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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. |
One-fourth of a sphere of radius R is removed as shown in figure. An electric field E exists parallel to the xy plane . Find the flux through the remaining curved part. A. `piR^(2)E`B. `sqrt(2)piR^(2)E`C. `pir^(2)E//sqrt(2)`D. `2piR^(2)E` |
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Answer» Correct Answer - 3 `phi=vec(E).(vec(A_(1))+vec(A_(2)))` |
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| 2. |
A cylinder of radius `R` and length `L` is placed in the uniform electric field `E` parallel to the cylinder axis.The total flux from the two flat surface of the cylinder is given byA. `2piR^(2)E`B. `(piR^(2))/(E)`C. `(piR^(2)-piR)/(E)`D. zero |
| Answer» Correct Answer - 4 | |
| 3. |
A bob of a simple pendulum of mass `40gm` with a positive charge `4xx10^(-6)C` is oscillating with a time period `T_(1)`.An electric field of intensity `3.6xx10^(4)N//C` is applied vertically upwards.Now the time period is `T_(2)` the value of `(T_(2))/(T_(1))` is `(g=10//s^(2))`A. `0.16`B. `0.64`C. `1.25`D. `0.8` |
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Answer» Correct Answer - 3 `T=2pisqrt((l)/(g_(eff))` |
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| 4. |
The inward and outward electric flux for a closed surface unit of `N-m^(2)//C` are respectively `8xx10^(3)` and `4xx10^(3)`. Then the total charge inside the surface is [where `epsilon_(0)=` permittivity constant]A. `4xx10^(3)`B. `-4xx10^(3)`C. `-(piR^(2)-piR)/(E)`D. `-4xx10^(3)epsilon_(0)` |
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Answer» Correct Answer - 4 `phi_(E)=(q)/(epsilon_(0))` |
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| 5. |
A hemispherical body of radius R is placed in a uniform electric field E. What is the flux linked with the curved surface if, the field is (a) parallel to the base, (b) perpendicular to the base.A. `2piR^(2)E`B. `piR^(2)E`C. `4piR^(2)E`D. `6piR^(2)E` |
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Answer» Correct Answer - 2 `phi=bar(E).bar(A)` |
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| 6. |
A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at `x=+1cm ` and C be the point on the y-axis at `y=+1`cm. then the potetial at the points A,B and C satisfya. `V_AltV_B`, b. `V_AgtV_B` c. `V_AltV_C` d. `V_AgtV_C`A. `V_(A)lt V_(B)`B. `V_(A)gt V_(B)`C. `V_(A)lt V_(C)`D. `V_(A)gt V_(C)` |
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Answer» Correct Answer - 2 Along the field direction potential decrease |
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| 7. |
A cube of arranged such that its length ,breadth ,height are along`X,Y,Z` directions one of its corners is situated at the origin Length of each side of the cube is `25cm` .The components of electric field are `E_(x)=400sqrt(2)N//C` E_(y)=0` and `E_(z)=0` respectively.Flux coming out of the cube at one end will beA. `25sqrt(2)Nm^(2)//C`B. `5sqrt(2)Nm^(2)//C`C. `250sqrt(2)Nm^(2)//c`D. `25Nm^(2)//C` |
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Answer» Correct Answer - 1 `bar(E)_(1).bar(A)_(1)+bar(E)_(2).bar(A)_(2)+bar(E)_(3)bar(A)_(3)=phi` |
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| 8. |
`A,B,C` are three points as a circle of radius `1cm`.These points form the corners an equilateral triangle .A charge `2C` is placed at the centre of circle .The work done in carrying a charge of `0.1muc` form `A` to `B` isA. zeroB. `18xx10^(11)J`C. `1.8xx10^(11)J`D. `54xx10^(11)J` |
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Answer» Correct Answer - 1 Equipotnetial surface |
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| 9. |
Point charge of `3xx10^(-9)C` are situated at each of three corners of a square whose side is `15cm`. The magnitude and direction of electric field at the vacant corner of the square isA. `2296V//m` along the diagonalB. `9622V//m` along the diagonalC. `22.0V//m` along the diagonalD. Zero |
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Answer» Correct Answer - 1 `E=E(sqrt(2)+1//2)E=(1)/(4piepsilon_(0))(q)/(r^(2))` r =length of the side |
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| 10. |
Three charges each`20muC` are placed at the corners of an equilateral triangle of side `0.4m` The potential energy of the system isA. `18xx10^(-6)J`B. `9J`C. `9xx10^(-6)J`D. `27J` |
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Answer» Correct Answer - 4 `U=(1)/(4piepsilon_(0))((q_(1)q_(2))/(r_(12))+(q_(2)q_(3))/(r_(23))+(q_(3)q_(1))/(r_(13)))` |
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| 11. |
Four idential charges each of charge `q` are placed at the corners of a square .Then at the centre of the square the resultant electric intensity `E` and the net electric potential `V` areA. `E!=0,V=0`B. `E=0,V=0`C. `E=0,V!=0`D. `E!=0,V!=0` |
| Answer» Correct Answer - 3 | |
| 12. |
Three charge `-q,+q` and `-q` are placed at the cornors of an equilateral triangle of side `a` The resultant electric froce an a charge `+q` placed at a the centroid `O` of the triangle isA. `(3q^(2))/(4piepsilon_(0)a^(2))`B. `(q^(2))/(4piepsilon_(0)a^(2))`C. `(q^(2))/(2piepsilon_(0)a^(2))`D. `(3q^(2))/(2piepsilon_(0)a^(2))` |
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Answer» Correct Answer - 4 `F=(1)/(4piepsilon_(0))(q_(1)q_(2))/(r^(2))` |
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| 13. |
Two charge `+Q` each are placed at the two verticles of an equilateral triangle of side `a` .The intensity of electric field at the third vertex is |
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Answer» `E^(1)=sqrt(E^(2)+E^(2)+2E E cos theta)` `=sqrt(2E^(2)+2E^(2)costheta)` ` =sqrt(2E^(2)(1+costheta))` `=2Ecos""(theta)/(2),Esqrt(3)(1)/(4piepsilon_(0))(Q)/(a^(2))`. |
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| 14. |
The longer side of a rectangle is twice the length of its shorter side .A charge `q` is kept at one vertex.The maximum electric potential due to that charge at any other vertex is `V`, then the minimum electric potential at any other vertex will beA. `2V`B. `sqrt(3)V`C. `V//sqrt(5)`D. `sqrt(5)V` |
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Answer» Correct Answer - 3 long and short diagonal lengths are `sqrt(p^(2)+q^(2)+-pqcostheta)` |
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| 15. |
Which of the following pair is releated as in work and forceA. electric potential and electric intensityB. momentum and forceC. impulse and forceD. resistance and voltage |
| Answer» Correct Answer - 1 | |
| 16. |
Electric potential at the center ofa charged hollow spherical conductor isA. zeroB. twice as that on the surfaceC. half of that on the surfaceD. same as that on the surface |
| Answer» Correct Answer - 4 | |
| 17. |
A hollow metal sphere of radius ` 5 cm` is charged such that the potential on its surface is ` 10V`. The potential at the center of the sphere is -A. `0V`B. `10V`C. same as at point `5cm` away from the surfaceD. same as that on the surface |
| Answer» Correct Answer - 2 | |
| 18. |
Assertion. If a point charge `q` is placed in front of an infinite grounded conducting plane surface, the point charge will experience a force. Reason. This force is due to the induced charge on the conducting surface which is at zero potential |
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Answer» Correct Answer - A `E` is proportional to distance upto `x=R` so the sphere has uniformly distributed charge over its volume and because of this reason ,the sphere is made of non-conduting material. The electric field is same just inside and outside of the sphere.therefore dielectric constnats of the materials of sphere and surrounding are same .Potential at the centre is maximum. |
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| 19. |
If `E_(a)` be the electric field strength of a short dipole at a point on its axial line and `E_(e)` that on the equatorial line at the same distance, thenA. `E_(a)=E_(r)`B. `E_(a)=2E_(r)`C. `E_(r)=2E_(a)`D. `E_(r)=2E_(a)` |
| Answer» Correct Answer - 2 | |
| 20. |
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. |
| Answer» Correct Answer - 2 | |
| 21. |
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 - 4 `mg(sinalpha-mucosalpha)-muqsinalpha=qEcosalpha` |
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| 22. |
Due to the motion of a charge its magnitudeA. changesB. does not changeC. increase (or) decrease depends on its speedD. can not be predicted |
| Answer» Correct Answer - 2 | |
| 23. |
Equipotential sufaces are shown in figure a and b the field in A. a is uniform onlyB. b is uniform onlyC. a and b is uniformD. both are nonuniform |
| Answer» Correct Answer - 1 | |
| 24. |
An equipotential line and a line of force areA. perpendicular to each otherB. parallel to each otherC. in any directionD. at an angle of `45^(@)` |
| Answer» Correct Answer - 1 | |
| 25. |
When a positively charged conductor is placed near an earth connected conductor its potentialA. always increaseB. always decreaseC. may increase and decreaseD. remains the same |
| Answer» Correct Answer - 2 | |
| 26. |
A charged spherical conductor has a surface charge density of `0.7C//m^(2)` .When its charge is increased by `0.44C` the charge denstiy changes by `0.14C//m^(2)` the radius of the sphere isA. `5cm`B. `10m`C. `0.5m`D. `5m` |
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Answer» Correct Answer - 3 `Delta sigma=(Delta Q)/(A)` |
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| 27. |
The electric field in a region of space is given by `E=5hati+2hatjN//C`. The flux of E due ot this field through an area `1m^2`lying in the y-z plane, in SI units isA. `10`B. `20`C. `10sqrt(2)`D. `2sqrt(29)` |
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Answer» Correct Answer - 1 `phi_(E)=intvec(E).vec(ds)=Exxs=5xx2` |
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| 28. |
Electric flux at a point in an electric field isA. positiveB. negativeC. zeroD. positive and negative |
| Answer» Correct Answer - 3 | |
| 29. |
The electric field in a region is given by `E=alphaxhati`. Here `alpha` is a constant of proper dimensions. Find a. the total flux passing throug a cube bounded by the surface `x=l, x=2l, y=0, y=l, z=0, z=l`. b. the charge contained inside in above cube.A. `alphal^(3)`B. `2alphal^(3)`C. `3alphal^(3)`D. `4alphal^(3)` |
| Answer» Correct Answer - 1 | |
| 30. |
Electric flux over a surface in an electric fieldA. positiveB. negativeC. zeroD. positive and negative ,zero |
| Answer» Correct Answer - 4 | |
| 31. |
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))` |
| Answer» Correct Answer - 3 | |
| 32. |
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 | |
| 33. |
A particle of charge `-q` and mass m moves in a circle of radius r around an infinitely long line charge of linear charge density `+lamda`. Then, time period will be where , `k=1/4piepsilon_0)`A. `T=2pir sqrt((m)/(2klambdaq)`B. `T^(2)=(4pi^(2)m)/(2klambdaq)r^(3)`C. `T^(2)=(1)/(2pir)sqrt((2klambdaq)/(m))`D. `T=(1)/(2pir)sqrt((m)/(2klambdaq))` where `k=(1)/(4piepsilon_(0))` |
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Answer» Correct Answer - A We have centripetal force equation `q((2klambda)/(r))=(mv^(2))/(r) sov=sqrt((2kqlambda)/(m))`Now, `T=(2pir)/(v)=sqrt((m)/(2klambdaq))` where `k=(1)/(4piepsilon_(0))` |
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| 34. |
An oil drop carrying charge `Q` is held in equilibrium by a potential difference of `600V` between the horizontal plates.In order to hold another drop of radius in equilibrium a potential drop of `1600V` had to be maintained .The charge on the second drop isA. `(Q)/(2)`B. `2Q`C. `(3Q)/(2)`D. `3Q` |
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Answer» Correct Answer - 4 `(V_(1))/(V_(2))=((R_(1))/(R_(2)))^(3),(Q_(2))/(Q_(1))` |
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| 35. |
Two thin wire rings each having radius R are placed at distance d apart with their axes coinciding. The charges on the two are `+Q` and `-Q`. The potential difference between the centre so the two rings isA. `(Q.R)/(4piepsilon_(0)d^(2))`B. `(Q.R)/(2piepsilon_(0))((1)/(R)-(1)/(sqrt(R^(2)+d^(2))))`C. `(Q.R)/(4piepsilon_(0))((1)/(R)-(1)/(sqrt(R^(2)+d^(2))))`D. `0` |
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Answer» Correct Answer - 2 `E=(1)/(4piepsilon_(0))(qsinpi//2)/(r^(2)pi//2)` `E=(qsin pi//2)/(2pi^(2) epsilon _(0) r^(2)) (-overline(j))` |
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| 36. |
Two point metal charges`-q` and `+2q` are placed at a certain distance apart.Where should a third point charge be placed so that it is in equilibrium?A. on the line joining the two charges on the right of `+2q`B. on the line joining the two charge on the left of `-q`C. between `-q` and `+2q`D. at any point on the right bisector of the 1 line joining `-q` and `+2q` |
| Answer» Correct Answer - 2 | |
| 37. |
Two point charges of magnitude `4muC` and `-9muC` are `0.5` apart .the electric intensity is zero at a distance `xm` from `A` and `ym` form `B` .`x` and `y` are respectively A. `0.5m,1.0m`B. `1.0m,1.5m`C. `2.0m,1.5m`D. `1.5m,2.0m` |
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Answer» Correct Answer - 2 Distance of null point `x=(d)/(sqrt(Q_(2)/(Q_(1))+-1))` +ve for like charges -ve for unlike charges |
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| 38. |
a conducting disc of radius `R` roatates about its axis when an angular velocity `omega`.Then the potential difference between the centre of the disc and its edge is (no magnetic field is present ):A. zeroB. `(m_(e)omega^(2)R^(2))/(2e)`C. `(m_(e)omega^(2)R^(2))/(3e)`D. `(em_(e)omega^(2)R^(2))/(2)` |
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Answer» Correct Answer - BC `eE=m_(e)w^(2)r` `intEdr=(m_(e)omega^(2))/(e) int_(0)^(R)rdr V=(m_(e)omega^(2)R^(2))/(2e)` |
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| 39. |
Two infinitely long thin styraight wires having uniform linear charge densities `lambda` and `2lambda` are arranged parallel to each other at a distance `r` apart.The intensity of the electric field at a point midway between them isA. `(2lambda)/(piepsilon_(0)`B. `(lambda)/(piepsilon_(0)r)`C. `(lambda)/(2piepsilon_(0)r)`D. `(3lambda)/(2piepsilon_(0)r)` |
| Answer» Correct Answer - 2 | |
| 40. |
A charged bead is capable of sliding freely through a string held veticlaly in tension .an electric field is applied parallel to the string so that the bead stays at rest at the middle of the string .If the electric field is switched off momentarily and switched onA. the bead moves downwards and stops as soon as th efield is switched onB. the bead moved downwards when the field is switched off and moves upwards when the field isC. the bead moves downwards with constant acceleration till it reaches the bottom of the stringD. the bead moves downwards with constant velocity till it bottotm of the string |
| Answer» Correct Answer - 4 | |
| 41. |
Two charges are placed at a distance apart if a glas slab is placed between them force between them willA. be zeroB. increaseC. decreaseD. remains the same |
| Answer» Correct Answer - 3 | |
| 42. |
An infintie by long plate has surface change density `sigma` As shown in the fig .apoint charge `q` is moved form `A` to `B` .Net work done by electric field is A. `(sigmaq)/(2epsilon_(0))(X_(1)-X_(2))`B. `(sigmaq)/(2epsilon_(0))(X_(2)-X_(1))`C. `(sigmaq)/(epsilon_(0))(X_(2)-X_(1))`D. `(sigmaq)/(epsilon_(0))(2pir+r)` |
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Answer» Correct Answer - A `W_("net")=qvec(E).vec(d)` where `vec(E)=(sigma)/(2epsilon_(0))hat(i):.W_("net")=q(sigma)/(2epsilon_(0)).(x_(1)-x_(2))` |
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| 43. |
A solid metallic sphere has a charge `+3Q`. Concentric with this sphere is a conducting spherical shell having charge `-Q`. The radius of the sphere is `a` and that of the spherical shell is `b(bgta)` What is the electric field at a distance `R(a lt R lt b)` from the centreA. `(Q)/(2piepsilon_(0)R)`B. `(3Q)/(2piepsilon_(0)R)`C. `(3Q)/(4piepsilon_(0)R)`D. `(4Q)/(2piepsilon_(0)R^(2)` |
| Answer» Correct Answer - 3 | |
| 44. |
An infinitely long thin straight wire has uniform linear charge density of `1//3 "coul" m^(-1)` .Then the magnitude of the electric intensity at a point `18cm` away isA. `0.33xx10^(11)N//C`B. `3xx10^(11)NC^(-1)`C. `0.66xx10^(11)NC^(-1)`D. `1.32xx10^(11)NC^(-1)` |
| Answer» Correct Answer - 1 | |
| 45. |
If `r` and `T` are radius and surface tension of a spherical soap bubble respectively then find the charge needed to double the radius of bubble. |
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Answer» For smalle bubble `P_(1)=(P_(0)+(4T)/(r))` and `V_(1)=(4)/(3)pir^(3)` for larger bubble `P_(2)=P_(0)+(4T)/(R)-(sigma^(2))/(2epsilon_(0))` and `V_(2)=(4)/(3)piR^(2)` where `sigma=(q)/(4piR^(2))` for air inthe bubble `P_(1)V_(1)=P_(2)V_(2) (P_(0)+(4T)/(r))r^(3)=[(P_(0)+(4T)/(R))-(q^(2))/(16pi^(2)R^(4)xx2epsilon_(0))]R^(-3)` `P_(0)[R^(3)-r^(3)]+4T[R^(2)-r^(2)]P_(0)[R^(3)-r^(3)]+4T[R^(2)-r^(2)]=(q^(2))/(32pi^(2)epsilon_(0)R)=0` But `R=2r` `P_(0)[7r^(3)]+4T[3r^(2)]-(q^(2))/(32pi^(2)epsilon_(0)(2r))=0` `(q^(2))/(64pi^(2)epsilon_(0)r)=7P_(0)r^(3)+12Tr^(2)` `q^(2)=64pi^(2)epsilon_(0)r^(3)[7P_(0)r+12T]` `q=8pir[epsilon_(0)r(7P_(0)r+12T)]^(1//2)` |
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| 46. |
A soap bubble is given a negative charge, then its radiusA. DecreaseB. increaseC. remains unchangedD. Nothing can be predicted as information is insufficient. |
| Answer» Correct Answer - 2 | |
| 47. |
A charge `Q=(5)/(100)nC` is distributed over two concentric hollow spheres of radii `r=3` cm and `R=6cm` such that the densities are equal .Find the potential in volt at the common centre. |
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Answer» Correct Answer - 9 Let `q_(1)` and `q_(2)` be the respective charges distributed over two concentric spehres of radii `r` and `R` such that `q_(1)+q_(2)=Q` As surface densities are given to be equal therefore `sigma_(1)=sigma_(2)rArr (q_(1))/(4pir^(2))=(q_(2))/(4piR^(2))rArr (q_(1))/(q_(2))=(r^(2))/(R^(2))` `rArr (q_(1))/(q_(2))+1=(r^2)/(R^(2))+1 rArr (q_(1)+q_(2))/(q_(2))=(r^(2)+R^(2))/(R^(2))` using (1) &(2) we get, `(Q)/(q_(2))=(r^(2)+R^(2))/(R^(2))` this given `q_(2)=[(R^(2))/(r^(2)+R^(2))]+Q` Therfore `q_(1)=Q-q_(2)` ` rArr q_(1)=Q-((R^(2))/(r^(2)+R^(2)))Q=((r^(2))/(r^(2)+R^(2)))` ` rArr q_(1)=Q-((R^(2))/(r^(2)+R^(2)))Q=((r^(2))/(r^(2)+R^(2)))Q` the potential `V_(1)` at common centre due to charge `q_(1)` is given by `V_(1)=(1)/(4piepsilon_(0))(q_(1))/(r) rArr V_(1)=(1)/(4piepsilon_(0))((r^(2))/(r^(2)+R^(2)))(Q)/(r)` `rArr V_(1)=(1)/(4piepsilon_(0))(Qr)/(r^(2)+R^(2))` the potential `V_(2)` at common centre due to charge `q_(2)` is `V_(2)=(1)/(4piepsilon_(0))(QR)/(r^(2)+R^(2))` `:.` Net potential at common centre `V=V_(1)+V_(2)` `rArr V=(1)/(4piepsilon_(0))(QR)/((r^(2)+R^(2))[r+R]` `rArr V=(1)/(4piepsilon_(0))(Q(R+r))/(R^(2)+r^(2))` `rArr V=9xx10^(9)xx(5)/(100)xx10^(-9)xx(9)/(45)xx100` `rArr V=9V` |
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| 48. |
A graph of the x-component of the electric field as a function of x in a region of space is shown in figure. The `y- and z-components ` of the electric field are zero in this region. If the electric potential is 10 V at the origin, then the potential at x = 2.0 m is A. `10V`B. `40V`C. `-10V`D. `30V` |
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Answer» Correct Answer - D `V_(B)-V_(A)=-intE_(x)dx` `=-["Area under" `E_(x)-x` curve]` `V_(B)-10=-(1)/(2).2.(-20)=20 V_(B)=30V` |
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| 49. |
Three conducting concentric shell s`A,B, and C` of radius `a,2a,3a` are as shown in the figure.The charge on shell `B` is `Q`..The switch `S` is closed.Then after closing the switch If `sigma_(A),sigma_(B)` and `sigma_(C)` are the surface charge densities on `A,B and C` respectively then `sigma_(A):sigma(B):sigma(C)`isA. `9:9:1`B. `9:-9:1`C. `-9:9:1`D. `1:1:1` |
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Answer» Correct Answer - C `sigma_(A)=(-Q)/(4(4pia^(2)))sigma_(B)=(Q)/(4pi4a^(2))sigma_(C)=(Q)/(44pi9a^(2)) sigma_(A): sigma_(B): sigma_(C)=-(1)/(4):(1)/(4):(1)/(36)=-9:9:1` |
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| 50. |
Surface charge density of soap bubble of radius `r` and surface tension `T` is `sigma`.If `P` is excess pressure the value of `sigma` isA. `epsilon_(0)[(4T)/(r)-P]^(3/2)`B. `[2epsilon_(0)((4T)/(r)-P)]^(1/2)`C. `(4T)/(r )`D. `[4epsilon_(0)((2T)/(r)-P)]^(1/2)` |
| Answer» Correct Answer - 2 | |