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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.
| 301. |
The energy stored in a capacitor is given by(a) qV(b)1/2 qv(c) 1/2 qv(d) q/2c |
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Answer» Correct answer is (b) 1/2 qv |
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| 302. |
A sphere of radius R has a charge density `sigma`. The electric intensity at a point at a distance r from its centre isA. `E=(sigmaR^(2))/(in_(0)kr^(2))`B. `E=(sigma)/(in_(0)k)`C. `E=R^(2)`D. `E=in_(0)kr^(2)sigma^(2)` |
| Answer» Correct Answer - A | |
| 303. |
The electric potential at the centre of a charged conductor is-(a) zero (b) twice that on the surface (c) half that on the surface (d) same as that on the surface |
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Answer» (d) same as that on the surface |
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| 304. |
A charge +q is placed at the mid point of a cube of side L. The electric flux emerging from cube isA. `q/(in_(0))`B. `q/(6L^(2)in_(0))`C. `(6qL^(2))/(in_(0))`D. zero |
| Answer» Correct Answer - A | |
| 305. |
A point charge +q is placed at the midpoint of a cube of side l. The electric flux emerging ’ from the cube is-(a) \(\frac{q}{ε_0}\)(b) \(\frac{6ql^2}{ε_0}\)(c) \(\frac{q}{6l^2ε_0}\)(d) \(\frac{C^2V^2}{2}\) |
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Answer» Correct answer is (a) \(\frac{q}{ε_0}\) |
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| 306. |
The total electric flux for the following closed surface which is kept inside water- (a) 80q/ε0(b)q/40ε0(c) q/80ε0(d) q/40ε0 |
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Answer» Correct answer is (b) q /40ε0 |
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| 307. |
Rank the electrostatic potential energies for the given system of charges in increasing order(a) 1 = 4 < 2 < 3 (b) 2 = 4 < 3 < 1 (c) 2 = 3 < 1 < 4 (d) 3 < 1 < 2 < 4 |
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Answer» Correct answer is (a) 1 = 4 < 2 < 3 |
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| 308. |
Two identical conducting balls having positive charges q1 and q2 are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be-(a) less than before (b) same as before (c) more than before (d) zero |
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Answer» (c) more than before |
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| 309. |
Two small identical conducting balls each of radius r and mass m are placed on a frictionless horizontal table, connected by a light conducting spring of force constant K and un-deformed length L (L gtgt r). A uniform electric field of strength E is switched on in horizontal direction parallel to the spring. (a) How much charge will appear on the two balls when they are at separation L.(b) The system fails to oscillate if K lt `K_0`. Find `K_0`. (c) Assuming `K = 2K_0`, find the time period of oscillation after the electric field is switched on. |
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Answer» Correct Answer - (a). `+-2pi in_(0)ErL` (b). `k_(0)=2pi in_(0)rE^(2)` (c) `(1)/(E)=sqrt((pim)/(in_(0)r))` |
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| 310. |
The work done in carrying a charge `Q_(1)` once round a circle of radius `R` with a charge `Q_(2)` at the centre isA. `(Q_(1) Q_(2))/(4 pi in_(0) R^(2))`B. zeroC. `(Q_(1) Q_(2))/(4 pi in_(0) R)`D. infinite |
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Answer» Correct Answer - 2 Work done on equilibrium line/surface is zero. |
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| 311. |
On a horizontal table, there is a smooth circular groove of mean radius R. The walls of the groove are non conducting. Two metal balls (each having mass m and radius r) are placed inside the groove with their centers R apart. The balls just fit inside the groove. The two balls are given charge + 3q and – q and released from state of rest. Ignore the non-uniformity in charge distribution as the balls come close together and collide. The collision is elastic. Find maximum speed acquired by each ball after they collide for the first time. |
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Answer» Correct Answer - `V_(0)=sqrt((Kq^(2)(4R-7r))/(2mrR))` |
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| 312. |
Conisder a neutral conducting sphere. A poistive point charge is placed outisde the sphere. The net charge on the sphere is thenA. Negative and distributed uniformly over the surface of the sphereB. Negative and appears only at the point on the sphere closed to the point chargeC. Negative and distributed non-uniformly over the entire surface of the sphereD. zero |
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Answer» Correct Answer - D |
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| 313. |
Poistive and negative point charges of equal magnitude are kept at `(0, 0, a/2)` and `(0, 0, (-a)/(2))` respectively. The work done by the electric field when another poistive point charge is moved from `(-a, 0, 0)` to `(0, a, 0)` isA. positiveB. negativeC. zeroD. depends on the path connecting the initial and final positions |
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Answer» Correct Answer - C The two charge an electric dippole. The points (- a, 0, 0, ) and ( 0, a, 0 ) are at the same distance from the positive and negative charges of equal magnitud Hence the leectric potential si zero at these poits. Hence the work done is zero. |
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| 314. |
Conisder a neutral conducting sphere. A poistive point charge is placed outisde the sphere. The net charge on the sphere is thenA. negative and distributed uniformly over the surface of the sphere.B. negative and appears only at the point on the sphere closest to the point charge.C. negative and distributed non-uniformly over the entire surface of the sphere.D. zero. |
| Answer» Correct Answer - D | |
| 315. |
Poistive and negative point charges of equal magnitude are kept at `(0, 0, a/2)` and `(0, 0, (-a)/(2))` respectively. The work done by the electric field when another poistive point charge is moved from `(-a, 0, 0)` to `(0, a, 0)` isA. positiveB. negativeC. zeroD. depends on the path connecting the initial and final positions. |
| Answer» Correct Answer - C | |
| 316. |
A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inisde the emptied space is A. zero every whereB. is not zero but uniformC. nonuniformD. is zero at centre only |
| Answer» Correct Answer - B | |
| 317. |
A charge `q_0` is distributed uniformly on a ring of radius R. A sphere of equal radius R constructed with its centre on the circumference of the ring. Find the electric flux through the surface of the sphere.A. `(2 piR lambda)/(epsi_(0))`B. `(pi R lambda)/(epsi_(0))`C. zeroD. None of these |
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Answer» Correct Answer - D |
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| 318. |
A horizontal circular groove is made in a wooden board. Two positive charges (q each) are placed in the groove at a separation of `90^(@)` (see figure). Where shall we place (in the groove) a third charge and what shall be its magnitude such that all three of them remain at rest after they are released. Answer for two cases: (a) When the third charge is positive. (b) When the third charge is negative. Neglect friction and assume that the groove is very thin just wide enough to accommodate the particles. [Take `sin22.5^(@)=0.38,cos22.5^(@)=0.92]` |
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Answer» Correct Answer - (a). `Q=+3.14q` (b). `Q=-0.22q` |
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| 319. |
Two identical positive charges Q each are placed on the x axis at point (-a,0) and (a,0) A point charge of magnitude q is placed at the origin. For small displacement along x axis, the charge q executes simple harmonic motion if it is poitive and its time period is `T_(1)`. if the charge q is negative, it perform oscillations when displaced along y axis. in this case the time period of small oscillations is `T_(2).` find `(T_(1))/(T_(2))` |
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Answer» Correct Answer - `(1)/(sqrt(2))` |
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| 320. |
A ring of radius R has uniformly distributed charge q. A point charge Q is placed at the centre of the ring. (a) Find the increase in tension in the ring after the point charge is placed at its centre. (b) Find the increase in force between the two semicircular parts of the ring after the point charge is placed at the centre. (c) Using the result found in part (b) find the force that the point charge exerts on one half of the ring. |
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Answer» Correct Answer - (a). `DeltaT=(KQq)/(2piR^(2))` (b). `(KQq)/(piR^(2))` (c). `(KQq)/(piR^(2))` |
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| 321. |
A fixed non conducting smooth track is in the shape of a quarter circle of radius R in vertical plane. A small metal ball A is fixed at the bottom of the track. Another identical ball B, which is free to move, is placed in contact with ball A. A charge Q is given to ball A which gets equally shared by the two balls. Ball B gets repelled and ultimately comes to rest in its equilibrium position where its radius vector makes an angle `theta (theta lt 90^(@))` with vertical. Mass of ball is m. Find charge Q that was given to the balls. |
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Answer» Correct Answer - `Q=8Rsin((theta)/(2))sqrt(2piepsilon_(0)mgsin((theta)/(2)))` |
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| 322. |
Twelve charges have been placed at the centre of each side of a cube as shown in the figure. Find the magnitude of Electric force acting on a charge Q placed at the centre of the cube. Take the side length of the cube to be r. |
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Answer» Correct Answer - `(8KQq)/(r^(2))` |
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| 323. |
A positive charged particle when moves from potential to lower potentialA. its potential energy must decreaseB. it potential energy may decreaseC. its kinetic energy must increaseD. its kinetic energy may increase |
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Answer» Correct Answer - A::D |
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| 324. |
Electric field intensity at a point due to an infinite sheet of charge having surface charge density `sigma` is `E`.If sheet were conducting electric intensity would beA. `2E`B. `E`C. `E/2`D. none of these |
| Answer» Correct Answer - B | |
| 325. |
Two short electric dipole are placed as shown. The energy of electric interaction between these dipole will be A. `(2k P_(1) P_(2) cos theta)/r^(3)`B. `(-2k P_(1)P_(2) cos theta)/r^(3)`C. `(-2k P_(1)P_(2) sin theta)/r^(3)`D. `(-4k P_(1)P_(2) cos theta)/r^(3)` |
| Answer» Correct Answer - B | |
| 326. |
A charge `Q`is placed at the mouth of a conical flask. The flux of the electric field through the flask isA. zeroB. `Q//epsilon_(0)`C. `Q/(2epsilon_(0))`D. `lt Q/(2epsilon_(0))` |
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Answer» Correct Answer - C |
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| 327. |
Charge `q` on a small conducting sphere `S_(1)` is placed inside a large hollow metallic sphere `S_(2)` having a charge `Q` as shown in figure. The sphere is connected to shell by a conducting wire. The charge on `S_(1)` will then be A. `Q - q`B. `(qQ)/(2)`C. `((Q + q))/(2)`D. Zero |
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Answer» Correct Answer - 4 The whole charge remains on outer surface. |
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| 328. |
A positive point charge, which is free to move, is placed inside a hollow conducting sphere with negative charge, away from its centre. It willA. move towards the centreB. move towards the nearer wall of the conductorC. remain stationaryD. oscillate between the centre and the nearer wiall |
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Answer» Correct Answer - C |
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| 329. |
Assetrion: Electric lines of force never cross each other. Reason: Electric field at a point superimpose to give one resultant electric fieldA. intersect with each otherB. divergingC. convergingD. parallel to each other |
| Answer» Correct Answer - A | |
| 330. |
Assetrion: Electric lines of force never cross each other. Reason: Electric field at a point superimpose to give one resultant electric fieldA. If both Assertion and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are correct but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is falseD. If both Assertion and Reason are false |
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Answer» Correct Answer - D |
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| 331. |
An infinite sheet of charge has a surfaces charge density of `10^-7C/m^2`. The separation between two equipotential surfce whose potentials differ by `5 V` isA. ` 0. 88 cm`B. ` 0. 88 mm`C. ` 0 .88m`D. ` 5 xx 10^(-7) m` |
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Answer» Correct Answer - B Field due to sheet = `(sigma)/(2 in_0)` and ` C=- Ed =- (sigma) / (2 in_0 ) d` ` 5 = ( 10^(-7))/( 2 xx 8. 85 xx 10^(12)) xx d` ` d= 8.8 xx 10 10^(-4) m = 0. 88`. |
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| 332. |
Eletrostatic forces are much stronger than gravitatinal forces. Give one example. |
| Answer» A charged glass rod can lift a piece of a paper against the gravitational pull of earth on this piece. | |
| 333. |
The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electronstatic field inside a conductor zero? |
| Answer» As is known, electrostatic field is caused by excess charges. On the inside of an isolated conductor, there is no excess charge. Therefore, electrostatic field inside a conductor is zero. | |
| 334. |
A non-conductiong ring of radius ` -0.5 m` camies a total of ` 1. 11 xx 10 ^(-10) C` distributed non-unifrmly on its circumfernce producing an electric field ` vec E ` every where in space . The value of the line integral ` int_(t=infty)^(l=0) -vecE- vec Ed vec l (l=0)` being centre of the ring ) in volts is -A. ` +2`B. ` -1`C. ` -2`D. zero` |
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Answer» Correct Answer - A Potential at the centre of ring ` =- int_((l00))^(l-0) vec E. vec (Dl)` d ` - int_(l=infty)^(l=0) vec E. vec (dl) = (Kq)/R =2` volt. |
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| 335. |
Two point charge `Q_(a)` and `Q_(b)` are positional at point A and B. The field strength to the right of charge `Q_(b)` on the line that passes through the two charges varies according to a law represented schematically in fig. (without employing a definite scale). The field strength is assumed to be positive if its direction coincides with the positive direction of the x-axis. The distance between the charges is `l=21 cm`. (a) Find the sign of the charges. (b) Find the ration between the absolute value of charge `Q_(a)` and `Q_(b)`. (c ) Find the coordinate x of the point where the field strength is maximum. |
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Answer» Correct Answer - A::B::C Over charge `Q_2` field intensity is infinite along negatve `x` -axis. Therefore `Q_2` is negative. Beyon `xgt(l+a)`, field intensity is positive. Therefore `Q_1` is positive b. At `x=l+a`, field intensity is zero. `:. (kQ_1)/((l+a)^2)=(kQ_2)/a^2` or `|Q_1/Q_2|=((l+a)/a)^2` c. Intensity at distance `x` from charge `2` would be `E=(kQ_1)/((x+l)^2-(kQ_2)/x^2` For `E` to be maximum `(dE)/(dx)=0` or `-(2kQ_1)/((x+l)^3)+(2kQ_2)/x^3=0` or `(1+l/x)^3=Q_1/Q_2=((l+a)/a)^2` or `1+l/x=((l+a)/a)^(2//3))` or 1+l/x=((l+a)/a)^(2//3)` or `x=l/(((l+a)/a)^(2//3)-1)` or `b=l/(((l+1)/a)^(2//3)-1)` |
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| 336. |
Two particles have equal masses of 5.0 g each and opposite charges of `+4.0 xx10^(-5) C.` They are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reducced to 50 cm. |
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Answer» Here, `m_(1) = m_(2) = 5.0g = 5xx10^(-3) kg`. `r_(1) = 1.0 m, r_(2) = 50 cm = (1)/(2) m` From symmetry , `v_(1) = v_(2) = v = ?` As increase in K.E. = loss in P.E, - initial P.E. `:. 2xx(1)/(2) mv^(2) = (| q_(1)|| q_(2)|)/(4pi in_(0)) [(1)/(r_(2)) - (1)/(r_(1))]` `5xx10^(-3) v^(2) = 9xx10^(9) (4xx10^(-5))^(2) [(1)/(1//2) - 1]` `= 9xx16xx10^(-1) (2-1)` `v^(2) = (1.44)/(5xx10^(-3)) = 2.88xx10^(3)` `v = 53.7 m//s` |
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| 337. |
Two charges `10 muC and - 10 muC` are placed at points `A. and B` separated by a distance of 10 cm. Find the electric. field at a point P on the perpendicular bisector of `AB` at. a distance of 12 cm from its middle point.A. `16.4xx10^(6) NC^(-1)`B. `28.4xx10^(6) NC^(-1)`C. `8.2xx10^(6) NC^(-1)`D. `4.1xx10^(6) NC^(-1)` |
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Answer» Correct Answer - D |
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| 338. |
Four charges are arranged at the corners of a square ABCD pf side d, as shown in Fig. Find the work required to put together this arrangement (b) A charge `q_(0)` brought to the center E of the square, the four charges being held fixed at the corners . How much extra work in needed to do this ? |
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Answer» (a) Here, AB = BC = CD = DA = d `AC = BD = sqrt(d^(2) + d^(2)) = d sqrt(2)` Total work done = P.E. of system fo 4 charges `= (1)/(4pi in_(0)) sum_("all pairs") (q_(j) q_(k))/(r_(jk))` `= (1)/(4pi in_(0)d) [q(-q) + (-q) (+q)` `+q(-q) + (-q) (+q)] + (1(q^(2) + q^(2)))/(4pi in_(0) d sqrt(2))` `= (-4 q^(2))/(4pi in_(0) d) + (2q^(2))/(4pi in_(0) d sqrt(2))` `= (-q^(2))/(4pi in_(0) d) (4 - sqrt(2))` (b) Extra work needed = charge `xx` potential at E due to charges at four corners. `W = q_(0) xx ((q-q+q-q)/(4pi in_(0) (d sqrt(2)/(2)))) = zero` |
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| 339. |
Four point `+ve` charges of same magnitude`(Q)` are placed at four corners of a rigid square frame as shown in figure. The plane of the frame is perpendicular to `z`-axis. If a `-ve` point charge is placed at a distance `z` away from the above frame `(z lt lt L)` then A. negative charge oscillates along the Z-axisB. it moves away from the frameC. it moves slowly towards the frame and stays in the plane of the frameD. it passes through the frame only once |
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Answer» Correct Answer - A |
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| 340. |
Four charges are arranged at the corners of a square ` ABCD`, as shown. The force on a =ve charge kept at the centre of the square is - A. ZeroB. along diagonal `AC`C. along diagonal ` BD`D. perpendicular to the side ` AB` |
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Answer» Correct Answer - D ` vec E_A + vec E_C = ` will be along `OA` ` vec E_B + vec E_E = ` will be along ` OB` ` | vec E_A | vec E_C |= | vec E_B + _D|` So`E_o` will be perpendicular to side ` AB`. |
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| 341. |
Four charges are arranged at the corners of a square `ABCD`, as shown in the adjoining figure, The force on a positive charge kept at the centre `O` is A. zeroB. along the diagonal `AC`C. along the diagonal `BD`D. perpendicular to side `AB` |
| Answer» Correct Answer - 3 | |
| 342. |
Four charges equal to `-Q` are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q isA. `(-Q)/(4) (1 + 2 sqrt2)`B. `(Q)/(4) (1 + 2 sqrt2)`C. `(-Q)/(2) (1 + 2 sqrt2)`D. `(Q)/(2) (1 + 2 sqrt2)` |
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Answer» Correct Answer - B |
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| 343. |
Two point charges `+8q` and `-2q` are located at `x=0` and `x=L` respectively. The location of a point on the x axis at which the net electric field due to these two point charges is zero isA. (a) `L/4`B. (b) `2L`C. (c) `4L`D. (d) `8L` |
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Answer» Correct Answer - B `(-K2q)/(x-L)^2+(K8q)/(x^2)=0implies(1)/((x-L)^2)=(4)/(x^2)` or `(1)/(x-L)=2/ximpliesx=2x-2L` or `x=2L` |
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| 344. |
A charged oil drop is suspended in a uniform filed of `3xx10^4 v//m` so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge `=9.9xx10^-15kg` and `g=10m//s^2`)A. `3.3xx10^(-18)C`B. `3.2xx10^(-18)C`C. `1.6xx10^(-18)C`D. `4.8xx10^(-18)C` |
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Answer» Correct Answer - A F = mg = qE `therefore" "q=(mg)/E=(9.9xx10^(-15)xx10)/(3xx10^(4))` `=3.3xx10^(-18)C` |
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| 345. |
A charged oil drop is suspended in a uniform filed of `3xx10^4 v//m` so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge `=9.9xx10^-15kg` and `g=10m//s^2`)A. (a) `1.6xx10^-18C`B. (b) `3.2xx10^-18C`C. (c) `3.3xx10^-18C`D. (d) `4.8xx10^-18C` |
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Answer» Correct Answer - C At equilibrium, electric force on drop balances weight of drop. `qE=mgimpliesq=(mg)/(E)=(9.9xx10^-15xx10)/(3xx10^4)=3.3xx10^(-18)C` |
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| 346. |
A and B are two points on the axis and the perpendicular bisector, respectively, of and electric dipole. A and B are far away from the dipole and at equal distance from it. The fields at A and B are `vecE_(A)` and `vecE_(B)`. ThenA. `vecE_(A)=vecE_(B)`B. `vecE_(A)=2vecE_(B)`C. `vecE_(A)=-2vecE_(B)`D. `|E_(B)|=1/2|E_(A)|`, and `vecE_(B)` is perpendicular to `vecE_(A)` |
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Answer» Correct Answer - C |
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| 347. |
A large flat metal surface has a uniform charge density `+sigma`. An electron of mass `m` and charge `u` and returns to it at point `B`. Disregard gravity . The maximum value of `AB` isA. `(u^(2)m epsilon_(0))/(sigma e)`B. `(u^(2)e epsilon_(0))/(m sigma)`C. `(u^(2)e)/(epsilon_(0)sigmam)`D. `(u^(2) sigma e)/(epsilon_(0)m)` |
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Answer» Correct Answer - A |
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| 348. |
A spring - block system undergoes vertical oscillation above a large horizontal metal sheet with uniform positive charge. The time period of the oscillation will be is given a charge `Q` , its time period of oscillation will beA. `T`B. `gt T`C. `lt T`D. `gt T` if `Q` is positive and `lt T` if `Q` is negative |
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Answer» Correct Answer - A |
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| 349. |
A spring - block system undergoes vertical oscillation above a large horizontal metal sheet with uniform positive charge. The time period of the oscillation will be is given a charge `Q` , its time period of oscillation will beA. `T`B. `gt T`C. `ltT`D. `gtT` if `Q` is `+ve ` and `lt T` if `Q` is `-ve` |
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Answer» Correct Answer - 1 The time - period `T = 2pi sqrt((m)/(k))` of spring pendulum remins same. |
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| 350. |
A spherical shell of radius r carries a uniformly distributed surface charge q on it. A hemispherical shell of radius R(gt r) is placed covering it with its centre coinciding with that of the sphere of radius r. The hemisphere has a uniform surface charge Q on it. The charge distribution on the sphere and the hemisphere is not affected due to each other. Calculate the force that the sphere will exert on the hemisphere. |
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Answer» Correct Answer - `(Qq)/(4piepsilon_(0)R^(2))(1-(1)/(sqrt(2)))` |
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