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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.
| 201. |
If the electric field is given by `vec(E) = 8 hat(i) + 4 hat(j) + 3 hat(k) NC^(-1)`, calculate the electric flux through a surface of area `100 m^(2)` lying in X-Y plane. |
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Answer» Correct Answer - `300 Nm^(2) C^(-1)` `vec(E) = 8 hat(i) + 4 hat(j) + 3 hat(k) NC^(-1), vec(S) = 100 hat(k) m^(2)` `phi_(E) = vec(E). Vec(S) = (8 hat(i) + 4 hat(j) + 3 hat(k)) - (100 hat(k))` `= 300 Nm^(2) C^(-1)` |
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| 202. |
A rectangular surface of sides `10 cm and 15 cm` is palaced inside a uniform electric field fo `25 Vm^(-1)`, such that normal to the surface makes an angle of `60^(@)` with the direction of electric field. Find the flux of electric field through the rectangular surface. |
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Answer» Correct Answer - `1.8xx10^(-1) Nm^(2) C^(-1)` Here, `E = 25 Vm^(-1)`, `s = 10xx15 cm^(2) = 150xx10^(-4) m^(2)` `theta = 60^(@) , phi = ?` `phi = Es cos theta = 25xx150xx10^(-4) cos 60^(@)` `phi = 1.8xx10^(-1) Nm^(2) C^(-1)` |
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| 203. |
A point charge of 3 mC is situated in air at the centre of a sphere of radius 10 cm. Find the T.N.E.I. through a portion of the surface of the sphere which subtends an angle of `pi//2` steradian at its centre.A. 0.375 mCB. 0.573 mCC. 0.735 mCD. 0.473 mC |
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Answer» Correct Answer - A `q=3mC,R=10cm,domega=pi/2" steradian",T.N.E.I.=?` `T.N.E.I.=q/(4pi)xxdw` `=3/(4pi)xxpi/2=3/8=0.375mC` |
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| 204. |
A spherical Gaussian surface encloses a charge of `8.85xx10^(-8) C` (i) Calculate the electric flux passing through the surface (ii) If the radius of Gaussian surface is doubled, how would the flux change ? |
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Answer» Correct Answer - `10^(4) Nm^(2) C^(-1)`, No chang (i) `phi_(E) = (q)/(in_(0)) = (8.85xx10^(-8))/(8.85xx10^(-12)) = 10^(4) Nm^(2) C^(-1)` (ii) flux will remain the same as chagre enclosed by the surface is same as in the case (i). |
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| 205. |
A cubical Gaussian surface encloses electric flux of `30C` per unit permittivity of a charge ,the electric flux through each face of the cube per unit permittivity isA. 30 CB. 5 CC. 1 CD. 0.5 C |
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Answer» Correct Answer - B The electric flux through each surface of the cube is given by, `phi=1/6" total flux"=30/6=5` |
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| 206. |
The charges on two spheres are `+7muC` and `–5muC` respectively. They experience a force F. If each of them is given an additional charge of `–2muC`, the new force of attraction will beA. FB. `F//2`C. `F//sqrt3`D. `2F` |
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Answer» Correct Answer - A |
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| 207. |
A Gaussian surface in the form of a cube is centred on a point charge Q. What is the flux through one face of the cube? |
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Answer» Since the point charge is at the centre of the cube, the electric flux through each face is the same. Therefore, the electric flux through one face is1/6 th of the electric flux originating from or terminating at the point charge. If ε is the permittivity of the medium, the electric flux through one face = \(\cfrac{Q}{6ε}\). |
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| 208. |
A particle of mass `10^(-4) kg` and charge `5muC` id thrown at a speed of `20m//s` against a uniform electric field of strength `2xx10^(5) NC^(-1)`. How much distance will it travel before coming to rest momentarilly ? |
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Answer» Here, `m = 10^(-4) kg ,` `q = 5 muC = 5xx10^(-6) C , v = 20 ms^(-1),` `E = 2xx10^(5) N//C` Force on charged particle `F = qE = 5xx10^(-6)xx2xx10^(5) = 1N` As particle is thrown against the field, so `a = - (F)/(m) = - (1)/(10^(-4)) = - 10^(4) m//s^(2)` From `v^(2) - u^(2) = 2as` `:. (0)^(2) -(20)^(2) = 2 (-10^(4))s` `:. s = 0*023m` |
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| 209. |
Four point charges `+8muC,-1muC,-1muC` and , `+8muC` are fixed at the points `-sqrt(27//2)m,-sqrt(3//2)m,+sqrt(3//2)m` and `+sqrt(27//2)m` respectively on the y-axis. A particle of mass `6xx10^(-4)kg` and `+0.1muC` moves along the x-direction. Its speed at `x=+ infty` is `v_(0)`. find the least value of `v_(0)` for which the particle will cross the origin. find also the kinetic energy of the particle at the origin in tyhis case. Assume that there is no force part from electrostatic force. |
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Answer» Correct Answer - `(v_(0))_(min)=3m//s,K=3xx10^(4)J` |
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| 210. |
Careful measurements of the electric field at the surface of a box inidcates that the net outward flux through the surface of box is `60xx10^(3) Nm^(2) C^(-1)`. Find (i) the net charge inside the box ? (ii) If the net outward flux through the surface of box were zero, could you conclude that there were no charges inside the box ? Explain your answer. |
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Answer» (i) Here, `phi_(E) = 16xx10^(3) Nm^(2) C^(-1)` Acc. To Gauss theorem `phi_(E) = (q)/(in_(0))` So `q = phi_(E) xx in_(0) = 16xx10^(3) xx8.85xx10^(-12)` `= 0.14xx10^(-6) C = 0.14 muC` (ii) No, we can not say that there are no charges at all inside the box. We can only say that algebraic sum of charges inside the box is zero. |
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| 211. |
A bollow cylindrical box of length 1m and area of cross section `25 cm^(2)` is placed in a three dimensional co-ordinate system as shown in Fig, The electric field in the region is given by `vec(E) = 50 x hat(i)` , where E is in `NC^(-1)` and x is in meter. Find (i) Net flux through the cylinder (ii) Charge enclosed by the cylinder. |
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Answer» Here, `vec(ds) = 25 cm^(2)` `= 25xx10^(-4) m^(2)`, along X-axis `vec(E) = 50xx hat(i), phi = ?, q = ?` On the left end of the cylinder, `vec(E_(1)) = 50xx1 NC^(-1)` along X-axis `:. Phi_(1) = E_(1) ds cos 180^(@) = 50xx25xx10^(-4) (-1)` `= -0.125 NC^(-1) m^(2)` On the right end of the cylinder, ` vec(E_(2)) = 50xx2 = 100 NC^(-1)` along X-axis `:. phi_(2) = E_(2) = ds cos 0^(@) = 180xx25xx10^(-4) (1)` `=0.250 NC^(-1) m^(2)`. Net flux through the cyclinder `phi = phi_(1) + phi_(2) = 0.125 + 0.250` `= 0.125NC^(-1) m^(2)` `q = in_(0) phi = 8.85xx10^(-12) xx 0.125` `= 1.106xx10^(-12) C` |
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| 212. |
A metal wire is bent in a circle of radius 10 cm. It is given a charge `200 muC` which is spread on it uniformly. Calculate the electric potential at its center. |
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Answer» Here, `q = 200 muC = 2xx10^(-4) C` `r = 10cm = 0.10m` Electric potential at the center, `V = (1)/(4pi in_(0)) (q)/(r )` `= (9xx10^(9)xx2xx10^(-4))/(0.1) = 18xx10^(6)V` |
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| 213. |
Two similar very small conducting spheres having vharges `40 muC` and `-20 muC` are some distance apart. Now they are touched and kept at the same distance. The ratio of the initial to the final force between them is :A. `8 : 1`B. `4 : 1`C. `1 : 8`D. `1 : 1` |
| Answer» Correct Answer - A | |
| 214. |
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of `180.0 muC//M^(2)` (ii) Find the charge on the sphere. (ii) what is the total flux leaving the surface of the sphere ? |
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Answer» Here, `R = (2.4)/(2) = 1.2 m` `sigma = 180.0 muC//m^(2) = 180xx10^(-6) Cm^(-2)` (i) Charge on the sphere `q = 4pi R^(2) sigma` `= 4xx3.14xx(1.2)^(2)xx180xx10^(-6)` `= 3.53xx10^(-3) C` (ii) Electric flux, `phi_(E) = (q)/(in_(0)) = (3.53xx10^(-3))/(8.85xx10^(-12))` `= 3.9xx10^(8) Nm^(2) C^(-1)` |
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| 215. |
Two small conducting spheres of equal radius have charges `+ 10 muC` and `-20 muC` respectively and placed at a distance R from each other. They experience force `F_(1)`. If they are brought in contact and separated to the same distance, they experience force `F_(2)`. The ratio of `F_(1)` to `F_(2)` isA. `1 : 8`B. `-8 : 1`C. `1 : 2`D. `-2 : 1` |
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Answer» Correct Answer - B |
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| 216. |
What equal charges would have to be placed on earth and moon to neutralize their gravitational force of attraction? Given that mass of earth `= 10^(25) kg` and mass of moon `= 10^(23) kg` |
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Answer» Correct Answer - `8.6xx10^(13)C` Use `(Gm_(1) m_(2))/(r^(2)) = k (qq)/(r^(2))` |
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| 217. |
Electrical potential `V` in space as a function of co-ordinates is given by `,V=(1)/(x)+(1)/(y)+(1)/(z)` then the electric field intensity at `(1,1,1)` is given by:A. `hati+hatj+hatk`B. `hati+hatj-hatk`C. `hati-hatj+hatk`D. `-hati+hatj+hatk` |
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Answer» Correct Answer - B `V=(1/x+1/y+1/z)` `vec(E)=-(delV)/(delx) hati-(delV)/(dely) hatj-(delV)/(dz) hatk=1/(x^(2)) hati+1/(y^(2)) hatj+1/(z^(2)) hatk` `:. vec(E)(1,1,1)=hati+hatj+hatk` |
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| 218. |
The p.d. across the capacitance of `2muF` in the figure along with is A. 10 VB. 60 VC. 28 VD. 56 V |
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Answer» Correct Answer - B From figure, equivalent capacity of `3muF, 6muF and 3muF` condensers is `C_(p)=12muF`. Now, `C_(p)` and condenser of `2muF` are in series. Thus, p.d. across `2muF` condenser is, `V=((VC_(p))/(C_(1)+C_(p)))=(70xx12xx10^(-6))/(14xx10^(-6))=60V` |
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| 219. |
Which of the following instruments works on the principle of action of sharp points ?A. Van de Graff generatorB. CyclotronC. DynamoD. Induction coil |
| Answer» Correct Answer - A | |
| 220. |
Van de Graaff generator isA. high voltage generatorB. low voltage generatorC. ac generatorD. dc generator |
| Answer» Correct Answer - A | |
| 221. |
In van de Graaff generator, potential difference is of the order ofA. `10^(7)V`B. `10^(3)V`C. `10^(2)V`D. `10^(12)V` |
| Answer» Correct Answer - A | |
| 222. |
Van de Graaff generator is used forA. carry out radioactive disintegrationB. produce ac voltageC. convert ac into dc voltageD. produce total internal reflection |
| Answer» Correct Answer - A | |
| 223. |
In a Van de graaf type genertor a sphrical metal shell is to be a `15xx10^(6)` volt electrode. The dielectric strength of the gas surrounding the elctrode is `5xx10^(7) Vm^(-1). ` What is the minimum radius of the spherical shell required ? [you will learn form this exercise why one cannot build an electrostatic generator using a very small shell, which requires a small charge to acquire a high potential.] |
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Answer» Here, `V = 15xx10^(6) vol t`, Dielectric strength `= 5xx10^(7) Vm^(-1)`, Minimum radius, `r = ?` Max. electric field, `E = 10%` dielectric strength `E = (10)/(100) xx5xx10^(7) = 5xx10^(6) Vm^(-1)`, As `E = (V)/(r ) :. r = (V)/(E) = (15xx10^(6))/(5xx10^(6)) = 3m` Obviously, we cannot build an electrostatic generator, using a very small sheet. |
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| 224. |
Assertion. A sphrical equipotential surface is not possible for a point charge. Reason. A spherical equipotential surface is possible inside a spherical capacitor.A. both, Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. both, Assertion and Reason are true, but Reason is not the correct explanation of the Asserrtion.C. Assertion is true, but the Reason is false.D. both, Assertion and Reason are false. |
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Answer» Correct Answer - d Here, Assertion is wrong and Reason is also wrong. |
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| 225. |
From what distance should a 100 eV electron be fired towards a large metal plate having a surface charge of `-2.0 xx 10^(-6) Cm^(-2)`, so that it just fails to strike the plate ?A. 0.50 mmB. 0.44 mC. 0.60 mmD. 0.77 mm |
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Answer» Correct Answer - B |
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| 226. |
A thin spherical shell of metal has a radius of 0.25 m and carries a chaarge of 0.2 mC. The electric indtensity at a point for outside the shell will beA. `2.88 xx 10^(4) N//C`B. `3.4 xx 10^(4) N//C`C. `325 xx 10^(4) N//C`D. `3.88 xx 10^(4) N//C` |
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Answer» Correct Answer - A |
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| 227. |
A clock face has charges `-q, -2q, ,.....-12q` fixed at the position of the corresponding numerals on the dial. The clock hands do not disturb the net field due to point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial. |
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Answer» Correct Answer - `9.30` |
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| 228. |
There charges q, 3q and 12q are to be placed on a straight line AB having 12 cm length. Two of the charges must be placed at end points A and B and the third charge can be placed anywhere between A and B. Find the position of each charge if the potential energy of the system is to be minimum. In the position of minimum potential energy what is the force on the smallest charge? |
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Answer» Correct Answer - 12q and 3q at the two ends and q at a distance 8 cm from 2q. Force on q is zero. |
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| 229. |
A charge of `5xx10^(-10)C` is given to a metal cylinder of length 10 cm, placed in air. The electric intensity due to the cylinder at a distance of 0.2 m from its axis isA. `150V//m`B. `250V//m`C. `350V//m`D. `450V//m` |
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Answer» Correct Answer - D `E=1/(4piin_(0)k)(2q)/(rL)=(9xx10^(9)xx2xx5xx10^(-10))/(1xx0.2xx0.1)` `=450V//m` |
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| 230. |
Charge `q_(2)` of mass m revolves around a stationary charge `q_(1)` in a circulare orbit of radius r. The orbital periodic time of `q_(2)` would beA. `[(4 pi^(2) mr^(3))/(k q_(1)q_(2))]^(1//2)`B. `[(k q_(1) q_(2))/(4 pi^(2) mr^(3))]^(1//2)`C. `[(4 pi^(2)mr^(4))/(k q_(1) q_(2))]^(1//2)`D. `[(4 pi^(2) mr^(2))/(k q_(1) q_(2))]^(1//2)` |
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Answer» Correct Answer - A |
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| 231. |
A uniformly charged sphere has charge Q. An electron (charge – e, mass m) revolves around it in a circular orbit of radius r. (a) Write the total energy (i.e., sum of its kinetic energy and electrostatic potential energy in the field of the sphere) of the electron. (b) If the time period of revolution of the electron in circular orbit of radius r is T, then find the time period if the orbital radius is made 4r. |
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Answer» Correct Answer - (a). `-(Qe)/(8pi in_(0)r)` (b). 8T |
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| 232. |
Statement-1. In a series combination of capacitors, charge on each capacitor is same. Statement-2. In such a combination, charge cannot move only along one route.A. Statement-1. is true , Statement-2 is true , Statement-2 is correct explanation of Statement-1.B. Statement-1. is true , Statement-2 is true , Statement-2 is not a correct explanation of Statement-1.C. Statement-1. is correct and Statement-2 is false.D. Statement-1. is false and Statement -2 is true. |
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Answer» Correct Answer - a Both the statements are true, and Statement-2 is correct explanation of the Statement -1 |
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| 233. |
A charge moves with a speed `v` in a circular path of radius `r` around a long uniformly charged conductor.A. `v prop r`B. `v prop 1/r`C. `v prop 1/(sqrt(r))`D. `v` is independent of `r` |
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Answer» Correct Answer - D |
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| 234. |
Statement-1. For a charged particle moving from pont `P` to point `Q`, the net work done by an electrostatic field on the particle is independent of the path connecting point `P` to point `Q` Statement-2. The net work done by a conservatie force on an object moving along a closed loop is zero.A. Statement-1. is true , Statement-2 is true , Statement-2 is correct explanation of Statement-1.B. Statement-1. is true , Statement-2 is true , Statement-2 is not a correct explanation of Statement-1.C. Statement-1. is correct and Statement-2 is false.D. Statement-1. is false and Statement -2 is true. |
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Answer» Correct Answer - b Statement-1 and 2 both are true, but Statement-2 is not a correct explanation of statement-1 |
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| 235. |
A charge ` +Q` is uniformluy distributed over a thin ring with radus ` R. a` negative point charge -Q and mass `m` starts from rest at a pont far away from the centre of the ring and moves towardsn then centre. Find the velocity of this particle at the moment it passes through the centre of the ring . |
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Answer» Correct Answer - ` sqrt (2 k q^2)/( mR)` From energy conservation Loss of `PE` = gain of `KE` ` Q. (KQ)/R = 1/2 mv^2 rArr v= sqrt((2 KA^@ )/(mR))`. |
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| 236. |
Two balls with charges `5 muC` and `10m C` are at a distance of `90m` from each other. In order to reduce the distance between them to `45m`, the amount of work to be performed in joule is : |
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Answer» Correct Answer - 5 Here, `q_(1) = 5 mu C = 5xx10^(-6) C` `q_(2) = 10m C = 10xx10^(-3) C` `r_(1) = 90m, r_(2) = 45m, W = ?` `W = U_(2) - U_(1) = (q_(1) q_(2))/(4pi in_(0) r_(2)) - (q_(1) q_(2))/(4pi in_(0) r_(1))` `= (q_(1) q_(2))/(4pi in_(0)) [(1)/(r_(2)) - (1)/(r_(1))]` `(5xx10^(-6))xx(10xx10^(-3)) xx9xx10^(9) [(1)/(45) - (1)/(90)]` `W = 450xx (1)/(90) = 5 J` |
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| 237. |
Two unequal masses, `m_1 = 2m` and `m_2 = m` have unequal positive charge on them. They are suspended by two mass-less threads of unequal lengths from a common point such that, in equilibrium, both the masses are on same horizontal level. The angle between the two strings is `theta=45^(@)` in this position. Find the Electrostatic force applied by `m_1` on `m_2` in this position. |
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Answer» Correct Answer - `((sqrt(17)-3)/(2))mg` |
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| 238. |
When a conducting disc is made to rotate about its axis, the centrifugal force causes the free electrons to be pushed toward the edge. This causes a sort of polarization and an electric field is induced. The radial movement of free electrons stops when electric force on an electron balance the centrifugal force. Calculate the potential difference developed across the centre and the edge of a disc of radius R rotating with angular speed `omega`. [This potential difference is sometimes called as sedimentation potential]. Take mass and charge of an electron to be m and e respectively |
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Answer» Correct Answer - `(mR^(2)omega^(2))/(2e)` |
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| 239. |
A uniform electric field is along x-axis. The potential difference `V_(A)-V_(B)=10 V` is between two points A (2m, 3m) and (4m, 8m). Find the electric field intensity. |
| Answer» `E=(Delta V)/(Delta d)=10/2=5 V//m`. (It is along + ve x-axis) | |
| 240. |
A charge q is moved from a point A above a dipole moment p to a point B below the dipole on equatorial plane without acceleration. Find the work done in the process. |
| Answer» At any point on equatorial line of an electric dipole, electric potential is Zero. Therefore, work done in moving a charge q from A to B is zero. | |
| 241. |
The electric field at a point on equatorial of a dipole and direction of the dipole momentA. `180^(@)`B. `0^(@)`C. `45^(@)`D. `90^(@)` |
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Answer» Correct Answer - A |
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| 242. |
The net charge given to a solid insulating sphere :A. must be distributed uniformly in its volumeB. may be distributed uniformly in its volumeC. must be distributed uniformly on its surfaceD. the distribution will depend upon whether other charges are present or not. |
| Answer» Correct Answer - B | |
| 243. |
An insulating rod carries some net charge, and a copper sphere is neutral. The rod and the sphere do not touch. Can there be force of attration/repulsion between the two? |
| Answer» When the charged rod is brought near the neutral sphere, the sphere gets oppositely charged by induction. Therefore, the sphere the the rod will attract each other. However force of repulsion between the two is not possible. | |
| 244. |
The electrostatic potential due to the charge configuration at point P as shown in figure for bltlta is A. `(2q)/(4piepsilon_0a)`B. `(2qb^2)/(4piepsilon_0a^3)`C. `(qb^2)/(4piepsilon_0a^3)`D. zero |
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Answer» Correct Answer - C `V_p=2(1/(4piepsilon_0)q/a)-2[1/(4piepsilon_0).q/sqrt(b^2+a^2)]` `=q/(4piepsilon_0a)[2-2(1+b^2/a^2)^(-1//2)]` Since `bltlta`,we can apply binoial expansion `:. V_p=q/(4piepsilon_0a)[2-2(1-b^2/(2a^2))]` =`(qb^2)/(4piepsilon_0a^3)` |
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| 245. |
A solid conducting sphere of radius a having a charge q is surrounded by a concentric conducting spherical shell of inner radius `2a` and outer radius `3a` as shown in figure. Find the amount of heat porduced when switch is closed `(k=1/(4piepsilon_0))` |
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Answer» Correct Answer - 4 |
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| 246. |
A conducting sphere of radius R carries a charge Q. It is enclosed by another concentric spherical shell of radius 2R. Charge from the inner sphere is transferred in infinitesimally small installments to the outer sphere. Calculate the work done in transferring the entire charge from the inner sphere to the outer one. |
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Answer» Correct Answer - `-(Q^(2))/(16pi in_(0)R)` |
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| 247. |
Five identical charges, `q_1` each, are placed at the vertices of a regular pentagon having side length `l_1`. The net electrostatic force on any of the charges due to other four is `F_1`. Find the electrostatic force `F_2` on any one of the five identical charges, `q_2` each, placed at the vertices of a regular pentagon having side length `l_2`. |
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Answer» Correct Answer - `F_(2)=F_(1)((q_(2)^(2)l_(1)^(2))/(q_(1)^(2)l_(2)^(2)))` |
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| 248. |
Two identical positive charges are fixed on the y-axis, at equal distances from the origin O. A particle with a negative charges starts on the x-axis at a large distance from O. moves along the x-axis passes through O, and moves far away from O on the other side. Its acceleration a is taken as positive along its direction of motion. Plot acceleration a of the particle against its x-coordinate.A. B. C. D. |
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Answer» Correct Answer - B |
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| 249. |
Two identical simple pendulums, `A` and `B` are suspended from the same point. The bobs are given positive charges, with `A` having more charge than `B` making angles `theta_(1)` and `theta_(2)` with the vertical respectively. Which of the following is correct?A. `theta_(1) gt theta_(2)`B. `theta_(1) lt theta_(2)`C. `theta_(1)=theta_(2)`D. The tension in `A` is greater than that in `B` |
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Answer» Correct Answer - C |
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| 250. |
Two ideantical point charges are placed at a separation of `d`. `P` is a point on the line joining the charges, at a distance `x` from any one charge. The field at `P` is `E, E` is plotted against `x` for value of `x` from close to zero to slightly less then `d`. Which of the following represents the resulting curveA. B. C. D. |
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Answer» Correct Answer - D |
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