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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.
| 1251. |
For the situation shown in the figure, select the correct statements (s). |
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Answer» Potential of the conductor is `q//4piepsilon_0(d+R)` As no charge is present inside the conductor potential at any point inside the conductor is same as that of the potential or conductor. So, potential of the conductor `=` potential at the center `=V_(q)=V_("induced charges")` `V_("conductor")=(q)/(4piepsilon_(0)(d+R))+0` As the total induced charge at conductor's surface is equal zero and HENCE to the potential at center due to the INDUCTOR charge. For point `B`, `V_("conductor")=V_("at point"B)=V_(q)+V_("induced charges")` or `V_("induced charges")=(q)/(4piepsilon_(0)(d+R))-(q)/(4piepsilon_(0)d)` `=(-qR)/(4piepsilon_(0)d(d+R)` |
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| 1252. |
Assertion (A) The effective resistance of the network between P and Q is (4)/(5)r.Reason (R ) Symmetry can be applied to be network with respect to centre. |
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Answer» If both ASSERTION and REASON are TRUE and Reason is CORRECT EXPLANATION of Assertion. |
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| 1253. |
Dr. Zakir Hussain entered the office of the president in a spirit of |
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Answer» TOTAL faith |
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| 1254. |
From a supply of identical capacitors rated 8muF, 250V , the minimum number of capacitors required to form a composite 16muF, 1000Vis : |
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Answer» 2 |
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| 1255. |
To find the resistance of a galvanometer by the half deflection method the following circuit is used with resistances R_1 =9970 Omega , R_2 = 30 Omegaand R_3= 0. The deflection in the galvanometer is d. With R_3, =107Omegathe deflection changed to (d) /(2) the galvanometer resistance is approximately: |
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Answer» `107Omega ` |
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| 1256. |
Two amplifiers are connected one after the other in series (cascaded) . The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20 . If the input signal is 0.01 volt, calculate the output are signal. |
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Answer» SOLUTION :`A_(1) = 10 , A_(2)= 20 , V_(i) = 0.01 V , V_(o) = ?` `A_(V) = A_(1) xx A_(2) = 10 xx 20 = 200` `A_(V) = (V_(o))/(V_(i)) , V_(B) = A_(V) xx V_(i) = 200 xx 0.01 = 2 V` |
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| 1257. |
Carbon, silicon and germanium have four valence electrons each. These are characterisedby valence and conduction bands separated by energy band gap respectively equal to (E_(g))_(C ), (E_(g))_(Si) and (E_(g))_(Ge). Which of the following statements is true? |
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Answer» `(E_(g))_(Si) lt (E_(g))_(Ge) lt (E_(g))_(C )` Among the given ELEMENTS the band gap ofSi and Ge is 5.4 eV, 1.1 eV and 0.7 eV respectively. Hence, OPTION (c ) is CORRECT. |
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| 1258. |
A projectile has the maximum range of 400m. If it is pushed up smooth inclined plane of angle 30^@ with the same speed, the distance travelled by it along the inclined plane will be |
| Answer» ANSWER :C | |
| 1259. |
Two metallic solid spheres of radii R and 2R are changed such that both of them have same charge density sigma. If the spheres are located for away from each other and connected by a conducting wire, the new charge density on bigger sphere is : |
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Answer» `5 sigma` |
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| 1260. |
1 curie is equal to |
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Answer» `3xx10^(10)BQ` |
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| 1261. |
The particle P of mass m is attached to two light, rigid rods AP and BP of length l each. A and B are hings on a fixed vertical axis. The system APB can rotate freely about this axis. The angle ABP = the angle BAP = theta. The tensions in AP and BP are T_(1) and T_(2) respectively. When the system is made to rotate about AB with angular velocity omega, which of the following is not correct? |
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Answer» `T_(1)` will always be greater than `T_(2)`. |
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| 1262. |
The particle P of mass m is attached to two light, rigid rods AP and BP of length l each. A and B are hings on a fixed vertical axis. The system APB can rotate freely about this axis. The angle ABP = the angle BAP = theta. The tensions in AP and BP are T_(1) and T_(2) respectively. The system is now rotated by 90^(@) so that must be imparted to P, normal to the plane of the figure, such that it moves in a complete circular path in a vertical plane with AB as the axis? |
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Answer» `2sqrt(gl sin THETA)` |
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| 1263. |
The drift speed of an electron in a metal is of the order of |
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Answer» `10^(-13)m//s` |
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| 1264. |
Double - convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of cavature required if the focal length is to be 20 cm? |
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Answer» Solution :Here, `n=1.55, f=+20 cm` and `|R_(1)| =|R_(2)| = R` SAY) MOREOVER, as per sign convention, for a CONVEX lens `R_(1)` is +ve and `R_(2)` is `-ve` hence, `1/f =(n-1) [1/R -1/(-R)] =(n-1) (1/R + 1/R) =(2(n-1))/R` `rArr R =2(n-1) f = 2 xx (1.55 -1) xx 20 = 2 xx 0.55 xx 20 = 22 cm` |
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| 1265. |
A wire of resistance 'R' is cut into four equal parts and they are bundled together side by side to form a thicker wire. The resistance of the bundle is |
| Answer» Answer :A | |
| 1266. |
Who is the author of "His First Flight"? |
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Answer» LIAM O' Flaherty |
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| 1267. |
In a semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called :- |
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Answer» ACCEPTOR |
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| 1268. |
Which of the following system of unit is also called a British system? |
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Answer» CGS |
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| 1269. |
A square surface of side L meter in the pane of the paper is placed in a uniform electric field E (volt/m) acting along the same plane at an angle theta with the horizontal side of the square as shown in figure. The electric flex linked to the surface, in units of volt/m is |
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Answer» `EL^2` |
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| 1270. |
If me and m,are magnifications of objective and microscope, then magnification of microscope is ....... |
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Answer» `m_0+m_e` |
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| 1271. |
Distinguish between half-wave and full-wave rectifiers. |
| Answer» Solution :`{:(,"Half wave rectifier","Full wave rectifier"),(1.,"Single diode is used. ",1."TWO diodes are used"),(2.,"Only halfwave is rectified",2."Full wave is rectifed"),(3.,"Rectifier efficiency "eta= ( 0.406R_(L))/(r_(f)+R_(L)),3."Rectifier efficiency "eta= (0.812R_(L))/(r_(f)+R_(L))),(4.,"Efficiency of half wave rectifier is 40.6%",4."Efficiency of full wave rectifier is 81.2%"),(5.,"OUTPUT is discontinuous and pulsative.",5."Output is continuous and pulsative."):}` | |
| 1272. |
Two sound waves of wave length 0.8m and 0.81m produces 8 beats per second in air. The respective frequencies of the two waves are |
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Answer» 320HZ and 312Hz |
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| 1273. |
Only draw a figure which explains the reflection and refraction of light. |
Answer» SOLUTION :
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| 1274. |
A Cube made of material having density of 0.9 xx 10^(3) kg m^(-3) floats between water and a liquid of density 0.7 xx 10^(3) kg//m^(3) which is immiscible with water. What part of cube is immersed in water. |
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Answer» Solution :Let a be the SIDE of the cube and x be the length dipping in WATER and (a -x) in the liquid `thereforea^(3)xx0.9xx10^(3)g=a^(2)xx x xx 1000g+a^(2)(a-x)xx0.7xx10g` `axx0.9=x+(a-x)0.7` `0.3x=0.2a` `x/a=2/3` Thus correct choice is (c). |
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| 1275. |
What is Peltier effect ? |
| Answer» SOLUTION :When an electric current is passed through a circuit of a thermocouple, HEAT is EVOLVED at one junction and ABSORBED at the other junction. This is known as Peltier effect. | |
| 1276. |
A point sound source is situated in a medium of bulk modulus 1.6 xx 10^5 N//m^2. An observer standing at a distance 10 m from the source writes down the equation for the wave asy = A sin (15pi x - 6000 pi t). Here y and x are in metres and t is in second. The maximum pressure amplitude tolerable by the observer's ear is 24 pi Pa. Find : the density of the medium. |
| Answer» SOLUTION :`1 kg//m^3` | |
| 1277. |
What is the relation between pole strength m , area of cross-section A and the intensity of magnetization M ? |
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Answer» M = mA |
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| 1278. |
A radioactive nucleus emits beta particle. The parent and daughter nuclei are |
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Answer» isotopes |
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| 1279. |
The magnetic flux through a coil perpendicular to its plane is varying according to the relation phi_(B)= (5t^(3) +4t^(2) + 2t - 5) weber. Calculate the induced current through the coil at t = 2 second. The resistance of the coil is 5 Omega. |
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Answer» Solution :`phi=5t^(3)+4t^(2)+2t-5 IMPLIES |E| =(d phi)/(DT)=15t^(2)+8t+2` at t=2sec, e = -78V `implies e=iR=15t^(2)+8t+2` `I xx 5=15 xx 4+8 xx 2 implies i=15.6A` |
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| 1280. |
Which of the following physical quantities represent the dimensions of b/a in the relaction 1, where p is power, x is distance and t is time. |
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Answer» Power |
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| 1281. |
Conductors used in high-voltage equipments are designed in such a way that there are no sharp edges or points on the conductors. Give a reason for this precaution taken while designing the conductors. |
| Answer» Solution :A sharp point or EDGE on a charged conductor would IMPLY a large electric field in that REGION. An electric discharge can POSSIBLY take place at that sharp point or edge. | |
| 1282. |
(A) : It is necessary to use satellites for long distance TV transmission. (R) : Television signals are not properly reflected by the ionosphere, therefore reflection is effected by satellites. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 1283. |
The magnetic moment vectors mu_s and mu_l associated with the intrinsic spin angular momentum vecS and orbital angular momentum vecl respectively, of an electron are predicated by quantum theory (and verified experimentally to a hing accuracy ) to be given by : mu_s = -((e)/(m)) vecS , mu_I =-((e)/(2m))vecI Which of these relations is in accordance with the result expected classically ?Outline the derivation of the classical result . |
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Answer» Solution :Of the two RELATIONS GIVEN here, the relation `mu_I = -((e)/(2M)) vecI` is in accordance with the classical theory. According to basic definitions, magnetic moment due to orbital motion of electron `mu_I = I A (e/T) * pi R^2 , ` where T = time of revolution of electron and R = radius of orbit and `|vecl| = MVR = m * (2pi R)/(T)* R = (m2pi R^2)/(T)` `therefore (mu_l)/(l) = e/(2m)` As charge on electron is negative , `mu_l and vecl` must be ANTIPARALLEL and hence we can write `mu_l =-((e)/(2m))* vecl` |
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| 1284. |
Which of the following is dimensionless (a) Boltzmann constant(b) Planks constant (c) Poissons ratio(d)Relative density |
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Answer» Both A & B |
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| 1285. |
A : Workdoneby thegravitational force is positivewhen the twopoint massesare brought from infinityto any two pointsin space . R : Gravitationalpotential energyincreases during the above process . |
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Answer» If both Assertion & Reasonare true . Andthe reasonis the CORRECT explanationof theassertion , then mark (1) |
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| 1286. |
A junction diode is forward biased. If the forward voltage is increased from 0.4 V to 0.7 V the forward current changes by 1.5 mA. What is the forward resistance of the junction diode ? |
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Answer» `50OMEGA` |
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| 1287. |
Find the momentum of an electron having wavelength 2A^(0)(h=6.62xx10^(-34)Js) |
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Answer» `1xx10^(-24)kg ms^(-1)` |
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| 1288. |
If n moles of a gas of molar mass M contains Nmolecules each of mass m_(0) and N_(A) is the Avogadro constant, then N and M are respectively equal to |
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Answer» `nN_(A),m_(0)N_(A)` |
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| 1289. |
Figure shows electric field lines in which an electric dipole vecp is placed as shown. Which of the following statements is correct? |
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Answer» The dipole will not EXPERIENCE any FORCE. |
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| 1290. |
An n-p-n transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of 4 mA. The terminal of a 8 V battery is connected to the collector through a load resistance R_L and to the base through a resistance R_B The collector-emitter voltage V_(CE)= 4 V, base-emitter voltage V_(BE)= 0.6 V and base current amplification factor B_(d.c) = 100. Calculate the values of R_L and R_B |
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Answer» Solution :Potentialdifferenceacross ` R_L` ` (##AKS_NEO_CAO_PHY_XII_V02_C_C06_SLV_039_S01.png" width="80%"> ` = 8 VV_(CE)` `= 8V - 4V ` `= 4 V ` Now` I_CR_L= 4V ` `R_L = (4)/( 4 x 10^(-3)) = 10^3 Omega= 11 kOmega ` Furtherfor baseemitterequation, ` V_(C C )= I_BR_B+V_(BE)` or ` I_B R_B `= potential differenceacross`R_B ` ` V_(C C ) =- V_(BE)= 8-06= 7.4V ` Again ` ,I_B= (I_C )/(beta) = ( 4 xx 10 ^(-3))/( 100 )= 4 xx 10^(-5) A` ` thereforeR_B= ( 7.4)/( 4 xx 10^(-5)) = 1.85 xx 10^5Omega= 185k Omega ` |
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| 1291. |
The activity of a radioactive source initially is 1600 Bq and after 8 s the activity falls to 100 Bq. What is its activity at time t=6s ? |
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Answer» 400 Bq |
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| 1292. |
What is the name for information sent from robot sensors to robot controllers? |
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Answer» temperature |
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| 1293. |
In the Example 6 find the force per unit length on the wire if the wire is placed south to north? |
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Answer» |
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| 1294. |
Five lumen/watt is the luminous efficiency of a lamp and its luminous intensity is 35 candela . The power of the lamp is |
| Answer» Answer :c | |
| 1295. |
How many types of wavefronts are there ? |
| Answer» SOLUTION :THREE, SPHERICAL, cylindrical and plane WAVEFRONTS, | |
| 1296. |
The figure shows a constant deviation prism ABCD. The incident ray is PQ and the emergent ray is ST. Although it is made up of one piece of glass, but it is equivalent to two 30^(@)-60^(@)-90^(@) prism and one 45^(@)-45^(@)90^(@) prism.The angle theta_(1) is the angle of incidence on face AB. The path of the ray inside the prism is indicated in the figure. for this prism mu=2sintheta_(1). Q. The total deviation of incident ray when it emerges out of the prism is |
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Answer» `60^(@)` `mu=(sintheta_(1))/(sinr),sintheta_(1)=(mu)/(2)impliessinr=(1)/(2),r=30^(@)` considering ABC as a triangular prism `r+r^(')=A` `30^(@)+r^(')=75^(@)impliesr^(')=45^(@)` THUS ray is incident on AB at `45^(@)` it is reflected at the same angle on to CD. From the figure `angleRSD=360^(@)-(45^(@)+135^(@)+60^(@))=120^(@)` (considering the quadrilateral) `because` angle of incidence at `CD=120^(@)-90^(@)=30^(@)` `impliestheta_(1)=theta_(2)` `because` total deviation of ray `=(theta_(1)-30^(@))+(180^(@)-90^(@))-(theta_(2)-30^(@))=90^(@)` |
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| 1297. |
A five-pointed regular star as shown in Fig 26.3. has been soldered together from uniform wire. The resistance of the section EL is r. Find the resistance of the section FL. |
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Answer» |
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| 1298. |
A shell of mass m is fired from a cannon at angle q to horizontal with a velocity V.The shell at the heighest point breaks into two fragments having masses in the ratio 2:3. The lighter fragment has a velocity zero immediately after explosion and falls vertically downward. the velocity of the other fragment just after explosion is |
| Answer» Answer :D | |
| 1299. |
A soap film of 5xx10^(-15)cm thick is viewed at an angle of 35^(@) to the normal. Find the wavelength of light in the visible spectrum which will be absent from reflected light (mu=1.3). Given wavelength range of the visible spectrum is 3900 Å to 7800 Å. |
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Answer» |
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| 1300. |
(a) Define electric dipole moment. Is it a scalar or a vector ? Derive the expression for the electric field of a dipole at a point on theequatorial plane of the dipole. (b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero. |
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Answer» Solution :(a) Electric dipole MOMENT : The strength of an electric dipole is measured by the quantity electric dipole moment. Its magnitude is equal to the product of the magnitude of either charge and the distance between the two charges. Electric dipole moment, `p = q xx d` It is a vector quantity. In vector form, it is written as `vec(p)=q xx vec(d)`, where the direction of `vec(d)` is form negative charge to positive charge. Electric field of dipole at points on the equatorial PLANE : The magnitude of the electric field due to two charges + q and -q are given by `E_(+q)=(q)/(4pi E_(0))(1)/(r^(2)+a^(2))""...(i)` `E_(-q)=(q)/(4pi E_(0))(1)/(r^(2)+a^(2))""...(ii)` `therefore""E_(+q) = E_(-q)` The direction of `E_(+q) and E_(-q)` are SHOWN in figure :  The COMPONENTS normal to the dipole axis cancel away. The components along the dipole axis add up. `therefore` Total electric field `E=-(E+(+q)+E_(-q))cos theta hat(p) rArr E =-(2qa)/(4 pi in_(0)(r^(2)+a^(2))^(3//2))hat(p)""...(iii)` At large distance `(r gt gt a)`, this reduce to `E = -(2qa)/(4 pi in_(0)r^(3))hat(p)""...(iv)` `vec(P)=q xx 2 vec(a) hat(p)` `E = (-vec(p))/(4 pi in_(0)r^(3))(r gt gt a)` (b) Potential at all points in equatorial plane is zero everywhere `V_(net) = V_(A) + V_(B)` `=(-Kq)/((r^(2)+a^(2)))+(Kq)/((r^(2)+a^(2)))=0` 
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