This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
2. Joseph jogs from one end A to the other end B of a strai300 m road in 2 minutes 30 seconds and then turnsand jogs 100 m back to point C in another 1 minute. What areJoseph's average speeds and velocities in jogging (a) from A toB and (b) from A to C?aroun |
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| 2. |
1. State the effect of bulk density of coarse aggregate on the compressive strength ofconcrete. |
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Answer» Concrete is a mixture of cementious material, aggregate, and water. Aggregate is commonly considered inert filler, which accounts for 60 to 80 percent of the volume and 70 to 85 percent of the weight of concrete. Although aggregate is considered inert filler, it is a necessary component that defines the concrete’s thermal and elastic properties anddimensional stability. Aggregate is classified as two different types, coarse and fine. Coarse aggregate is usually greater than 4.75 mm (retained on a No. 4 sieve), while fine aggregate is less than 4.75 mm (passing the No. 4 sieve). The compressive aggregate strength is an important factor in the selection of aggregate. When determining the strength of normal concrete, most concrete aggregates are several times stronger than the other components in concrete and therefore not a factor in the strength of normal strength concrete. Lightweight aggregate concrete may be more influenced by the compressive strength of the aggregates. Other physical and mineralogical properties of aggregate must be known before mixing concrete to obtain a desirable mixture. These properties include shape and texture, size gradation, moisture content, specific gravity, reactivity, soundness and bulk unit weight. These properties along with the water/cementitious material ratio determine thestrength,workability, anddurabilityof concrete. The shape and texture of aggregate affects the properties of fresh concrete more than hardened concrete. Concrete is more workable when smooth and rounded aggregate is used instead of rough angular or elongated aggregate. Most natural sands and gravel from riverbeds or seashores are smooth and rounded and are excellent aggregates. Crushed stone produces much more angular and elongated aggregates, which have a higher surface-to-volume ratio, better bond characteristics but require more cement paste to produce a workable mixture. The surface texture of aggregate can be either smooth or rough. A smooth surface can improve workability, yet a rougher surface generates a stronger bond between the paste and the aggregate creating a higher strength. The grading or size distribution of aggregate is an important characteristic because it determines the paste requirement for workable concrete. This paste requirement is the factor controlling the cost, since cement is the most expensive component. It is therefore desirable to minimize the amount of paste consistent with the production of concrete that can be handled, compacted, and finished while providing the necessary strength and durability. The required amount of cement paste is dependent upon the amount of void space that must be filled and the total surface area that must be covered. When the particles are of uniform size the spacing is the greatest, but when a range of sizes is used the void spaces are filled and the paste requirement is lowered. The more these voids are filled, the less workable the concrete becomes, therefore, a compromise between workability and economy is necessary. Themoisture contentof an aggregate is an important factor when developing the proper water/cementitious material ratio. All aggregates contain some moisture based on the porosity of the particles and the moisture condition of the storage area. The moisture content can range from less than one percent in gravel to up to 40 percent in very porous sandstone and expanded shale. Aggregate can be found in four different moisture states that include oven-dry (OD), air-dry (AD), saturated-surface dry (SSD) and wet. Of these four states, only OD and SSD correspond to a specific moisture state and can be used as reference states for calculating moisture content. In order to calculate the quantity of water that aggregate will either add or subtract to the paste, the following three quantities must be calculated: absorption capacity, effective absorption, and surface moisture. Most stockpiled coarse aggregate is in the AD state with an absorption of less than one percent, but most fine aggregate is often in the wet state with surface moisture up to five percent. This surface moisture on the fine aggregate creates a thick film over the surface of the particles pushing them apart and increasing the apparent volume. This is commonly known as bulking and can cause significant errors in proportioning volume. The density of the aggregates is required in mixture proportioning to establish weight-volume relationships. Specific gravity is easily calculated by determining the densities by the displacement of water. All aggregates contain some porosity, and the specific gravity value depends on whether these pores are included in the measurement. There are two terms that are used to distinguish this measurement; absolute specific gravity and bulk specific gravity. Absolute specific gravity (ASG) refers to the solid material excluding the pores, and bulk specific gravity (BSG), sometimes called apparent specific gravity, includes the volume of the pores. For the purpose of mixture proportioning, it is important to know the space occupied by the aggregate particles, including the pores within the particles. The BSG of an aggregate is not directly related to its performance in concrete, although, the specification of BSG is often done to meet minimum density requirements. For mixture proportioning, the bulk unit weight (a.k.a. bulk density) is required. The bulk density measures the volume that the graded aggregate will occupy in concrete, including the solid aggregate particles and the voids between them. Since the weight of the aggregate is dependent on the moisture content of the aggregate, a constant moisture content is required. This is achieved by using OD aggregate. Additionally, the bulk density is required for the volume method of mixture proportioning. The most commonclassification of aggregateson the basis of bulk specific gravity is lightweight, normal-weight, and heavyweight aggregates. In normal concrete the aggregate weighs 1,520 – 1,680 kg/m3, but occasionally designs require either lightweight or heavyweight concrete. Lightweight concrete contains aggregate that is natural or synthetic which weighs less than 1,100 kg/m3and heavyweight concrete contains aggregates that are natural or synthetic which weigh more than 2080 kg/m3. Although aggregates are most commonly known to be inert filler in concrete, the different properties of aggregate have a large impact on the strength, durability, workability, and economy of concrete. These different properties of aggregate allow designers and contractors the most flexibility to meet their design and construction requirements. can u explain in short Cement is a glue. It holds aggregates to withstand compression. 100 % aggregate would theoretically limit the compression to that of aggregate itself. In reality, good compaction of concrete can not be done with larger aggregates in over abundance using traditional vibratory methods . There would be large number of air pockets as these aggregates get stuck due to their sizes. Therefore, one needs an optimal distribution of aggregates of different sizes, which would slip seamlessly into the smaller and smaller gaps. When everything flows correctly during compaction, one can obtain a dense concrete with least amount of water:cement ratio, with little voids. That's the secret of obtaining higher compressive strength. W:C ratio is so important, that a small number of air voids and bug holes (at outer surface) are considered acceptable, removing them by increasing water content would actually make the concrete weaker. The answer to your question lies in proportion of the coarse aggregate in overall mix (assuming it has uniform size), and it's relation to size distribution of smaller ‘particles’ . An overabundance of coarse aggregate would generally produce a weaker concrete as it would be difficult to compact in real life using vibration. This should not apply as rigidly when using super plasticizers, they are otherwise reserved for crowded places using higher content of smaller aggregate or sand. Primary reason for limiting their use to such applications is cost. I would hypothesize, using super plasticizers one should be able to use a mix with larger aggregates in relative excess, especially in applications like making a floor. Pushing content of coarse aggrgates to extreme, one may be able to harness superior compression approaching closer and closer to that of coarse aggregate itself, but this material would not flow well around densely set rebars, thus may end up being weaker if applied without forethought. On another note, if I have an excess of large aggregates, and the project is a small homely one, there is a way to obtain optimal results making something like a floor. At the very bottom pour a good mix, then, manually sprinkle large aggregate or chunks (such as pieces of old concrete) on the top, bury this with good concrete, srinkle again, and repeat as needed. Top with good concrete. Here we are visually ensuring that larger aggregate is getting surrounded by good concrete. Excellent compaction can be achieved with vibration. If cost is not an issue, such as spare superplasticizer laying around, it can make it very easy utilizing any remaining excess of coarse aggregate. |
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| 3. |
Joseph jogs from one end A to the other end B of a st300 m road in 2 minutes 30 seconds and then turns aand jogs 100 m back to point C in another 1 minute. WhJoseph's average speeds and velocities in jogging (a) fromB and (b) from A to C? |
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| 4. |
2. Joseph jogs from one end A to the other end B ol a straigi300 m road in 2 minutes 30 seconds and then turns aroundand jogs 100 m back to point C in another 1 minute. What areJoseph's average speeds and velocities in jogging (a) fromB and (b) from A to C? |
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Answer» Velocity = dispacement / timeSpeed = distance / time a) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 s velocity = 300/150 = 2 m/sspeed = 300/150 = 2m/s b)when he jogs from A to B and turns back to C,displacement = 300-100 = 200mdistance = 300+100 = 400mtime = 3 minute 30 second = 210 s velocity = 200/210 = 20/21 m/sspeed = 400/210 = 40/21 m/s |
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| 5. |
Joseph jogs from one end A to the other end B of a straight300 m road in 2 minutes 30 seconds and then turns aroundand jogs 100 m back to point C in another 1 minute. What areJoseph's average speeds and velocities in jogging (a) from A toB and (b) from A to C? |
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| 6. |
11. निम्नलिखित समुच्चयों में बतलाइए AB है अथवा नहीं[In the following sets state whether A=B or not:O AD 14,8,12,16), B= [8,4.16.18)O AD [xx10 का गुणज है),B+[10.15,20,25,30..O AD (2,4,6,8,10)-B = [xxx सम धन पूर्णांक है और x510)(iv) A= (a,b.c.d), B= [d.c.b,a)(M AD12.3). B= {x: P+5x+6%D0) |
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Answer» I)A=Bit)A=Biii)A is not=Bif)A=Bv)A is not=B 1. A=B 2.A=B 3.A is not B 4.A=B 5. A is not B |
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| 7. |
24. A projectile is thrown with speed 40 ms at angie6from horizontal. It is found that projectile is at sameheight at 1s and 3 s. What is the angle of projection?-arran-la)(4) tan(2)tan-1(72)(1)11 ㅡㅡ(3) tan-1(3) |
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| 8. |
A body is thrown vertically upwards with a velocityto the true statement from the following K erala(aj both velocity and acceleration are zero at its highestpoint(5) Velocity is maximum and acceleration is zero at thehighest point(0) Velocity is maximum and acceleration isdownwards at its highest pointVelocity is zero at the highest point and maximumheight reached is u?/2g62. |
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Answer» option d is write answer |
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| 9. |
At what point of the path of the projectile speed isminimum ? At which point is it maximum? |
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| 10. |
what happened to kinetic energy if the mass is tripled |
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| 11. |
For a given current and time, what is the change inthe heat produced in a conductor if the resistanceof the conductor is a. tripled b. reduced to one-third |
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Answer» H=i^2Rthence heat produced is directly proportional to resistancehence if resistance is tripled heat will be tripledand if resistance is one third then heat will be one third |
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| 12. |
B) The resistance of a wire of 0.01 cm radius is 10 olm. If the resistivity of the material ofthe wire is 50 x 103 ohm meter, find the length of the wire? |
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| 13. |
Lolbal is the cong tular) velocity.cloeiLbol |
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Answer» Angular velocity of earth is 2πr/time of day in seconds r = radian subtended by radius of earth 2π r/ 86400 = 7.25 × 10 ^ -5 Radians / sec |
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| 14. |
9 How will the resistivity of a conductor change when its length is tripled by stretching it? |
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Answer» It stays the same, no matter how long or thick your conductor is. That is not to say thatresistivitycannotchange. Itvarieswith temperature. So there is nochange with lengthinresistivitybut resistancechangesin direct proportion withlengthof the conductor |
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| 15. |
A copper wire has a diameter of 0.5mm and resistivity of 1.6 into 10 to the power minus 8 ohm meter?a. what will be the length of this wire to make resistance 10ohm?b.How much does the resistance change if diameter is double? |
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Answer» 10 = 1.6 * 10^-8 * l/(π*25*10^-8)l=250π/1.6=2500π/16=625π/4 m b. R is proportional to 1/d^2thus Resistance is one fourth |
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| 16. |
1.A wire of length L and resistance R is stretched so that itslength is doubled and area of cross section is halved. How willits (i) resistance change? (ii) resistivity change? |
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Answer» i) The Resistance Will increase by 4ii) There Will be no change in resistivity because resistivity does not depend on leng th and area of cross section. Resistance=resistivity*length/areahence if length is doubled are area halvedresistance will become 4 timesand resistivity will become 1/4th times |
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| 17. |
1.When a fast moving bus is suddenly stopped, a stand-ing passenger tends to fall in the forward direction |
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Answer» It is because of the inertia of motion. The lower part of our body is in contact with the bus. As the bus is moving our whole body is moving along with it. As the bus stops, our lower body also stops with it as it is in contact whereas our upper body tends to remain in the state of motion due to inertia of motion and we fall forward. By the way nice question can answer this question in malayalam no we answer only in English ok tnkzzz nice answer how we identify it is inertia of mation or inertia of rest what u didn't understand...what to identify about how will identify the question is inertia of motion or inertia of rest can u answe this question in manglish |
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| 18. |
calculate de strength of curreain a wire cal 1200 coulomb atchange padser te moeghacivencross section at the loire in 20min. |
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| 19. |
If source and observer are moving away,appearent frequency-(a) (increase) (b) & (Decrease)(c) अचर (Constant) (d) कोई नहीं () |
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Answer» c is the right answer |
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| 20. |
2. The initial speed of a bullet fired from a rifle is 630 m/s. The rifle is fired at the centre ofa target 700 m away at the same level as the target. How far above the centre of thetarget the rifle must be aimed in order to hit the target?(a) 1.0 m(d) 9.8 m(b) 4.2 m(c) 6.1 mnstant force R-召^+ R ^ has velocity at t=0 s |
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Answer» option (c) is correct. can I please elaborate |
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| 21. |
1. A paramagneticagnetic salt has a magnetic moment of5.0 A m2 at a temperature of 4 K in a magnetic fieldof 1.0 T. Find its magnetic moment at a temperatureof 3 K in a magnetic field of 1.5 TAns. 10 A m |
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| 22. |
Example 5.1 In Fig. 5.4(b), the magnetic needle has magnetic momentX 10-2 Am" and moment of inertia = 7,5 x 106 kg m2. It performs10 complete oscillations in 6.70 s. What is the magnitude of the6.7magneticfield? |
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| 23. |
71. If the distance between two masses is increasedby a factor of 5, by what factor would the massof one of them have to be altered to maintainthe same gravitational force? Would this be anincrease or decrease in the mass ? |
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Answer» thank u |
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| 24. |
13. In figure, the vertices of square DEFG are onthe sides of △ABC.LA-90°. Then prove thatDE BD x EC.Hint : Show that AGBD is similar to ACFE. Use |
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| 25. |
Write in detail about the methods for determination ofage and volume of felled and standing tree.8. |
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Answer» Tree age measurement- When you make cross section of the trunk you'll see some rings. This annual ring formation is so regular, for this reason the age of a tree may be accurately calculated simply by counting the rings "each ring represents one year of growth". Because the vascular cambium suspends its activity during fall and winter when the tree becomes dormant, ring growth occurs only during spring and summer. Each annual ring consists of several layers of large xylem cells, referred to as springwood, followed by progressively smaller cells, the summerwood. Tree volume measurement-1-Tree climbers can physically measure the height and circumference of a tree using a tape. The distance from the highest climb point and the top of the tree is measured using a pole that extends from the tree top to the anchor point of the tape. This height is noted and the diameter of the tree is measured at that point. The climber then rappels down the tree measuring the trunk circumference by tape wrap at different heights with the height of each measurement referenced to the fixed tape running down the trunk. 2-Remote measurements of trunk volume are usually made from a position on the ground where the observer has a clear view of the entire length of the trunk. Measurements may be made using professional surveying equipment such as a total station thank you so much mam mam I need a notes |
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| 26. |
Express the Kinetic Energy of ans Object in the form of an equation |
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| 27. |
Q. 5. Find the position of the object placed infront of a concave mirror of focal length 20 cm ifimage is formed at the distance of 30 em from themirror |
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| 28. |
1. (a) What is meant by power of a lens ? Define its S.I. unit.(b) You have two lenses A and B of focal lengths +10 em and-10 em respectively.State the nature and power of each lens. Which of the two lenses will form a virtualand magnified image of an objeet placed 8 em from the lens ? Draw a ray diagramto justify your answer.ur of a lrns is defined as the reciprocal of its focal length. It is |
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Answer» Thepower of a lensis the reciprocal ofitsfocal length in meters. It is represented by the letter P. ThepowerP of alensof focal length f is given by f = 1/P. TheSI unitofpoweris dioptre . |
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| 29. |
An angular magnification of 30 X is desired using anive of focal length 1-25 cm and an eyepiece of focalngth 5 em. How will you set up the compoundle |
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| 30. |
Give three drawbacks of Mendleev's Periodic Table.P |
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Answer» Drawbacks/Demerits of Mendeleev's periodic table :- 1) Hydrogen's positionHydrogen resembled the properties of both alkali metals(like lithium) and also that of halogens(like iodine). Hence, the position of hydrogen(whether hydrogen is to be placed with halogens or alkali metals) was not specified. 2) Position for IsotopesIsotopes are those with same atomic no: and different mass no:s.Mendeleev'speriodic table was based on arranging elements in increasing order of atomic masses. But isotopes were not included in his periodic table. 3) Certain elements were arranged in reverse order. Elements having higher atomic mass were placed in front(or before) the elements with less atomic mass. Example - Cobalt and NickelCobalt being more heavier than Nickel ,was placed before Nickel , in the Mendeleev's periodic table. hit like if you find it useful |
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| 31. |
velocity of the electron at A is zero, find the velocity at B. (Assume mass of electron 9.1 x 103and charge 1.6 x 1 0-19C).ron moves from point A where the polential is zero to a point B where potential is 60 V. IF helv . 4.6x 106 ms-1] |
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| 32. |
15. Define G. How does its value vary in space?16. Why does a body reach the ground quicker atpoles than at the equator when dropped fromthe same height. |
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| 33. |
mass of Jupiter is 1.899×10^27kg and radius 7.14×10^7m find the acceleration due to gravity on the surface of Jupiter |
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| 34. |
(a) What do you mean by impulse of force? Show that impulse of a force is equalto the change in momentum produced by the force.(b) A ball of mass 200 g hits a wall at an angle of 45° with a velocity of 15 m/s.Ifthe ball rebounds at 90° to the direction of incidence, calculate the impulsereceived by the ball. |
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Answer» b) |
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| 35. |
Q. 3. Obtain an expression for maximum safety speed with which vehicle can be safely driven along curved bankedroad0:4 show that te angle of bankine is indenendent of mass of vehicle. |
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| 36. |
10 Derive an expression for the maximum speed with which a vehicle can be safely driven alongthe banked curve of a road.he centripetal force |
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| 37. |
The first manned space vehicle to land on the moon |
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Answer» Lunar Exploration Module (LEM) nick-named ‘Eagle’ |
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| 38. |
30. What is the maximum acceleration of a vehicle on the horizontal road? Given that coefficient of static frictiobetween the road and the tyres of the vehicle is μ. |
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| 39. |
Name the device used to measure the distance travelled by vehicle?of vehicle |
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Answer» the answer is odometer The answer will be Odometer. the correct answer is odometer |
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| 40. |
s. An automobile vehicle has a mass of 1500 kg. What mustbe the force between the vehicle and road if the vehicle isbe stopped with a negative acceleration of 1.7 m s?9. What is the momentum ot anohi |
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| 41. |
Write 3 features of Rutherford nuclear model of an atom |
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Answer» Ans :- The major features of the model are as follows. 1 All protons are present inside the nucleus, which is situated at the center of the atom. 2. Electrons reside outside the nucleus and revolve around the nucleus in well-defined orbits. 3. The size of the nucleus is very small in comparison to the size of an atom. As per Rutherford’s calculations, the size of the nucleus is 105times smaller than an atom. |
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| 42. |
FORCE ON A CURRENT-CARRYING CONDuIN A MAGNETIC FIELD |
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Answer» Force on a current carrying conductor in the presence of a magnetic field is explained using Fleming's Left Hand Rule According toFleming’s left-hand rule, if we stretch the thumb,forefinger and middle finger of our left handsuch that they are mutually perpendicular to each other,If the first finger points in thedirection of magnetic field and the secondfinger in the direction of current, then thethumb will point in the direction of motion (the force acting on the conductor.) |
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| 43. |
Is the electric field inside a conductor always zero |
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Answer» In electrostatics free charges in a goodconductorreside only on the surface. So the free chargeinsidetheconductoris zero. So thefieldin it is caused by charges on the surface. Since charges are of the same nature and distribution is UNIFORM, theelectricfields cancel each other. |
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| 44. |
5)Is the electric field inside a conductor always zero ? |
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Answer» In electrostatics free charges in a goodconductorreside only on the surface. So the free chargeinsidetheconductoris zero. So thefieldin it is caused by charges on the surface. Since charges are of the same nature and distribution is UNIFORM, theelectricfields cancel each other. |
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| 45. |
work done to move a charge inside the equipotential surface is zero. justify |
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Answer» When a charge is moved from one point to another along some equipotential surface, it is true that the work done on the chargeby the electric fieldis zero. Additional work done on the particle over this path depends on how the particle is moved. If the particle is initially moving in the correct direction, then obviously no work is required. This is why the Moon is able to maintain its motion along an (approximate) gravitational equipotential without an army of angels diligently pushing it on it's path. But if you are moving a particle from one position at rest to a second position at rest, along the equipotential, some external agent supplying work is needed to start and stop the particle. However, you in principle you could make this motion arbitrary slow over arbitrarily long times, so that at any point along the path it's kinetic energy is indistinguishable from zero, hence zero change in kinetic energy along the path, hence zero work necessary. |
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| 46. |
10A stone of mass 1 kg bed to a light inextensible string of length L m, whiting in a circular path in avertical plane. The ratio of maximum tension in the string to the minimum tension in the string is 4, If gis taken to be 10 m/s, the speed of the stone at the highest point of the circle is: |
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| 47. |
Magnetic Field due to a Current in a Solenoid |
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| 48. |
A lte Biot- Savart law in vector fothe magnetic field due to an elementcurrent I at a distance i from the elemsen |
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| 49. |
A w re of resistance 8 Ω is bent into a circle. Find the resistance across the diameter |
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| 50. |
A wire of length carries a steady current. It isbent first to form a circular plane loop of one turn.The magnetic field at the centre of the loop is B、The same length is now bent more sharply to givea double loop of smaller radius. The magnetic fieldat the centre caused by the same is |
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