx 10^(3) kgm^(-3)` (a) Absolute pressure, `P=P_(u)+h rho g = 1.01 xx 10^(5)+500xx(1/03xx10^(3))xx10` or, `P=1.01 xx 10^(5) + 51.5 xx 10^(5)=52.51 xx 10^(5)-(52.51 xx 10^(5))/(1.01 xx 10^(5)) = 52 atm` (b) Gauge pressure `P_(G)=P-P_(a)=(p_(a)+h rho g)-P_(a)` `=h rho g = 500 xx (1.03 xx 10^(3)) xx 10 = 51.5 xx 10^(5) Pa= (51.5 xx 10^(5))/(1.01 xx 10^(5))~~51 atm`. |
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| 277. |
Water is flowing continuously from a tap having an internal diameter `8xx10^(-3)`m. The water velocity as it leves the tap is `0.4 ms^(-1)`. The diameter of the water stream at a distance `2xx10^(-1)`m below the tap is close to `(g=10m//s^(2))`A. `7.5xx10^(-3)m`B. `9.6xx10^(-3)m`C. `3.6xx10^(-3)m`D. `5.0xx10^(-3)m` |
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Answer» Correct Answer - C Velocity of stream at 0.2 m below tap `v_(2)^(2) =v_(1)^(2) +2as =(0.4)^(2) +2xx10xx0.2=4.16` or `v_(2) =sqrt(4.16) =2.04m//s =2 m//s ~~ 2m//s` Ac cording to equation of continuity `A_(1)v_(1) =A_(2)v_(2)` or `pi r_(1)^(2)v_(1) =pi r_(2)^(2)v_(2)` or `r_(2)= sqrt((r_(1)^(2)v_(1))/(v_(2))) =sqrt(((4xx10^(-3))^(2)xx0.4)/(2))` `~~1.8xx10^(-3)m` Diameter of the water stream `=2 r_(2) =2xx1.8xx10^(-3)m =3.6xx10^(-3)m` |
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| 278. |
A square plate of 20 cm side moves over another plate with velocity `5 cm s^(-1)`, both plates immersed in water. If the viscous force is `0.205 gf` and viscosity of water is 0.01 poise, what is the separation between the plates? |
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Answer» Area of plate, `A= 20 xx 20 ` `= 400 cm^(2)` `d upsilon = 5 cm s^(-1), F=0.205 gf` `= 0.205 xx 980 dyn e`, `eta = 0.01 poise , dx = ?` Now `F= eta A (d upsilon)/(dx) or dx = (eta A d upsilon)/(F)` `:. Dx = (0.01 xx 400 xx 5)/(0.205 xx 980) = 0.1 cm`. |
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| 279. |
From what heigh should a piece of ice fall so that it melts completely ? Only one quarter of the heat produced is absorbed by the ice. Latent heat of ice is `3.4xx10^(5) j kg^(-1)` and `g=10 ms^(2)`. |
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Answer» Correct Answer - `[136 km]` Heat produced on falling the ice `=(1)/(4)xx` decrease in P.E. `=(1)/(4) mgh` Heat required to melt the ice =mL `:. (mgh)/(4)=mL` or `h=(4L)/(g) =(4xx3.4xx10^(5))/(10) =136xx10^(3)m` |
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| 280. |
A metallic cube whose each side is 10 cm is subjected to a shearing force of 100 kgf. The top face is displaced through 0.25 cm with respect to the bottom ? Calculate the shearing stress, strain and shear modulus. |
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Answer» Correct Answer - `9.8 xx 10^4 Nm^(-2), 0.025, 3.92 xx 10^6 Nm^(-2)` Here, `L = 10 cm = 10 xx 10^(-2)m` `F = 100 kg f= 100 xx 9.8 N` `Delta L = 0.25 cm = 0.25 xx 10^(-2) m,` Shearing stress ` = (F)/(L_2) = (100xx9.8)/((10xx10^(-2))` Sheraing strain ` = (Delta L)/(L) = (0.25xx10^(-2))/(10xx10^(-2))` = 0.025 Shear Modulus of elasticity, `G = ("Shearing stress")/("Shearing strain") = (9.8 xx 10^4)/(0.025)` ` =3.92xx 10^6 Nm^(-2)` |
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| 281. |
A `0.05 m` cube has its upper face displaced by `0.2 cm` by a tangential force of `8N`. Calculate the shearing strain, shearing stress and modulus of rigidity of the material of the cube. |
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Answer» `L=5 cm =5 xx 10^(-2) m, Delta L=0.2 cm =0.2 xx 10^(-2) m`, `F=8N` , Shearing strain `=(Delta L)/(L) = (0.2)/(5) = 0.04`, Shearing stress = `(F)/(L xx L) = (8)/((5 xx 10^(-2))^(2))` `= 3200 N//m^(2)` Modulus of rigidity, `G = ("shearing stress")/("shearing strain") =(3200)/(0.04)` `=80000 N//m^(2)` `=8 xx 10^(4)N//m^(2)`. |
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| 282. |
A cylinder of radius `R` made of a material of thermal conductivity `K_(1)` is surrounded by cylindrical shell of inner radius `R` and outer radius `2R` made of a material of thermal con-ductivity `K_(2)` The two ends of the combined system are maintained at two differnet tem-peratures There is no loss of heat across the cylindrical surface and system is in steady state What is the effective thermal conductivity of the system . |
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Answer» Correct Answer - `[(K_(1)+3K_(2))/(4)]` At steady state, `(dQ)/(dt)=(dQ)/(dt)+(dQ_(2))/(dt)` or `(K pi(2R)^(2)(T_(1)-T_(2)))/(l)=(K_(1)pi R^(2)(T_(1)-T_(2)))/(l)+(K_(2)pi(2R)^(2)-pi R^(2)(T_(1)-T_(2)))/(l)` or `4 K=K_(1)+3 K_(2)` or `K+((K_(1)+3K_(2))/(4))` |
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| 283. |
Calcuate the heat required to convert `0.6` kg of ice at - `20^(@)C`, kept in a calorimeter to steam at `100^(@)C` at atmospheric pressure. Given the specific heat capacity of ice`=2100 j kg^(-1) K^(-1)`, specific heat capacity of water `=4186 j kg^(-1) K^(-1)` latent heat ice `=3.35xx10^(5) j kg^(-1)` and latent heat of steam `=2.256xx10^(6)j kg^(-1)` |
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Answer» Correct Answer - `[1.8xx10^(6)j]` Here, `m=0.6 kg, T_(1)=20^(@)C T_(2)=100^(@)C` `s_(i)2100 j kg^(-1) K^(-1), K^(-1), s_(w)=4186 j kg ^(-1) K^(-1)`, `L_(i)=3.35xx10^(5) j kg ^(-1), L_(s)=2.256xx10^(6) j kg ^(-1)`, Heat required to convert ice at `-20^(@)C` to ice at `0^(@)C` `Q_(1)=ms_(i)Delta T_(1)=0.6xx3.35xx10^(5)=2.01xx10^(5)j =201000 j` Heat required to convert water at `0^(@)C` to `100^(@)C` `Q_(3)=ms_(w)Delta T_(2)=0.6xx4186xx(100-0)=251160 j` Heat required to convert water at `100^(@)C` into steam at `100^(@)C` `Q_(4)mL_(s)=0.6xx2.256xx10^(6)=1353600 j` Total heat required to convert `0.6` kg of ice at `-2^(@)C` into steam at `100^(@)C` is `Q=Q_(1)+Q_(2)+Q_(3)+Q_(4)` `=25200+201000+251160+1353600` `=1830960 j =1.83xx10^(6)j` |
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| 284. |
The property of a body by virtue of which it tends to regain its original conifguration as soon as the deforming forces applied on the body are removed is called elasticity. Coefficient of elasticity, `E = (stress)/("strain") = (F//a)/(Delta l//l) = (F l)/(pi r^2(Delta l))` The value of E depends upon nature of meterial. A meterial with higher value of E is said to be more elastic. Read the above passage and answer the following question : (i) Which is more elastic steel or rubber ? Why ? (ii) What values of life so you learn from this concept ? |
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Answer» Steel is more elastic then rubber. This is because `E prop (1)/((Delta l))` Under a given deforming force `F, (Delta l)_(steel)lt (Delta l)_(rubber) :. E_(steel) gt E_(rubber)` (ii) in day to day life, elasticity corresponds to adjustability of the person, which depends on his nature. From `E =(Strees)/("strain"),` we find, strian ` =(stress)/(E )` For a given stress in life, strain will be minimum when E is maximum. it implies that a person with adjustable//flexible nature will undergo minimum strain(tension) due to stresses (ups and donws) of life. |
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| 285. |
Air is streaming past a horizontal air plane wing such that its speed is `120 ms^-1` over the upper surface and `90 ms^-1` at the lower surface. If the density of air is `1.3 kg m^-3`m find the difference in pressure between the top and bottom of the wing. If the wing is `10 m` long and has an average width of `2 m`, calculate the gross lift of the wing. |
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Answer» Correct Answer - [`4095 Nm^(-20; 81900N`] Here, `v_(1)=120 ms^(-1), v_(2)=90 ms^(-1)`, `rho=1.3 kf m^(-3), l=10.0m , b=2.0m ,` Let `P_(1)` and `P_(2)` be the pressure of air at the upper and lower faces of plane respectively, then `(P_(1))/(rho)+(1)/(2)v_(1)^(2)=(P_(2))/(rho)+(1)/(2)v_(2)^(2)` `:. P_(2)-P_(1)=(1)/(2)rho (v_(1)^(2)-v_(2)^(2))` `=(1)/(2)xx1.3xx(120^(2)-90^(2))=4095 Nm^(-2)` Gross lift `=(P_(2)-P_(1))xxA=(A=(P_(2)-P_(1))xxlxxb` `=4095xx10xx2=81900 N` |
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| 286. |
Each of the two wings of an aeroplane has area `30m^(2)`. The speed of the air on the upper and lower surfaces of theon the wing of aeroplane are `90 ms^(-1)` and `70 ms^(-1)` respectively. If the plane is in level flight at constant speed, find the uplift and the mass of the aeroplane. Given density of air `=1.29 kg m^(-3)`. |
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Answer» Here, Area of each wing, `A=30m^(2)`, `v_(1)90 ms^(-1), v_(2)=70 ms^(-1), rho=1.29 kg m^(-3)` Since the aeroplane is in level flighth, so `P_(1)+(1)/(2)rho v_(1)^(2)=P_(2)+(1)/(2)rho v_(2)^(2)` or `P_(2)=P_(1)=(1)/(2)rho(v_(1)^(2)-v_(2)^(2))=(1)/(2)xx1.29(90^(2)-70^(2))` `2064 Nm^(-2)` Upward from both the wings of aeroplane `F_(1)=(P_(2)-P_(1))xx2A=2064xx2xx30=123840 N` `=1.24xx10^(5) N` Mass of the plane, `M=(F)/(g)=(1.24xx106(5))/(9.8)` `=1.26xx10^(4) kg` |
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| 287. |
The …….. at a point in a liqud is the difference of total pressure at that point and atmospheric pressure. |
| Answer» Correct Answer - gauge pressure | |
| 288. |
Two tubes A and B of length 80 cm and 40 cm have radii 0.1 mm and 0.2 mm respectively are connected in series end to end. If a liquid passing through tow tubes is entering A at a pressure of 82 cm of mercury and leaving B at a pressure of 76 cm of mercury. Find the pressure at the junction of A and B. |
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Answer» Correct Answer - 76.18 cm of Hg Let `P_(j)` (in cm of Hg) be the pressure at the junction of A and B. For of flow of liquid through tube A, `Q_(1)=(pi(P_(A)-P_(j))r_(1)^(4))/(8 eta l_(1))=(pi(82-P_(j))(0.01)^(4))/(8 etaxx80)` For tube `B:l_(2)=40 cm, r_(2)=0.02 cm`, Pressure at which liquid leaves tube B, `P_(B)=76` cm of Hg Rate of flow of liquid through tube B `Q_(2)=(pi(P_(j0-P_(B))r_(2)^(4))/(8 eta l)=(pi(P_(j)-76)(0.02)^(4))/(8 etaxx40))` Here, `:. (pi(82-P_(j))xx(0.01)^(4))/(8 etaxx80) =(pi(P_(j)-76)(0.02)^(4))/(8 etaxx40)` or `(82-P_(j))=(P_(j)-76)xx32` On solving, `P_(j)=76.18` cm of Hg |
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| 289. |
The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. What is the ratio between the volume of the first and the second bubble? |
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Answer» Given ` (4S)/(r_1) = (3 xx 4S)/(r^(2)) or r_(2) = 3 r_(1)` , `(V_1)/(V_2) = ((4//3) pi r_(1)^(3))/((4//3) pi r_(2)^(3)) = ((r_1)/(r_2))^(3) = (1/3)^(3) = 1/27`. |
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| 290. |
The maximum load that a wire can sustain is W. If the wire is cut to half its value, the maximum load it can sustain is |
| Answer» Since breaking load = breaking stress `xx` area, is free from the length of elastic wire, therfore, if the cable is cut to half of its original length, there is no charge in its area of cross section and breaking stree. Hence there is no effect on the maximum load ( breaking load), the cable can support. | |
| 291. |
The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically up, but tends to narrow down when held vertically down. Explain how? |
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Answer» As the stream falls, its speed `upsilon` will increase and hence its area of cross-section a will decrease, ac cording to equation of continuity, i.e., `aupsilon = a consta nt`. That is why the stream will become narrow. When the stream will go up, its speed will decrease, hence its area of crosssection will increase, ie. it will become broader and spreads out like a fountain. |
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| 292. |
Which type of strain is three, when a spiral spring is stretched by a force? |
| Answer» Longitudinal strain and shear strain. | |
| 293. |
STATEMENT-1: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down. STATEMENT-2: In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.A. both Assertion and Reason are true and the Reason in the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false. |
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Answer» Correct Answer - A Here, both Assertion and Reason are true and the Reason is the correct explanation of Assertion, because from equation of continuuity, `Av = a` constant or `A prop (1)/(v)`. |
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| 294. |
Assertion. A bubble comes from the bottom of a lake to the top. Reason. Its radius increases.A. both Assertion and Reason are true and the Reason in the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false. |
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Answer» Correct Answer - B Here, both Assertion and Reason are true but reason is not correct explanation of Assertion. When bubble moves from the bottom to top (i.e., from higher pressure to lower pressure) pressure on it decreases and its volume and hence radius increases. |
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| 295. |
As air is a bad conductor of heat, whu do we not feel warm without clothes? |
| Answer» This is because when we are without clothes, air carries away heat from our body due to convection and hence we feel cold. | |
| 296. |
Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is `rho` and L is its latent heat of vaporization.A. `rho L//S`B. `sqrt(S//rho L)`C. `S//rho L`D. `2S//rho L` |
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Answer» Correct Answer - D Conside a liquid drop of radius r. When a layer of lilquid drop of thickness dr is evaporated and temperature remains unchanged, then change in surface energy = change in surface area x surface tensiion `=d(4 pi r^(2))xxS=(8pi r dr)S` Energy required to evaporate the liquid drop layer of thickness dr `=(4pi r^(2))xxS=(4pi r^(2) dr)rho L` The process of evaporation only start if change in surface energy is just sufficient to evaporate the layer of liquid drop. Therefore, `(4pi r^(2) dr)rho L=(8pi r dr)S` or `r=(2S)/(rho L)` |
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| 297. |
What will the effect on the temperature, if the number of small drops fo mercury coalesce adiabatically to form a single drop? |
| Answer» When large number of drops coalesce, the surface area will decrease. The energy will be liberated. Due to adiabatic conditions, its temperature will increase. | |
| 298. |
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 `=4 VT [(1)/(r)-(1)/(R)]` is releasedB. Energy `3 VT[(1)/(r)+(1)/(R)]` is absorbedC. Energy `=3 VT[(1)/(r)-(1)/(R)]` is releasedD. Energy is neither relesed nor absorbed. |
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Answer» Correct Answer - C Let there be n small drop each of radius r, which when coalesce from a big drop of radius R. Then `V=nxx(4)/(3)pi r^(3) =(4)/(3)pi R^(3)` or `n=(R^(3))/(r^(3))` Energy released = surface tension x decrease in surface area `=T[4pi r^(2)xxn-4pi R^(2)]` `=T[4pi r^(2)xx(R^(3))/(r^(3)) - (4pi R^(3))/(R)]` `=3T[(4)/(3)pi R^(3)xx(1)/(r) - (4)/(3)pi R^(3)xx(1)/(R)]` `=3T[Vxx(1)/(r)-Vxx(1)/(R)] =3VT[(1)/(r) - (1)/(R)]` |
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| 299. |
The breaking force is independent of the ………of the wire till its……….. Remains constant. |
| Answer» length , area of cross section | |
| 300. |
The bulk modulus for an incompresssible liquid is |
| Answer» Correct Answer - Infinite. | |