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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.
| 1. |
Two semicircle wires ABC, and ADC, each of radius R are lying on xy and xz planes, respectively as shown in fig. if the linear charge density of the semicircle parts and straight parts and straight parts is `lambda`, Find the electric field entensity `vec(E)` at the origin . |
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Answer» Correct Answer - `-(lambda)/(2 pi epsilon_(0)R)(hat j+hat k)`. Electric field due to MA: `vecE_(1)=(lambda)/(4 pi epsilon_(0)R)(hati -hatk)` Electric field due to ADC: `vecE_(2)=(lambda)/(2 pi epsilon_(0)R)(-hatk)` Electric field due to ABC: `vecE_(3)=(lambda)/(2 pi epsilon_(0)R)(-hatj)` Electric field due to NC: `vecE_(4)=(lambda)/(4 pi epsilon_(0)R)(hati-hatk)` Net electric field is `vecE_(0)=vecE_(1)+vecE_(2)+vecE_(3)+vecE_(4)=-(lambda)/(2 pi epsilon_(0)R)(hat j+hat k)`. |
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| 2. |
Two identical conduction spheres, one having an initial charge +Q and the other initially uncharged, are brought into contact. (a) What is the new charge on each sphere? (b) While the spheres are in contact, a positively chaged rod is moved close to one sphere, causing a redistribution of hte charges on the two spheres so the charge on hte sphere closest to the rod has a charge-Q. What is the charge to the other sphere? |
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Answer» (a) This is due to the reason that the spheres are identical land they share the total charge equally. (b) The charge on other sphere will be +Q. Because of induction the charge on other sphere will be +Q. |
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| 3. |
A charged rod attracts bits of dry paper, which after touching the rod often jucp away from it rapidly, Expalin |
| Answer» When the paper comes in contact with the rod. The paper also gets similarly charged. Hence, repulsion takes place. | |
| 4. |
A person standing on an ainsulating stool touches a charged insulatd conductor. Will the conductor get completely discharged? |
| Answer» Conductor will not get completely discharged, because the person is standing on the insulating stool. Some charge of the rod may transfer to the person to make the potencial of the both same. | |
| 5. |
Two negative charges of a unit magnitude and a positive charge q are placed along a straight line. At what position and value of q will the system be in equilibrium? (Negative charges are fixed). |
| Answer» The charge q can be in equilibrium if it is placed at the middle of negative charges. As the electric field is zero at the middle of negative charges. The charge q can have any value. | |
| 6. |
When an electron moves in a circular path around a stationay nucles charge at the centerA. the acceleration of the electron changesB. the velocity of the electon changeC. electric field due to the nucleus at the electron changesD. none of these |
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Answer» Correct Answer - A::B::C Accelaeration, velocity, and electric field all change in direction althrough they remain same in magnitude |
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| 7. |
Figure shows the tracks of three charged particles in a unform electrostatic field projected parallel to a plate with the same velocity. Give the signs of the three charges. Which of the three charges. Which of the three particles has the highest charge-to-mass ratio? |
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Answer» (1) and (2) are negative while (3) is positive. Acceleration of each particle is `a=(qE)/(m)` Deflection suffered by a particle is `y=ut+(1)/(2) at^(2)=0+(1)/(2)(qE)/(m)t^(2)` or `y prop(q)/(m)` Hence, (3) will have the highest `q//m` ratio. |
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| 8. |
Two small identical balls lying on a horizontal plane are connected by a weightless spring. One ball (ball 2) is fixed at O, and the other (ball 1) is free. The balls are charged identically as a result of which the spring length increases `eta=2` times. Determine the change in frequency. |
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Answer» When the balls are unchangeed `v_(0)=(1)/(2 pi)sqrt((k)/(m))` where k is the force constant of the spring and ma is the mass of the oscillating ball (ball 1). When the balls are charged, ball 1 will oscillate about the new equilibrium position. At the equilibrium position of ball 1. `(1)/(4 pi epsilon_(0))(q^(2))/((eta l)^(2))=k(etal-l)=k(eta-1)` or `l^(3)=(q^(2))/(4 pi epsilon_(0)eta^(2)(eta-1)k)` ...(i) When ball 1 is displaced by a small distance from the equilibrium position to the right. the unbalanced froce to the reight is given by `F=(1)/(4 pi epsilon_(0))(q^(2))/((eta l+x)^(2))-k(eta l+x-1)` From Newton law, we have `m(d^(2)x)/(dt^(2))=(1)/(4 pi epsilon_(0))(q^(2))/(eta^(2)l^(2))[1+(x)/(eta l)]^(2)-kl(eta-1)-kx` `=(1)/(4 pi epsilon_(0))(q^(2))/(eta^(2)l^(2))[1-(2x)/(eta l)]-kl(eta-1)-kx` `=-[(1)/(4 pi epsilon_(0))(2q^(2))/(eta^(3)l^(3))+k]x` From Eqs (i) and (ii), `m(d^(2)x)/(dt^(2))=-[(2(eta-1))/(eta)k+k]x=-(3 eta-2)/(eta)kx` or `(d^(2)x)/(dt^(2))=-(3 eta-2)/(eta)(k)/(m)` or `a =-[(3eta-2 k)/(meta)]x` Comparing with `a=-omega^(2)x`, (i) `omega^(2)=(3 eta-2)/(eta)(k)/(m)` or `v=(1)/(2pi)sqrt(((3 eta-2)/(eta))(k)/(m))` (ii) or `(v)/(v_(0))=sqrt((3 eta-2)/(eta))` (iii) Thus the frequency is increased `sqrt((3 eta-2)//eta)` times . Hence `eta=2` and so frequency increases `sqrt(2)` times . |
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| 9. |
Apolythene piece rubbed will wool is found to have a negative charge of `3.0xx10^(-7)C`. (a) Estimate the number of electrons transferred (from which to which )? (b) Is there a transfer of mass from wool to polythene? |
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Answer» (a) Since the polythene piece acquires a negative charge (q) on rubbing with wool, electrons are transferred from wool to polythene. If n is the number of electrons transferred, then `q="ne"` or `n=(q)/(e )=(3xx10^(7)C)/(1.6xx10^(-19)C)=1.875xx10^(12)` (b) Mass transferred from wool to polythene is `(1.875xx10^(12))(9.1xx10^(-31)kg)=1.7xx10^(-18)Kg` of course, this mass is negatively small. |
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| 10. |
Two small identical conducting balls A and B of charges `+10 mu C` and `+30 mu C` respectively, are kept at a separation of 50 cm. these balls have been connected by a wire for a short time The final charge on each of the balls A and B and force of interaction will beA. `28.8N`B. `32.6N`C. `14.4N`D. `72N` |
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Answer» Correct Answer - C Charge on each ball is `(10+30)/(2)=20 mu C` `F=9xx10^(6)xx((20 xx10^(-6))^(2))/((0.50)^(2))=14.4N`. |
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| 11. |
Two identical small equally charged conducting balls are suspended from long threads secured at one point. The charges and masses of the balls are such that they are in equilibrium when the distance between them is a(the length of thread `Lgtgta`). Then one of the balls is discharged. what will be the distance `b(bltltL)` between the balls when equilibrium is restored? |
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Answer» Correct Answer - 4 `T=sin theta=F-(1)/(4 pi epsilon_(0)) (q^(2))/(a^(2))`…(i) `T cos theta=mg`….(ii) Form (i) and (ii), `tan theta=(q^(2))/(4 pi epsilon_(0)a^(2)mg)` or, `(a//2)/(L)=(q^(2))/(4 pi epsilon_(0)a^(2)mg)` (since for small `theta, tan theta~~a//2L`) or, `(a^(3))/(L) =(q^(2))/(2 pi epsilon_(0)mg)` ...(iii) When one of the balls is discharged, the balls come closer and touch each other and again separated due to repulsion. The charge on each ball after touching each other is `q//2`. Replacing q with `q//2` in (iii), we get `(b^(3))/(L)=((q//2)^(2))/(2 pi epsilon_(0)mg)`...(iv) from (iii) and (iv) , `(b^(3))/(a^(3))=(1)/(4)` or `(a^(3))/(b^(3))=4`. |
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| 12. |
In 1 g of a solid, there are `5xx10^(21)` atoms. If one electrons is removed fron each of 0.01% atoms of the solid, find the charge gained by the solid (given that electronic charge is `1.6xx10^(19)C`). |
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Answer» To calculate charge, we will apply formula `Q="ne"`, for this , we must have number of electrons. Here, number of electrons is `n=0.01% of 5xx10^(21)` i.e. `n=(5xx10^(21)xx0.1)/(100)=5xx10^(21)xx10^(-4)=5xx10^(17)` so, `Q=5xx10^(17)xx1.6xx10^(-19)=8xx10^(-2)=0.08C`. Since electrons have been removed, charge will be positive i.e. `Q=+0.08C` |
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| 13. |
Five balss numbered 1,2,3,4,and 5 are suspended using separated threads. The balls (1,2),(2,4) and (4,1) show electrostatic attraction while balls (2,3) and (4,5) show repulsion. Therefore, ball 1 must beA. negatively chargedB. Positively chargesC. neutralD. made of metal |
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Answer» Correct Answer - C `{:(1,2,3,4,5),(+,-,,+,):}` If 1 is positively charged, then 2 should be negatively charged, then 4 should be positively charged. Now 1 and 4 cannot attract. It means ball 1 should be neutral (2 and 4 cannot be neutral because they are showing repulstion with 3 and 5, respectively). |
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| 14. |
In fig, two particles, each of charge -q are arranged symmetrically about the y-axis, each producing an electric field at point P on the y-axis. . (a) Are the magnitudes of the fields at P equal? (b) Is each electric field directed towards or away form the charge producing it ? (c ) Is the magnitude of the net electric field at P equal to the sum of the magnitudes E of the two field vectors (is it equal to 2E)? (d) Do the x-components of those two field vectors add or cancel? (e ) Do their y-components add or cancel? (f) Is the diretion of the net field at P that of the canceling components or the adding components? (g) What is the direction of the net field? |
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Answer» (a) yes. (b) toward (c ) No, it is less than 2E. (d) Cancel (e ) Add (f) Adding components (g) Negative y. |
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| 15. |
An infinte wire having linear charge density `lambda` is arranged as shown in fig. A charge particle of mass m and charge q is released charged q is released from point P. Find the initial acceleration of the particle just after the particle is released. |
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Answer» Correct Answer - `-(lambda)/(2 pi epsilon_(0)R)(-hat j)`. Electric field due to straight wires will cancel out. The net electric field will be due to semicircular wire only. Hence, `vecE_("net")=vecE_("circular")=(lambda)/(2 pi epsilon_(0)R)(-hat j)` so, acceleration is ` vec(a) = (q vecE_("net"))/(m)=(q lambda)/(2 pi epsilon_(0)mR)(hat j)`. |
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| 16. |
Four identical point charges are placed at the corners of a square, A fifth point charge placed at the center of the square experiences zero net force . Is this a stable equilirium for the fifth charge? Explain. |
| Answer» No, Only for very special displacements, the electrostatic force acts in a direction that points back toward the equilibrium position. For a general displacment, the electrostatic force does not point toward the equilibrium position, and the fifth charge moves farther from equilibrium, making the system unstable. | |
| 17. |
There is and electric field E in the +x direction. If the work done by the electric field in moving a charges `0.2C` through a distance of 2m along a line making an angle of `60^(@)` with the x-axis is `1.0J`, what is the value of E in `NC^(-1)`? |
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Answer» Correct Answer - 5 `W=qE xx s cos theta or 1 = 0.2E xx 2 cos 60^(@)` or, `E=5NC^(-1)`. |
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| 18. |
If only one charge is availabel, can it be used to obtain a charge many times greater than itself in magnitude? |
| Answer» Correct Answer - By induction. | |
| 19. |
A rod of length l, has a uniform positive charge per unit length and a total charge Q. calculate the electric foeld at a point P located along the long axis of the rod and at a distance a from one end |
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Answer» The field `dvec(E )` at P due to each segment of charge on the rod is in the negative x direction because every segment carries a positive charge. Because the rod is continous we are evaluating the field due to a continuous charge distribution rather than a group of individual charges. Because every segment of the negative x direction, the sum of their contributions can be handled without the need to add vectors. Let us assume the rod is lying along the x-axis , dx is the length of one small segment and dq is the charge on that segment. Beacuse the rod has a charge per unt length 1, the charge dq on the small segment is `dq=lambda dx`. The magnitude of the electric field at P due to one segment of the rod having a charge dq is `dE=k_(e)(dq)/(x^(2))=k_(e)(lambda dx)/(x^(2))` The total field at P is `E=int_(a)^(l+a)K_(e )lambda(dx)/(x^(2))` If `K_(e )` and `lambda = Q//l` are constants and can be removed from the integral, them `E=int_(a)^(l+a) k_(e)lambda (dx)/(x^(2))[-(1)/(x)]_(a)^(l+a)` `=k_(e) Q/l(1/a-1/(l+a))=(k_(e)Q)/(a(l+a))` If a to b, which corresponds to sliding the bar to the left until its left end is at the origin, then `E to oo` that represents the condition in wich the observation point P is at zero distance from the charge at the end of the rod, so the field become infinite. |
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| 20. |
A particle has a charge of `+10^(-12)C`. (a) Does it contain more or less number of electrons as compared to the netual state?(b) Calculate the number of electrons transferred to provide this charge. |
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Answer» (a) Since the particle has positive charge, it has deficiency of electrons, so it has less number of electrons compared to the neutral state. (b) `q="ne"` `10^(-12)=nxx1.6xx10^(-19)` `n=6.25xx10^(6)`. |
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| 21. |
A block of mass m containing a net negative chage `-q` is placed on a froctionless horizontal tabe and is connected to a wall through an unstreached spring of spring constant k. If the horizontal electric field E parallel to spring is switched on , then the maximum compression of the spring is A. `sqrt(qE//k)`B. `2qE//k`C. `qE//k`D. zero |
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Answer» Correct Answer - B Force on the block: F=qE toward left. Let spring be compressed maximum by x. then `Fx=1/2 kx^(2)or qEx=1/2kx^(2)` or `x=(2qE)/(k)`. |
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| 22. |
Two identical particles are charged and held at a distance of 1m form each pther. They are found to be attraction each other with a force of 0.027N . Now they are connected by a conducting wire so that charge folws between them. Whe the charge flow stop, they are found to be repelling each other with a force of 0.009N. Find the initial charge on each particle. |
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Answer» Correct Answer - `pm3muC,pm1muC` As there is attraction (here we have assumed `Q_(1)` and `Q_(2)` in `muC`), so we get `(9xx10^(9)xxQ_(1)Q_(2)xx10^(-12))/(1^(2))=-0.027` or `Q_(1)Q_(2)=-3` Negative sign is due to attraction For repulsion : `(9xx10^(9)xx((Q_(1)+Q_(2))/(2))^(2)xx10^(-12))/(1^2)=0.009`. or `(Q_(1)+Q_(2))=4` or `Q_(1)+Q_(2)= +-2` Here, we have two sets of equations. (i) From` Q_(1)Q_(2)= -3` and `Q_(1)+Q_(2)=2` we get `3mu `C and `-1muC`. (ii) From ` Q_(1)Q_(2)= -3` and `Q_(1)+Q_(2)= -2` we get `-3muC` and `1muC`. Both are valid. |
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| 23. |
(a) How many electorns are in 1C of negative charge? (b) Which is the true test of electrification: attraction or repulsion? (c ) Can a body have a charge of `0.8xx10^(-19)C` ? |
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Answer» (a) `q="ne"` `n=(q)/(e )=(1)/(1.6xx10^(-19))=6.25xx10^(18)` (b) Repulsion. As charged body can attarct uncharged body. (c ) No, as charge is quantized. |
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| 24. |
An electron (mass `m_(e )`)falls through a distance d in a uniform electric field of magnitude E. , The direction of the field is reversed keeping its magnitudes unchanged, and a proton(mass `m_(p)`) falls through the same distance. If the times taken by the electrons and the protons to fall the distance d is `t_("electron")` and `t_("proton")` respectively, then the ratio `t_("electron")//t_("proton")`. |
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Answer» The acceleration experienced by a charged particle under the action of an electric field E is `a=(eE)/(m)` If it falls through a distance h, stating form rest, `h=(1)/(2)at^(2)` (as u=0) or, `t^(2)=(2h)/(a)=(2h)/(eE//m)=(2mh)/(eE)` or `t prop sqrt(m)` ...(i) From eq (i) the time of fall will be `(t_("electron"))/(t_("proton"))=sqrt((m_(e)/(m_(p)))` Now since `m_(p)gtm_(e)`, so `t_(p)gtt_(e)`, i.e., time of fall through the same distance is greater for a heavier particle, which is in constrant with the situation of free fall under gravity where the time of fall is independent of the mass of the body. |
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| 25. |
A particle of mass m carrying a charge `-q_(1)` starts moving around a fixed charge `+q_(2)` along a circulare path of radius r. Find the time period of revolution T of charge `-q_(1)`. |
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Answer» Electrostatic force on `-q_(1)` due to `q_(2)` will provide the necessary centripetal force, Hence, `(kq_(1)q_(2))/(r^(2))=(mv^(2))/(r ) implies v=sqrt((kq_(1)q_(2))/(mr))` Now, `T=(2 pi r)/(v)= sqrt((16 pi^(3)epsilon_(0)mr^(3))/(q_(1)q_(2))`. |
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| 26. |
A particle of mass m carrying a positive charge q moves simple harmonically along the x-axis under the action of a varying electric field E directed along the x-axis. The motion of the particle is confined between x=0 and x=2l. The angular frequency of the motion is `omega`. Then which of the following is correct ?A. `qE=-momega^(2)(x-1)`B. `qE=momega^(2)(x-1)`C. Electric field to the right of origin is directed along the positive x-axis for all balues of x.D. Electric field to the right of origin is directed along the negative x-axis for all balues of x. |
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Answer» Correct Answer - A Mean position : x=1 `F=-k(x-l)` where `k=m, omega^(2)` or `qE=-m omega^(2)(x-l)At x=l, E=0` To the right of x=l, E is negative, so toward left and to the left of x=l, E is positive. |
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| 27. |
A ring of radius R has charge `-Q` distributed uniformly over it. Calculate the charge that should be placed at the center of the ring such that the electric field becomes zero at apoint on the axis of the ring at distant R from the center of the ring. |
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Answer» Correct Answer - `(Q)/(2 sqrt(2))` Electric field at point P due to charge of ring is `E=(kQx)/((R^(2)+x^(2))^(3//2))` At `x=R, E=(kQ)/(2 sqrt(2)R^(2))` directed toward the center Electric fild at P due to charge at center , `kq//R^(2)`. For net field to be zero at P `(kq)/(R^(2))=(kQ)/(2 sqrt(2)R^(2))` or `q=(Q)/(2 sqrt(2))`. |
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| 28. |
A circular ring carries a uniformly distributed positive charge and lies in the xy plane with center at the origin of the cooredinate system. If at a point (0,0,z) the electric field is E, then which of the following graphs is correct?A. B. C. D. |
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Answer» Correct Answer - C The following two arguments shall leads us to right choice. (i) Electric field at the center of the ring zero. (ii) Electric field is directed away from the ring. |
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| 29. |
A point charge of `100mu C` is placed at `3 hat j+4 hat j m`. Find the electric field intensity due to this chargaes at a point located at `9 hat j+12 hat j m`.A. `8000Vm^(-1)`B. `9000Vm^(-1)`C. `2250Vm^(-1)`D. `4500Vm^(-1)` |
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Answer» Correct Answer - B `vec(r )=(9-3) hat i+(12-4)hat j=6 hat(i)+8 hat(j)` or, `r=sqrt(6^(2)+8^(2))=10m` `E=(9xx10^(9)xx100xx10^(-6))/(10^(2))=9000Vm^(-1)`. |
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| 30. |
An electron is projected with an initial speed `v_(0)=1.60xx10^(6)ms^(-1)` into the uniform field between the parallel plates as shown in fig. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. the electrons enters the field at a point midway between the plates. Mass of electron is `9.1xx10^(-31)kg`. The vertical displacement traveled by the proton as it exits the region between the plated is (mass pf proton is `1.67xx10^(-27)kg`).A. `124NC^(-1)`B. `364NC^(-1)`C. `224NC^(-1)`D. `520NC^(-1)` |
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Answer» Correct Answer - B Passing between the charged plates, the electron feels a force upward and just misses the top plate. The distance it travels in upward and just misses the to plate. The distance it travels in the y-direction is 0.005m. Time of flight is `t=(0.0200m)/(1.60xx10^(6)ms^(-1))` `=1.25zz10^(-8)s` the initial y-velocity is zero. Now, `y=v_(0y)t+1/2at^(2)` so, `0.005m=1/2a(1.25xx10^(-8)s)^(2)` or, `a=6.40xx10^(13)ms^(-2)` But also `a=(F)/(m)=(eE)/(m_(e))` `E=((9.1xx10^(-31)kg)(6.40xx10^(13)ms^(-2)))/(1.60xx10^(-19)C)` `=364NC^(-1)` Since the proton is more massive, it will acceletate less and not hit the plates. To find the vertical displacement when it exists the plates we use the kinematic equation again. `y=1/2at^(2)=1/2(eE)/(m_p)(1.25xx10^(-8)s)^(2)` `=2.73xx10^(-6)m` As mentioned in (b), the proton will not hit one of the plates because although the electric force felt by the proton is the same as the electrons felt, a smaller accelaration produced by the electric force is much greater than g, it is reasonable to ignore gravity. |
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| 31. |
A point charge `q = - 8.0 nC` is located at the origin. Find the electric field vector at the point `x = 1.2 m, y = -1.6 m. `A. `-14.4hat i+10.8hat j`B. `-14.4hat i-10.8hat j`C. `-10.8hat i+14.4hat j`D. `-10.8hat i-14.4hat j` |
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Answer» Correct Answer - C `vec(E ) = (1)/(4 pi epsilon_(0)) (q)/(r^(2)) hat r` `hat (r ) =(vec(r ))/(r )=((1.2-0)hat(i)+(-1.6-0)hat j)/(sqrt((1.2)^(2)+(1.6)^(2)))` `=(1)/(2)(1.2 hat(i)-1.6 hat(j))` `vec(E )=9xx10^(9)xx((-8xx10^(-9)))/(2^(2))xx(1)/(2)(1.2 hat(i)-1.6 hat(j))` `=-10.8 hat(i)+14.4hatj`. |
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| 32. |
An electron is projected with an initial speed `v_(0)=1.60xx10^(6)ms^(-1)` into the uniform field between the parallel plates as shown in fig. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. the electrons enters the field at a point midway between the plates. Mass of electron is `9.1xx10^(-31)kg`. The vertical displacement traveled by the proton as it exits the region between the plated is (mass pf proton is `1.67xx10^(-27)kg`).A. `1.6xx10^(-8)m`B. `3.25xx10^(-8)m`C. `5.25xx10^(-6)m`D. `2.73xx10^(-6)m` |
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Answer» Correct Answer - D Passing between the charged plates, the electron feels a force upward and just misses the top plate. The distance it travels in upward and just misses the to plate. The distance it travels in the y-direction is 0.005m. Time of flight is `t=(0.0200m)/(1.60xx10^(6)ms^(-1))` `=1.25zz10^(-8)s` the initial y-velocity is zero. Now, `y=v_(0y)t+1/2at^(2)` so, `0.005m=1/2a(1.25xx10^(-8)s)^(2)` or, `a=6.40xx10^(13)ms^(-2)` But also `a=(F)/(m)=(eE)/(m_(e))` `E=((9.1xx10^(-31)kg)(6.40xx10^(13)ms^(-2)))/(1.60xx10^(-19)C)` `=364NC^(-1)` Since the proton is more massive, it will acceletate less and not hit the plates. To find the vertical displacement when it exists the plates we use the kinematic equation again. `y=1/2at^(2)=1/2(eE)/(m_p)(1.25xx10^(-8)s)^(2)` `=2.73xx10^(-6)m` As mentioned in (b), the proton will not hit one of the plates because although the electric force felt by the proton is the same as the electrons felt, a smaller accelaration produced by the electric force is much greater than g, it is reasonable to ignore gravity. |
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| 33. |
A block of mass ma is suspended vertically with a spring of spring constant k. The block is made to oscillate in a gravitation field. Its time period is fould to be T. Now the space between the plates is made gravity free, and an electric field E is produced in the downward direction. Now the block is given a charge q. the new time period of oscillation is A. TB. `T+2 pi sqrt((qE)/(md))`C. `2 pi sqrt((qE)/(md))`D. none of the above |
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Answer» Correct Answer - A Time period is independent of a constant force acting on the block of spring-block system. Its time period will remain same as `T=2 pi sqrt((m)/(k))`. |
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| 34. |
Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of `10.0 g`. Silver has 47 electrins per atom, and its molar mass is `107.87 gmol^(-1)`. |
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Answer» Correct Answer - `2.62xx10^(24)` Number of atoms in 10g of silver is `n=(6.023xx10^(23)xx10)/(107.87)=5.58xx10^(22)` Number of electrons is `47n=2.62xx10^(24)`. |
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