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.
| 851. |
5. An electric dipole is placed in a uniform electricfield E with its dipole moment P parallel tothe field. Find) the work done in turning the dipole till its dipolemoment points in the direction opposite to Ethe orientation of the dipole for which the torqueacting on it becomes maximum. |
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| 852. |
Define dipole moment of an electric dipole. Is it a scalar quantity or a vector quantity? |
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| 853. |
1. Derive an expression for the torque acting on anelectric dipole kept in a uniform electric field and(2017, 14)hence define electric dipole moment. |
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Answer» The electric dipole moment is a measure of the separation of positive and negative electrical charges within a system, that is, a measure of the system's overall polarity. |
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| 854. |
s.Calculate electric dipole moment of an electron and a proton 4.30 nm apart. 2 |
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Answer» Magnitude of charge of electron and proton q = 1.6 x 10⁻¹⁹ C Distance between electron and proton 2a = 4 . 3 nm = 4.3 x 10⁻⁹ m Electric dipole moment = q x 2a = 1.6 x 10⁻¹⁹ X 4.3 x 10⁻⁹ = 6.88 x 10⁻²⁸ cm p =q*dSo, for this one, being e=1.6E-19 C the elementary chargep = 1.6E-19 * 4.30E-9p = 6.88E-28 C.m Also, p is a vector, so you need its orientation, which is from the electron to the proton. |
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| 855. |
Answer the following questions.1. Two metal bearings of the same size and weight are pushed with the same force ontwo different surfaces. One travels further than the other. When can this happen? |
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Answer» It happens if one surface is smooth and the other is rough. On smooth surface bearing travels faster due to less friction. |
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| 856. |
50. Electric dipole moment is a(a) vector quantity(b) scalar quantity(c) ratio(d) none of these |
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Answer» Electric dipole moment is defined as the product of the magnitude of one of the charges and the distance between them. It is a a vector quantity .its direction is from the negative charge to the positive charge. It's unit is coulomb meter. Thanks |
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| 857. |
A car travels in the tx - direction on straight level road. For the first 4 sec of it's motion. The average velocity of the car is Vavg =6.25 m/s. How far does the car travel in 4 seconds? |
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Answer» Distance travelled = Avg. Velocity * Time = 6.25*4=25m |
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| 858. |
92. One travels a distance of 250 km by car in4 hours. What is the speed of the car?99. |
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Answer» Formula: Speed=distance/timeGiven:distance =250 kmtime= 4 hrsspeed =250/4=62.5 km/hr |
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| 859. |
28 A spring with spring constants k when compressaby 1 em, the potential energy stored is U, If itfurther compressed by 3 cm, then change in ilspotential energy is1) 3u(2) 9U(3) sU(4) 15U |
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Answer» U is proportional to x^2 |
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| 860. |
en measured by meler scouealuervon in the meauurunnt |
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Answer» A meter scale has a least count of 1 mm so change in reading is +- 1 mm Now 1.41/1.4= 1.007 % |
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| 861. |
The ratio of angular speeds of minute hand and hour hand of a watch is - |
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Answer» The minute hand of a clock covers 360 deg in 60 min or 6 deg/min. The hour hand of a clock covers 30 deg in 60 min or 0.5 deg/min.The ratio of the angular speed of the minute to the hour hand = 6/0.5 = 12:1 |
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| 862. |
What will be the equivalent resistance between thetwo points A and D109210621062vor &vor}109210021052(1) 102(2) 2012(3) 30 S2(4) 402 |
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| 863. |
Parallelogram law of vector addition. |
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Answer» Parallelogram Law of Vector Addition states that when two vectors are represented by two adjacent sides of a parallelogram by direction and magnitude then the resultant of these vectors is represented in magnitude and direction by the diagonal of the parallelogram starting from the same point in need to know it's working and derivation |
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| 864. |
parallelogram law of vector addition |
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Answer» ParallelogramLaw ofVector Addition states that when twovectorsare represented by two adjacent sides of a parallelogramby direction and magnitude then the resultant of these vectorsis represented in magnitude and direction by the diagonal of the parallelogramstarting from the same point. |
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| 865. |
and prove parallelogram law of vector additionand determine magnitude and direction of resultantvector. |
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| 866. |
State and prove parallelogram law of vector additionand determine magnitude and direction of resultantvector.5) |
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| 867. |
*Q.16.State and prove parallelogram law ofvectors addition and determine magnitudeand direction of resultant vector. |
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Answer» If two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of two vectors is given by the vector that is a diagonal passing through the point of contact of two vectors. |
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| 868. |
head-tail method of vectoraddition also called Trianglelaw of vector addition.yes or not |
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Answer» Yes. It is called Triangle law of vector addition. |
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| 869. |
3)What is triangle law of vector addition? |
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Answer» Triangle law of vector additionstates that when twovectorsare represented by two sides of atrianglein magnitude and direction taken in same order then third side of the third side of thattrianglerepresents in magnitude and direction the resultant of thevectors. |
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| 870. |
20.A point charge q is situated at a distance r from one end of a thin conducting rod of lengthL having a charge Q (uniformly distributed along its length). The magnitude of electricforce between the two, is ..........X2kqQkqQkQ ID 7172kqQr(r +(C) Pir-L)r(r + LNr+L)distance of |
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| 871. |
Explain CPU |
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Answer» A central processing unit (CPU) is the electronic circuitry within a computer that carries out the instructions of a computer program by performing the basic arithmetic, logical, control and input/output (I/O) operations specified by the instructions. |
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| 872. |
Six charges, three positive and three negative of equamagnitude, are to be placed atthe vertices of a regular hexagonsuch that the electric field at Ois double the electric field when uonly one positive charge of samemagnitude ia placed at R. Whichof the following arrangements ofcharges is possible for P, Q, RS, T and U respectively ?[IIT 2004 |
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| 873. |
\left. \begin{array} { l l } { ( 10 / 3 ) m } & { ( 2 ) } \\ { 6 m } & { ( 4 ) } & { 4 m } \end{array} \right. |
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Answer» Use the formula for displacement in nth second Sn=u+a/2(2n-1) here u=0 as it is starting from rest, a=4/3m/s^2 S3=0+4/6(2(3)-1) S3=4/6(5) S3=20/6 S3=10/3 m |
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| 874. |
what the cestHa ena |
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Answer» energy =power *time E=400*8=3200 Wh/day for 30 days E=3200*30=96000Wh=96 kWh so total cost=96*3= Rs 288 |
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| 875. |
(D) π4. Acar Ale gving noth cest o 80her car B gg outh coet with a vey0 var Theveloolty d A relative to B makes an angde with the tnorth equal to75. Ahos |
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Answer» A and B make the base and the height of a right-triangle, of which the hypotenuse is sqrt(8²+6²)v or 10 kmphThe resultant is 10 at an angle 53.13 degree to BA or B are angled to the N-E and N-S, or at 45 deg. to the vertical axis, hence the resultant is at 53.13-45 or 8.13 degree to vertical axis. tanθ for 8.13 degree = 0.14285714285714 and the same value is tan^-1(1/7) . option D. |
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| 876. |
uist starts frown the centre O of a creular park of radhus 1 km, reaches the edge Pthe park, then cycles along the ctrcumference, and returns to the centre along 00s shoen h c 4.2t. tf the round trip takes 10 min, what is the (al net displacement.) avenaue rekeity, annd ld average speed of the cyelistFo. 4.21 |
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| 877. |
12. A resistance R draws power P when connected toan AC source. If an inductance is now placed inseries with the resistance, such that theimpedance of the circuit becomes Z, the powerAIPMT-2015drawn will be2(2) P(4) P Z(3) P |
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Answer» thanks |
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| 878. |
the given figure, AD = CB, AB = CD and EF bisects BD at G. Prat G is the mid-point of EF2ncept İSOSCELES TRIANGLESem 1: If two angles of a triangle are equal, then sides oppositem are also equal. |
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| 879. |
Very Short Answer Type uues ons " " … .. … . .. .......。boty is moving with an unilorm velecity 40 ms. Find its accelsmation afher 10 secsnds. |
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Answer» Initial velocity=40m/sfinal velocity=40m/stime=10 seconds v=u+at40=40+10a10a=0a=0m/s |
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| 880. |
l I ype Question:No.4:- Differentiate between a Bar Magnet and an Electromagnet. |
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| 881. |
showsmbolssindetanthediaandritingl letter, jouleshouldJ' forangathavsmalletcsed inthetheantwoFig. 1.1 Measurementof lengthtwo diametrically opposite points of the |
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Answer» You need the measurements for triangle? |
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| 882. |
1nFig 7.51, PR > PQ and PS bisects L QPR. Provethat LPSR PSQ. |
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| 883. |
A roller is made by joining together twocones at their vertices O. It is kept on tworails AB and CD which are placedasymmetrically (see figure), with its axisperpendicular to CD and its centre O atthe centre of line joining AB and CD (seefigure). It is given a light push so that itstarts rolling with its centre O movingparallel to CD in the direction shown. Asit moves, the roller will tend to: |
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Answer» If we take ‘r’ as the distance ofIAORIAORfrom the axis of rotation, then ‘r’ decreases on left side as the object moves forward So, for leftv=ωr′<ωrv=ωr′<ωr(for right point) So, the roller will turn to the left as it moves forward. |
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| 884. |
Given that C A+ B and C makes an angle α with andand ß with B. Which of the following options is correct?(1) α cannot be less than β (2) α<b, if A <B(3) α < β, if A >B.Ã(4) α < β, if A-B |
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Answer» C is the resultant of vectors A and B. The resultant is closer to the vector with greater magnitude. Hence, option C is correct. |
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| 885. |
an athlete completes one round of a circular track of diameter 200m in40sec.what will be the displacement at the end of 2minutes 20sec |
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Answer» Displacement is the shortest distance possible between initial and final position.As the athlete is in circular motion , displacement is zero. [ THis is because , the initial and final position is the same ] Distance covered in 1 round = Circumference of the circle Circumference of the circle = 2πr Diameter = 200 mRadius = 100 m Distance covered in 1 round[ 40 sec ] = 2 × 22/7 × 100 = 628.57 m Distance covered in 1 sec = 628.57 ÷ 40 = 15.71 m 2min 20 sec = 140 secDistance covered in 140 sec = 15.71 × 140 = 2199.4 m |
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| 886. |
an athlete completes one round of a circular track of radius R in 40 minutes what will be its displacement at the end of 2 minute 20 seconds |
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Answer» it's 40 min or sec Total time duration in consideration : 2 min 40 secs = 160 secs No of complete rounds in this time = 160/40 = 4 rounds Therefore he will be exactly at the point where he started. Hencehis displacement will be 0, although he will cover a distance of 4 (2πR) |
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| 887. |
What is a cyclic process? What is the change in internal energy of a system after it completes one cycle of such aprocess? |
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| 888. |
9. An athelete completes one round of a circulartrack of radius R in 40 sec. What will be hisdisplacement at the end of 2 min. 20 sec. |
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Answer» 2min 20 sec=140 sec140/40=3.5he will complete 3 and half roundsdisplacement is the shortest distance between the initial and the final points. so displacement will be 2R. 2min 20sec = 140seche completes one round in 40 secIn 140 sec, he completes 3and half round , So displacement = 2R |
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| 889. |
For ensuring disspation of same energy in all threeresistors (Ri, R: , Rs) connected as shown in figure,their values must be related as:(2005)R,R2R3Vin2.a. R1 = R2 = R3and R' = 4Rc. R2 R, and R, R 4 |
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| 890. |
An athlete completes one round of a cireular track of diam200 m in 40 s. What will be the distance covered anddisplacement at the end of 2 minutes 20 s? |
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| 891. |
0.1. An athlete completes one round of a circular track of diameter 200 min 40s. What will be the distancecovered and the displacement at the end of 2 minutes 207Tamula cds |
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Answer» Ans: Time taken = 2 min 20 sec = 140 sec. Radius, r = 100 m. InTime to complete one round = 40 sec So, Round completed in 140 sec = 140 ÷ 40 = 3.5 (three and a half) round. Or, distance covered = 2 x 22/7 x 100 x 3.5 = 2200 m. At the end, the athlete cavers half round and will be at a distance = diameter of circle. ie displacement = 200m. |
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| 892. |
athlete completes one round of a circular track of dia200 m in 40 s. What will be the distance covered and thedisplacement at the end of 2 minutes 20 s? |
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| 893. |
An athlete completes one round of a circular track ofradius R in 40 S. His displacement at the end of2 minutes will be:(A) 27TR(C) 2R(B) 6TR(D) zero |
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| 894. |
1. What are coherent sources of light? |
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Answer» Two sources are said to be coherent if they have exactly same frequency, and have zero or constant phase difference. Most of the light sources around us - lamp, sun, candle etc are combination of multitude of incoherent sources of light |
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| 895. |
Two cars A and B are at positions 50 m and100 m from the origin at t 0, They startsimultaneously with constant velocities 10 m/s and 5 m/s respectively in the same direction,Find the time at which they will overtake oneanother |
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Answer» Let S be the distance from the point car A start to the distance where they both meet and t be the time taken to by both car to meetCar A:S=10tCar B:S-100 = 5t dividing both equation we get2S-200=SS=200 ie... they both meet at distance of 200m from the start point of A ... which implies the distance from origin is 200+100 = 300mtime taken =>S=10t=20 secondsthey take 20 seconds to meet |
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| 896. |
andTwo cars start in a race with velocities u,u, and travel in a straight line with accelerationa' and b. If both reach the finish line at thesame time, the range of the race is |
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| 897. |
A point is situated at 6.5 cm and 6.65 cm fromtwo coherent sources. Find the nature ofillumination at the point if wavelength of light is5000 A(Ans: bright) |
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| 898. |
5. The expression for centripetal force depends upon massof body, speed of the body and the radius of circular path.Find the expression for centripetal force2mvmv(b) F =2r22, ,2mv(c) F =(d) F =2r |
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Answer» F=( mv^2)/r Where F=centripetal force m=mass of the object v=velocity r=distance from the centre uhhh. sry but i did'nt got it!! |
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| 899. |
(a) Define buoyant force. Name two factors on which buoyant force dependsb) What is the cause of buoyant force ?(c) When a boat is partially immersed in water, it displaces 600 kg of water. How much is the buacting on the boat in newtons ? (g 10 m s) |
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Answer» thank you for your help..☺ |
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| 900. |
State the factors on which buoyant force depends. |
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Answer» Buoyancy of a body depends on the following factors: Volume of the body submerged in the liquid or volume of the liquid displaced.Densityof the liquid. Acceleration due to gravity. |
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