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.
| 6251. |
the vectorIf A-. A cos θ i + Aperpendicular to A is given by,(1) Bcos θ i-Bsin θ J(2) B sin θ 1-B cos θ jsin θ j, then(4)Bsín θ 1 + B cosθ J |
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Answer» For a perpendicular vector.. the dot product should be 0. so only option is 2. as dot product is. (Acos∅)(Bsin∅) -(Asin∅)(Bcos∅) = ABsin∅cos∅ -ABsin∅cos∅ = 0 |
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| 6252. |
3Must be acceleratedNEWTON'S 2ND LAW OF MOTION18. A fore of 0.6N on a particde increases its velocity from 5 m/s to 6 m/s in 2 ses of theparticle is1)2.4kg4) 3.2kg2) 1.2kg3) 4.8kg |
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Answer» from the change in velocity , we find acceleration as 6-5/2 = 1/2m/s² now , the force is 0.6N so the mass is = 0.6/0.5 = 1.2kg tq |
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| 6253. |
equation of a standing wave is y=Asin(bx)cos(ct) find the dimension of bc |
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Answer» Dimension of bx and ct is null. This implies that dimension of b is the inverse of that of X, which is [L] and the dimension of c is the inverse of that of t, which is [T]. So, dimension of b is [L-¹] and dimension of c is [T-¹].Thus, dimension of bc is [L-¹ T-¹]. Please hit the like button if this helped you. |
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| 6254. |
A particle moves on circular path of radius 5 m with constant speed 5 m/s. Find the magnitude of its averageacceleration when it completes half revolution. |
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| 6255. |
explain schrondingers wave theory in detail? |
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Answer» The Schrodinger equation plays the role ofNewton's lawsandconservation of energyin classical mechanics - i.e., it predicts the future behavior of a dynamic system. It is a wave equation in terms of thewavefunctionwhich predicts analytically and precisely the probability of events or outcome. The detailed outcome is not strictly determined, but given a large number of events, the Schrodinger equation will predict the distribution of results. The Schrodinger equation is used to find the allowed energy levels of quantum mechanical systems (such as atoms, or transistors). The associated wavefunction gives the probability of finding the particle at a certain position. |
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| 6256. |
(How are wavefront and secondary wavelets defined ? Verify laws of reflection or laws ofrefraction on the basis of Huygen's wave theory.) |
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Answer» wavefront is termed as the locus of all the points joining together woth same phase, a line or curve in 2dimensional wheras a wavelet is defined as: a wave like oscillation with ampliyude expanding and contracting gradually and sequentially. the locus of all the points of the medium which receives disturbances simoultaneously and are in same state of vibration are called as wavefront .every point on a wavefront act as a new source called as secondary wavelets .wavelets are just a kind of daughter wave u can say .more precisely , wavelets is small wave like oscillation , which starts with zero displacement , increase to max . displacement , then decrease to zero again . |
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| 6257. |
14. The power of a biconcave lens having focal length 25cm is .............. |
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Answer» power = 1/ focal lengthpower = 1/ 25 = 0.04 dioptre |
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| 6258. |
derive equation of kinetic energy from equation of motion |
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| 6259. |
Q.21.A) Derive an equation for 2nd kinematical equation |
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| 6260. |
At what distance from a concave mirror of focallength 25 cm should an object be placed so thatthe size of image is equal to the size of theobject.Ans. 50 cm |
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| 6261. |
5. At what distance from a concave mirror of focallength 25 cm should an object be placed so thatthe size of image is equal to the size of theAns. 50 cmobject. |
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| 6262. |
Shots fired simultaneously from the top and foot ofa vertical cliff at elevations of 30° and 60respectively, strike an object simultaneously whichis at a height of 100 meters from the ground and at ahorizontal distance of 200/3 meters from the cliff.Find the height of the cliff, the velocities ofprojection of the shots and the time taken by theshots to hit the object. (g 10 m/sec2) [JEE-2000] |
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Answer» You need to use this procedure to solve. |
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| 6263. |
28. Let m,-1 kg, m2-2 kg and ma-3 kg in figure (5-E12).Find the accelerations of m, m2 and ma. The stringfrom the upper pulley to mi is 20 cm when the systemis released from rest. How long will it take before m,strikes the pulley? |
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| 6264. |
19.B) 3 kgIn the given figure, find mass of the block A, if it remains at rest, when the system is rele(C) 4 kg(D) 5 kgrest. Pulleys and strings are massless. I g = 10 m/s)(D) 3m(A) m(B) 2m(C) 2.5minn of the block C. |
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Answer» Here, it is given that the block A is at rest. So the pulley attached to masses m and 2m must also be at rest because string is ideal(i.e. inelastic and massless). So for this pulley, total downward force(mg+2mg) is equal to total upward force(tension T=mass of A*g).So, mass of A=(1+2)m=3m. |
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| 6265. |
A horizontal force of 1.2 kg is applied ona 1.5 kg block, which rests on a horizontalsurface. If the co-efficient of friction is 0.3,force of friction is:(A) 0.45 kgf (B) 1.2 kgf(C) 1.5 kgf (D) 0.3 kgf27. |
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| 6266. |
3. A horizontal force of 1O N is necessary to just holda block stationary against a wall. The coefficientof friction between the block and wall is 0.2. Theweight of the block is1ONWall(1120N(2)50 N(3) 100 N(4)2 N |
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| 6267. |
)100 kg m/s3.A vehicle of 100 kg is moving with a velocity of 5 m/sec. To stop it indirection is(a) 5000 Nsec, the required force in opposite1o(b) 500 N(c) SO N(d) 1000N |
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| 6268. |
A block of mass m is suspended through a spring ofspring constant k and is in equilibrium. A sharp blowgives the block an initial downward velocity v. How farbelow the equilibrium position, the block comes to aninstantaneous rest ? |
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| 6269. |
masses m, and m2 are connected by a spring of spring constant k and are placed on a smoothorizontal surface. Initially the spring is stretched through a distance 'd' when the system is releasedTwofrom rest. Find the distance moved by the two masses when spring is compressed by a distance 'd' |
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| 6270. |
A spring with spring constant k when stretched through 1 cm, the potential energy is U. Iftisstretched by 4 cm. The potential energy will be1) 4U2) 8U3) 16 U4) U |
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| 6271. |
Section (D): spring - mass systemwo masses mi and m2 are connected by a spring of spring constant k and are placed on a smoothspring is stretched through a distance 'd' when the system is releasedhorizontal surface. Initially theFind the distance moved by the two masses when spring is compressed by a distance 'd'. |
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| 6272. |
ction (D): springTwo masses m, ahorizontalfrom rest. Find the distance movn and m2 are connected by a spring of spring constant k and are placed on a stace. Initally the spring is stretched through a distance 'd when the system is released on a smoothed by the two masses when spring is compressed by a distance'd |
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Answer» please post clear image |
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| 6273. |
a capacitance is charged with a battery and thenits plate separated is increased without disconnecting the battery. What will be the change ina) charge stored in the capacitorb)energy stored in the capacitor |
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| 6274. |
what is semi conductor |
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Answer» A solid substance that allows heat or electricity to pass through it or along it in particular conditions Asemiconductormaterial has anelectrical conductivityvalue falling between that of aconductor, such as metallic copper, and aninsulator, such as glass. Itsresistancedecreases as its temperature increases, which is behaviour opposite to that of a metal. Its conducting properties may be altered in useful ways by the deliberate, controlled introduction of impurities ("doping") into thecrystal structure. Where two differently-doped regions exist in the same crystal, asemiconductor junctionis created. The behavior ofcharge carrierswhich includeelectrons,ionsandelectron holesat these junctions is the basis ofdiodes,transistorsand all modernelectronics. Some examples of semiconductors aresilicon,germanium,gallium arsenide, and elements near the so-called "metalloid staircase" on the periodic table. After silicon, gallium arsenide is the second most common semiconductor[citation needed]and is used in laser diodes, solar cells, microwave-frequency integrated circuits and others. Silicon is a critical element for fabricating most electronic circuits. Semiconductor devicescan display a range of useful properties such as passing current more easily in one direction than the other, showing variable resistance, and sensitivity to light or heat. Because the electrical properties of a semiconductor material can be modified by doping, or by the application of electrical fields or light, devices made from semiconductors can be used for amplification, switching, andenergy conversion. The conductivity of silicon is increased by adding a small amount (of the order of 1 in 108) of pentavalent (antimony,phosphorus, orarsenic) or trivalent (boron,gallium,indium) atoms. This process is known as doping and resulting semiconductors are known as doped or extrinsic semiconductors. Apart from doping, the conductivity of a semiconductor can equally be improved by increasing its temperature. This is contrary to the behaviour of a metal in which conductivity decreases with increase in temperature. The modern understanding of the properties of a semiconductor relies onquantum physicsto explain the movement of charge carriers in acrystal lattice.[1]Doping greatly increases the number of charge carriers within the crystal. When a doped semiconductor contains mostly free holes it is called "p-type", and when it contains mostly free electrons it is known as "n-type". The semiconductor materials used in electronic devices are doped under precise conditions to control the concentration and regions of p- and n-type dopants. A single semiconductor crystal can have many p- and n-type regions; thep–n junctionsbetween these regions are responsible for the useful electronic behavior. Some of the properties of semiconductor materials were observed throughout the mid 19th and first decades of the 20th century. The first practical application of semiconductors in electronics was the 1904 development of thecat's-whisker detector, a primitive semiconductor diode used in early radio receivers. Developments in quantum physics in turn allowed the development of thetransistorin 1947,[2]theintegrated circuitin 1958, and theMOSFET(metal–oxide–semiconductorfield-effect transistor) in 1959. solid substance that allows heat or electricity to pass through it or along it in particular conditions अर्धचालक (विशिष्ट दशाओं में ताप या विद्युत का चालन करने वाला एक ठोस पदार्थ); सेमिकंडक्टर |
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| 6275. |
br 120tal Pau(e Woud hove aiong a vertical lineFour identical spheres each of mass m are placed at the corners of square of side 2metre. Takingintersection of the diagonals as the origin, the co-ordinates of the centre of mass are(d) of these |
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Answer» Since, all the vertices have identical spheres, the centre of mass is at the intersection of diagonals. Hence, the centre of mass is at ORIGIN . option (a) is correct. |
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| 6276. |
magnetic field in center of an arc |
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| 6277. |
whatis diffexencecand b |
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Answer» Indirect current(DC), the electric charge (current) only flows in one direction. Electric charge inalternating current(AC), on the other hand, changes direction periodically. The voltage inACcircuits also periodically reverses because the current changes direction. thnx ACandDCare different types of voltage or current used for the conduction and transmission of electrical energy. ... According to Karl K. Berggren, professor of electrical engineering at MIT, the fundamentaldifference between ACandDCis the direction of flow.DCis constant and moves in one direction. |
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| 6278. |
on obfect is placed at soim conconemissor whose to cal length is 60cm,find the position of the inageand its magnification, a |
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Answer» u=-30cmf=-60cm1/v + 1/u = 1/f1/v + 1/-30 = 1/-601/v = -1/60 + 1/301/v = -1/60 + 2/601/v = 1/60v = 60cm |
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| 6279. |
4. A bullet is fired with a velocity u making an angle of 60°with the horizontal plane.he horizontal component of the velocity of the bullet when it reaches the maximumeights isWB-JEE 2016 |
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Answer» the horizontal component of velocity always remains constant which is equal to ucos60=u/2 |
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| 6280. |
10. A 100 W bulb B,, and two 60 Wbulbs B, and B, are connected to a 250 V source, as shownin the figure. Now W, W2 and W, are the output powers of the bulbs B,, B2 and B3, respectively.Then.B2B3s0250 V |
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| 6281. |
What is electrical resistivity? Derive its SI unit In a series electrical circuitcomprising a resistor made up of a metallic wire, the ammeter reads 100 mA. If thelength of the wire is doubled, how will the current in the circuit change? Justify youranswer |
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Answer» 1)Electrical resistivity is a fundamental property of a material that quantifies how strongly that material opposes the flow of electric current. A low resistivity indicates a material that readily allows the flow of electric current. Resistivity is commonly represented by the Greek letter ρ. 2)A lowresistivityindicates a material that readily allowstheflowof electriccurrent. Resistivityis commonly represented bytheGreek letter ρ (rho). TheSIunit of electrical resistivityistheohm-metre (Ω⋅m).3)In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 100mA. The reading of the ammeter decreases to half when the length of the wire is doubled. Hence according to ohms law for a given potential difference, the current reduces to half. Please hit a like if you find the answer useful |
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| 6282. |
What is electrical resistivity ? Derive its SI unit. In a series electrical circuitcomprising a resistor made up of a metallic wire, the ammeter reads 100 mA. If thelength of the wire is doubled, how will the current in the circuit change? Justify youranswer. |
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Answer» Electrical resistivity is a fundamental property of a material that quantifies how strongly that material opposes the flow of electric current. SI unit: Siemens per meterIn a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 100mA. The reading of the ammeter decreases to half when the length of the wire is doubled. ... Hence according to ohms law for a given potential difference, the current reduces to half. |
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| 6283. |
7.Whatiselectricalresistivity?Inaserieselectricalcircuitcomprising a resistor made up ometallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to halt whem tlength c the wire is doubled. VWhy? |
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Answer» Electrical resistivity quantifies how strongly a given material opposes the flow of electric current. A low resistivity indicates a material that readily allows the movement of electric charge. Ammeter is used to measure electrical charge within a circuit. Since resistivity is directly proportional to the length of wire, when the length of wire is doubled, the wire retards the flow of charge twice more forcefully. So electric current decreases due to decrease in rate of flow of charge. Hence, the reading in ammeter decreases to half. |
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| 6284. |
What is electrical resistivity? Derive itscomprising a resistor made up of a metallic wire, the ammeter reads 100 mA. If thelength of the wire is doubled, how will the current in the circuianswerSI unit. In a series electrical circuitt change ? Justify your |
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Answer» 1)Electrical resistivity is a fundamental property of a material that quantifies how strongly that material opposes the flow of electric current. A low resistivity indicates a material that readily allows the flow of electric current. Resistivity is commonly represented by the Greek letter ρ(Rho). 2.)TheSIunit of electrical resistivityistheohm-metre (Ω⋅m).3.)In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 100mA. The reading of the ammeter decreases to half when the length of the wire is doubled. ... Hence according to ohms law for a given potential difference, the current reduces to half. Like if you find it useful |
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| 6285. |
(b)Study the following circuit and find out:(i) Current in 12 (2 resistor.(ii) Difference in the readings of A and A2, if any.24 立24 212立A2 |
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Answer» Difference will be 0as both are in seriesand same current will flow so no difference in reading Current = V/R= 12+12/6= 24/6= 4 amp As in parallel = 24*24/24+24= 12 ohmnow 12+12= 24 ohm |
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| 6286. |
Example 9.6 Light from a point source in air falls on a sphericalglass surface (n 1.5 and radius of curvature 20 cm). The distanceof the light source from the glass surface is 100 cm. At what positionthe image is formed? |
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| 6287. |
2. Radius of curvature of a mirror is 20 cm.What type of mirror is it? |
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Answer» Theradius of curvatureispositive(20 cm) fora concave mirror. |
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| 6288. |
1. The focal length of a convex mirror is 15 cm. Find its radius of curvature. |
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Answer» f=+15 cm because its a convex lens R=2F R=30cm |
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| 6289. |
Subtract: 18 m 2 dm 5 cm from 43 m 9cm. |
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| 6290. |
Q.37. In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter readsS A. The reading of the ammeter decreases to half when the length of the wire is doubled Why? |
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| 6291. |
What is electrical resistivity? In a series electrical circuit comprising a resistor made up ofa metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to halfwhen the length of the wire is doubled. Why?9 |
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| 6292. |
1wo wires are made of the same metal. The length of the first wire is halfthat of the second and its diameter is double that of the second wire. If equalloads are applied on both wires, find the ratio of increase in their lengths? |
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| 6293. |
1. Two wires are made of the same metal. The lengthof the first wire is half that of the second wire andits diameter is double that of the second wire. Ifequal loads are applied on the wires, find the ratio ofincrease in their lengths. |
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Answer» D¹*D¹:D²*D² 4:1 kese hua..? |
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| 6294. |
4 Find the minimum resistance that can bemade using five resistors each of 1/5 立NCERT Exemplar |
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Answer» Minimum resistance is observed when they are connected in parallel (you can take it as a thumb rule that the resistance of a group of resistors connected in parallel will be less then the minimum resistance present in the group) thus net resistance 1/R = 1/R1 + 1/R2 + .... and so it can be calculated to be 1/25 |
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| 6295. |
11.A 9cm long object is placed at 90cm from a spherical mirror perpendicular to its principal axis. A real imageis formed at 30cm from the mirror. What is the radius of curvature of the mirror? {Ans: - 45 cm) |
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Answer» Height of object ho = 9 cm Focal lenght of convex lens = ? cm Object distance u = -90 cm Image distance v = 30 cm Image height hi = ? Lens formula: 1/v - 1/u = 1/f 1/f = 1/v - 1/u = 1/30 + 1/(-90) = -1/45 ⇒ v = - 45 cm The image distance is 45 cm on the same side of the lens as the object. Magnification m = v/u = (-45)/(-90) = 1/2 = hi/ho = hi/9 ⇒ hi = 9/2 cm |
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| 6296. |
five resistance au connected in the shape of Cettein sigue alongside. Detu mine the totalresistance of the circuit.325 M215V |
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Answer» 3 ohm+7 ohm= 10 ohm शिरीष ज्वाइंट1/10 ohm + 1/10 ohm= parallel joint1/5 ohm= 1/rr= 5 ohmthan5 ohm + 5 ohm + 5 ohm =15 ohm15 ohm ans.. 15 ohm is the correct answer answer of following is 15 ohm. |
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| 6297. |
five resistance,each of resistance 1 ohm, are connected in parallel . what is the total resistance . |
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Answer» total resistance will be 1 ohm. if five resistance are connected parallel. five resistance each of resistance 1 ohm are connected in parallel :-1/1+1/1+1/1+1/1+1/1= 5/1=5 |
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| 6298. |
3.A wire has resistance of 8 ohm ifit's length is made half by folding.Calculate its resistance after the freeends are connected each other. |
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| 6299. |
A wire has resistance of 8 ohm. If its length ismade half by folding, find its resistance after thefree ends are connected to each other.(Ans : 2ohm) |
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Answer» Resistance R = 8 ohmlet's take length as Land cross section area as A here we need to use the formula R= rho(L/A) here rhostands for resistivity as length is halved and r the radius is increased by two R is reduced to R/4=8/4=2 ohm so new resistance will be 2 ohm |
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| 6300. |
61.The figure below shows currents in a part of electric circuiTrickyThe current i is(a) 1.7 amp(b) 3.7 amp(c) 1.3amp(d) 1 amp1 amp2 amp1.3 amp2 amp |
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Answer» In starting 2amp current from both branch get added and 4 amp current flows through straight line which divided into two branches of 1 amp and 3 amp. Which further divided into 1.3 amp and 1.7amp Thus right ans is 1.7 amp |
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