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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.
| 51. |
Three concentric spherical metallic spheres `A,B` and `C` of radii `a , b` and `c(a lt b lt c)` have surface charge densities `sigma , -sigma` and `sigma` respectively.A. `(sigma)/(epsilon_(0))(a-b+c),(sigma)/(epsilon_(0))(a^(2)/(b)-b+c)`B. `(a-b+c),(a^(2))/(c )`C. `-(epsilon_(0))/(sigma)(a-b+c),(epsilon_(0))/(sigma)(a^(2)/(b)-b+c)`D. `-(sigma)/(epsilon_(0))((a^(2))/(c)-(b^(2))/(c)+c),(sigma)/(epsilon_(0))(a-b+c)` |
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Answer» Correct Answer - 1 `V_(A)=(1)/(4piepsilon_(0))((4pia^(2)sigma)/(a)-(4pia^(2)sigma)/(b)+(4pic^(2)sigma)/(c))` `rArr V_(A)=(sigma)/(epsilon_(0))(a-b+c)and ` `rArr V_(B)=(1)/(4piepsilon_(0))((4pia^(2)sigma)/(b)-(4pib^(2)sigma)/(b)+(4pic^(2)sigma)/(c))` `rArr V_(B)=(sigma)/(epsilon_(0))((a^(2))/(b)-b+c)` |
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| 52. |
the figure shows a charge `q` placed inside a cavity in an uncharged conductor, Now if na external electric field is switched on A. Only induced charge on outer surface will redistributeB. Only induced charge on inner surface will redistributeC. both induced charge on outer and inner surface will redistributeD. force on charge `q` placed inside the cavity will change. |
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Answer» Correct Answer - A The figure shows a charge `q` placed inside a cavity in an unchanged conductor,Now if an external electric field is switched on : (A) Only induced charge on outer surface will redistribute. Only induced charge on inner surface will redistribute (C ) both induced charge on outer and inner surface will redistribute (D) force on charge `q` placed inside the cavity will charge. |
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| 53. |
A spherical charged conductor has surface charge density `sigma`.The intensity of electric field and potential on its surface are `E` and `V` .Now radius of sphere is halved keeping the charge density as constant .The new electric field on the surface and potneial at the centre of the sphere areA. `4E,V`B. `E,V//2`C. `E,V`D. `2E,4V` |
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Answer» Correct Answer - 2 Pd between the two spheres is independent of charge on outer shell |
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| 54. |
Electric potential at some point in space is zero then at that pointA. electric intensity is necessarily zeroB. elecrtic intensity is necessarily non zeroC. electric intensity of may or may not be zeroD. electric intensity is necessarily infinte |
| Answer» Correct Answer - 3 | |
| 55. |
`X` and `Y` are large, parallel conducting plates close to each other. Each face has an area `A`. `X` is given a charge `Q`. `Y` is without any charge. Points `A,B` and `C` are as shown in the figure. A. The field at `B` is `(Q)/(2epsilon_(0)A)`B. The field at `B` is `(Q)/(epsilon_(0)A)`C. The field at `A,B` and `C` are of the same magnitude.D. The field at `A` and `C` are of the same magnitude but in opposite directions. |
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Answer» Correct Answer - A,C,D `E=(sigma )/(2epsilon_(0))=(q)/(2Aepsilon_(0))` at points `A,B,&C` |
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| 56. |
Assertion(A):`vec(E)` in the outside vicinity of a conductor depends only on the local charge density `sigma` and it si independent of the other charges present anywhere on the conductor. Assertion(R ):`vec(E)` in outside vicinity of a conductor is given by `(sigma)/(epsilon_(0))`A. Both `A` and `R` are true and `R` is the correct explanation of `A`B. Both `A` and `R` are True but `R` is not the correct explanation of `A`.C. `A` is true and `R` is falseD. `A` is false and `R` is true. |
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Answer» Correct Answer - D Electron being a negative charge, will be move from lower to highter potetial. |
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| 57. |
A point charge `q` si located at a distance `r` from the centre `O` of an uncharged conducting conducting spherical layer whose inside and outside radii are equal to `R_(1)` and `R_(2)` respectively. Find the potentail at the point `O` if `r lt R_(1)`.A. `((1)/(r )-(1)/(a)+(1)/(b))`B. `((1)/(a)-(1)/(r )+(1)/(b))`C. `((1)/(b)-(1)/(c )-(1)/(r))`D. `((1)/(a)-(1)/(b )-(1)/(r))` |
| Answer» Correct Answer - 3 | |
| 58. |
In the above question if the sheets were thick and codnucting .value of `E` in the space betwee the two sheets would beA. `2sigma//epsilon_(0)`B. `sigma//epsilon_(0)`C. zeroD. `4sigma//epsilon_(0)` |
| Answer» Correct Answer - 1 | |
| 59. |
The Gaussian surface for calculating the electric field due to a charge distribution isA. any closed surface around the charge distributionB. any surface near the charge distributionC. a spherical surfaceD. a closed surface at a every point of which electric field has a normal components which is zero or a fixed value |
| Answer» Correct Answer - 4 | |
| 60. |
Consider two large ,identical paralllel conducting plates having surface `X` and `Y` .facing eacg other.The charge per unit area on `X` is `sigma_(1)` and charge perunit surface area of `Y` is `simga_(2)` .Then,A. `sigma_(1)=-sigma_(2)` only if a charge is given to either `X` or `Y`.B. `sigma_(1)=sigma_(2)=0` if equal charges are given to both `X` and `Y`.C. `sigma_(1) gt sigma_(2)` if `X` is given a charge more than that given by `Y`D. `sigma_(1)=-sigma_(2)` in all cases. |
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Answer» Correct Answer - A,C Use `E=(sigma)/(epsilon_(0))` between the plates. |
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| 61. |
In the above problem the value of `E` in the space outside the sheets isA. `sigma//epsilon_(0)`B. `sigma//2epsilon_(0)`C. zeroD. `2sigma//epsilon_(0)` |
| Answer» Correct Answer - 3 | |
| 62. |
Two infinite parallel,non- conducting sheets carry equal positive charge density `sigma` ,One is placed on the other at distance `x=a` Take potential `V=0` at `x=0` thenA. For `0 le x le a` potential `V_(x)=0`B. for `x ge a` potential `V_(x)=(sigma)/(epsilon_(0))(x-a)`C. for `x ge a` potential `V_(x)=-(sigma)/(epsilon_(0))(x-a)`D. For `x le a` potential `V_(x)=-(sigma)/(epsilon_(0))x` |
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Answer» Correct Answer - A,B,D `0 lt xlt a` `V_(x)=[-int_(0)^(x)E_(x)dx]+V_(0)=0(as E_(x)=0)` `x gea,V_(s)=-int_(0)^(x)E_(x)dx+V_((a))` ` V_(x)=[-int_(a)^(x)(sigma)/(epsilon_(0))dx]+V_((a))=(sigma)/(epsilon_(0))(x-a)=(sigma)/(epsilon_(0))(x-a)` `x le a,V_(x)=` `-int_(0)^(x)E_(x)dx+V_((0))[:. E_(x)=(sigma)/(epsilon_(0))]` `V_((x))=((-sigma)/(epsilon_(0))x)+V_((0))=(sigma)/(epsilon_(0)x)` |
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| 63. |
Two infinite sheets of uniform charge density ` + sigma` and ` - sigma` are parallel to each other as shown in the figure . Electric field at the .A. points to the left or to the right of the sheets is zeroB. midpoint between the sheet is zeroC. midpoint of the sheets is `(sigma)/(epsilon_(0))` and is directed towards right.D. midpoint of the sheet is `2(sigma)/(epsilon_(0))` and is directed towards right |
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Answer» Correct Answer - A,C `E=(sigma )/(2epsilon_(0))` apply at the region `I,II`,&`III` |
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| 64. |
A metal ball of radius `R` is placed concentrically inside hollow metal sphere of inner radius `2R` and outer radius `3R`.The ball is given a charge `+2Q` and the hollow sphere a total charge `-Q` .The electrostatic potential energy of this system is A. `(7Q^(2))/(24piepsilon_(0)R)`B. `(5Q^(2))/(16piepsilon_(0)R)`C. `(5q^(2))/(8piepsilon_(0)R)`D. None |
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Answer» Correct Answer - A `U_("system")=U_("self")+U_("int")` |
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| 65. |
A hemispherical shell is uniformly charge positively .the electric field at point on a diameter away from the centre is directedA. perpendicular to the diameterB. parallel to the diameterC. at an angle tilted towards the centreD. at an angle of tiled away form the centre |
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Answer» Correct Answer - D The electric field at an internal point of a unformly charged spher is radially outward. |
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| 66. |
Two positively charged particles ` X` and `Y` are initially far away from each other and at rest , `X` begins to move towards `Y` with some initial velocity. The total momentum and energy of the system are `p` and `E`.A. If `Y` is fixed both `P` and `E` are conservedB. If `Y` is fixed ,`E` is conserved but not `P`.C. if both are free to move `p` is conserved but not `E`.D. if both are free,`E` is conserved but not `p` |
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Answer» Correct Answer - B `(A) v_(A)-V_(B)`=work done by electric field on `+1` coul.charge form `A` to `B` =`Erq` `:.v_(B)=V_(A)-E Rq=v-E Rq` |
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| 67. |
Six charges are placed at the vertices of a rectangular hexagon as shown in the figure. The electric field on the line passing through point O and perpendicular to the plane of the figure as a function of distance x from point O is (xgtgta) A. `(Qa)/(piepsilon_(0)x^(3))`B. `(2Qa)/(piepsilon_(0)x^(3))`C. `(sqrt(3)Qa)/(piepsilon_(0)x^(3))`D. zero |
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Answer» Correct Answer - 1 concept of force |
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| 68. |
A point mass `m` and charge `q` is connected with a spring of negligible mass with natural length `L`., Initially spring is in natural length .Now a horizontal unifrom electric field `E` is switched on as shown .Find (a) The maximum separation between the mass and the well (b) Find the separation of the point mass and wall at the equilbrium position of mass (c ) Find the energy stored in the spring at the equilirbium position of the point mass` |
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Answer» At ,maximum separation velocity of point mass is zero.From work energy theorem `W_("spring")+W_("field")=0` ` qEx_(0)-(1)/(2)kx_(0)^(2)(x_(0)` maximum elongation) `rArr x_(0)=(2qE)/(K),:.`separation =`L+(2qE)/(k)` (b) At equilateral position `EqEq=kxrArr x=(qE)/(k)` `rArr"separation"=L+(qE)/(k)` c )`U=(1)/(2)kx^(2)=(1)/(2)k((qE)/(k))^(2)=(q^(2)E^(2))/(2k)` |
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| 69. |
A ring of radius `R` is placed in the plane with its centre at origin and its axis along the `x-axis` and having uniformly distributed positive charge.A ring of radius `r( ltlt R)` and coaxial with the larger ring is moving along the axis with constant velocity then the variation of electrical flux `(phi)` passing through the smaller ring with position will be best represented by : A. B. C. D. |
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Answer» Correct Answer - C `f=vec(E),vec(ds)` since `r lt lt R` so we can consider electric field is constant throughout the surface of smaller ring. hence `f=E prop(x)/((R^(2)+x^(2))^(3//2))` |
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| 70. |
A dipole of dipole moment `P` is kept at the centre of a ring of radius `R` and charge `Q` the dipole moment has direction along the ring due to the dipole is :A. zeroB. `(kpQ)/(R^(3))`C. `(2kpQ)/(R^(3))`D. `(kpQ)/(R^(3))` only if the change is uniformly distributed on the ring . |
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Answer» Correct Answer - B Electric field at each point on the surface of ring due to the dipole is `E=(kp)/(R^(3))` in direction opposite to the dipole moments .(figure below) Hence net force on ring is `F=QE=(kpQ)/(R^(3))` Alternate solution Electric field due to ring at point`P` on its axis at a distance `x` from the centre `O` of the ring is `E=k.(Qx)/((x^(2)+R^(2))^(3//2)),(dE)/(dx)]_(at x=0)=(kQ)/(R^(3))` Force on dipole =`(dE)/(dx)=(kQp)/(R^(3))` |
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| 71. |
An electric dipole of moment `p` is kept at a distance `r` form an infinte long charged wire of linear charge density `lambda` as shown.Find the force acting on the dipole? |
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Answer» Fields intensity at a distance `r` form the line of charge is `e=(lambda)/(2piepsilon_(0)r)` The force on the dipole is `F=-p(dE)/(dr)` `=-p[(-lambda)/(2piepsilon_(0)r^(2))]=(plambda)/(2piepsilon_(0)r^(2))` Hence the net force on dipole due to the wire will be attractive. |
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| 72. |
The electric potential V at any point x, y, z (all in meters) in space is given by `V=4x^2` volts. The electric field at the point (1m, 0, 2m) is……………..`V//m`. |
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Answer» As electric field `E` is related to potential `V` through the relation `E=-(dV)/(dr)` `E_(x)=-(dV)/(dx)=-(d)/(dx)(4x^(2))=-8x` `E_(y)=(dV)/(dt)=-(d)/(dy)(4x^(2))=0` and `E_(z)=-(dV)/(dt)=-(d)/(dz)(4x^(2))=0` So `vec(E)=hat(i)E_(x)+hat(j)E_(y)+hat(k)E_(z)=-8xhat(i)` ltbr. i.e, it has magnitude `8V//m` and is directed along negative `x-axis` |
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| 73. |
A non-conducting ring of radius `0.5 m` carries a total charge of `1.11xx10^(-10)`C distributed non-uniformly on its circumference producing an electric field E everywhere is space. The value of the integral `int_(l=oo)^(l=0)-E.dI (l=0` being centre of the ring) in volt isA. `+2`B. `-1`C. `-2`D. zero |
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Answer» Correct Answer - 1 `V=(q)/(4piepsilon_(0))[(1)/(x_(0))-(1)/(2x_(0))+(1)/(3x_(0))+(1)/(4x_(0))+......]` `=(q)/(4piepsilon_(0)x_(0))[1-(1)/(2)+(1)/(3)+(1)/(4)+.....]` `=(q)/(4piepsilon_(0)x_(0))log(2)` |
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| 74. |
Two point charge `q_(1)=2muC` and `q_(2)=1muC` are placed at distance `b=1` and `a=2cm` from the origin on the `y` and `x` axes as shown in figure .The electric field vector at point `(a),(b)` will subtnd on angle `theta` with the "x-axis" given by A. `tantheta=1`B. `tan theta=2`C. `tan theta=3`D. `tantheta=4` |
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Answer» Correct Answer - 1 `Tan theta=(E_(2))/(E_(1))` |
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| 75. |
Four charges of `1mC,2mC,3mC` and `-6mC` are placed one at each corner of the square of side `1m`.The square lies in the `x-y` plane with its centre at the originA. The electric potential is zero at the originB. the electric potential is zero everywhere along the x-axis only on the sides of the square which are parallel to y-axisC. the electric potential is zero everywhere along the z-axis for any oritentation of the square in the x-y plane.D. The electric potential is not zero along the z-axis except at the origin |
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Answer» Correct Answer - A,C Verify electric field and electric potential at square |
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| 76. |
Which of the following statement are correct?A. both `a` &`c`B. only `c`C. both `c` & `d`D. all |
| Answer» Correct Answer - 2 | |
| 77. |
Which of the following transaction statement are correct.A. Electric lines of force are just imaginary linesB. Electric lines of force will be parallel to the surface of conductorC. If the lines of force will be parallel to the surface of conductorD. Electric lines of force are closed loops |
| Answer» Correct Answer - 1 | |
| 78. |
the property of the electric line of force (a) tangent to the line of force at any point is parllel to the parallel to the of force at any point (b) No two lines of force at any pointA. both `a` & `b`B. only `a`C. only `b`D. `a` or `b` |
| Answer» Correct Answer - 1 | |
| 79. |
The sketech below show cross secion so equipotential surfaces between two charged.Conductors that are shwon in solid black.Some points on the wquipotenital surfaces .near the conductors are marked as `A,B,C…. .` The arrangements lies in air (Take `epsilon_(0)=8.85xx10^(-12)C^(2)//Nm^(2)]` Surfaces charge density of the plate is equal toA. `8.85xx10^(-10)C//m^(2)`B. `-8.85xx10^(-10)C//m^(2)`C. `17.7xx10^(-10)C//m^(-2)`D. `-17.7xx10^(-10)C//m(2)` |
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Answer» Correct Answer - A `E=(40-10)/(0.3)=100V//m` (near the plate the electric field has to be uniform it is alomost due to the particle For conducting plate `E=(sigma)/(epsilon_(0))sigma=epsilon_(0)E` therefore `s=8.85xx10^(-12)xx100` `=8.85xx10^(-10)C//m^(2)` |
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| 80. |
A particle of mass `m_(2)` carrying a charge `Q_(2)` is fixed on the surface of the earth .Another particle of mass `m_(1)` and charge `Q_(1)` is positioned right above the first one at an altitude `h( ltlt R)`.R is radius of earth ,the charge `Q_(1)` and `Q_(2)` are of same sign ,then the magnitude of charge `Q_(2)` at which the velocity of `m_(1)` at an altitude `h_(2)` is zero is given byA. `Q_(2)=(piepsilon_(0)m_(1)ghh_(2))/(Q_(1))`B. `Q_(2)=(4piepsilon_(0)m_(1)ghh_(2))/(Q_(1))`C. `Q_(2)=(2piepsilon_(0)m_(1)ghh_(2))/(Q_(1))`D. `Q_(2)=(2piepsilon_(0)m_(1)ghh_(2))/(4Q_(1))` |
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Answer» Correct Answer - B `(1)/(4piepsilon_(0))(q_(1)q_(2))/(h)+m_(1)gh=m_(1)gh_(2)+(1)/(4piepsilon_(0))(q_(1)q_(2))/(h_(2))` `(:.KE` is zero at both positions) `rArr (1)/(4pepsilon_(0))q_(1)ql2((1)/(h_(2))-(1)/(h))=m_(1)g(h-h_(2))` `(1)/(4piepsilon_(0))q_(1)q_(2)((h-h_(2))/(hh_(2)))=m_(1)g(h-h_(2))` `rArr q_(2)=(4piepsilon_(0)m_(1)ghh_(2))/(q_(1))` |
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| 81. |
A particle of mass `m_(2)` carrying a charge `Q_(2)` is fixed on the surface of the earth .Another particle of mass `m_(1)` and charge `Q_(1)` is positioned right above the first one at an altitude `h( ltlt R)`.R is radius of earth ,the charge `Q_(1)` and `Q_(2)` are of same sign ,then At what altitude of `h_(3)` will object `m_(1)` be in equilibrium and what will be the nature of objects in motion if it is disturbed slightly from equilibriumA. `h_(3)=sqrt((Q_(1)Q_(2))/(4piepsilon_(0)m_(1)g))` periodic non oscillatoryB. `h_(3)=sqrt((Q_(1)Q_(2))/(2piepsilon_(0)m_(1)g))`,periodic and oscillatory.C. `h_(3)=sqrt((Q_(1)Q_(2))/(4piepsilon_(0)m_(1)g))`,periodic and oscillatory.D. `h_(3)=sqrt((Q_(1)Q_(2))/(4piepsilon_(0)m_(1)g))` non -periodic and non. |
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Answer» Correct Answer - C `(1)/(4piepsilon_(0))(q_(1)q_(2))/(h_(3)^(2))=mg h_(3)=sqrt(q_(1)q_(2))/(4piepsilon_(0)mg)` |
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| 82. |
The potential at a point x ( measured in `mu` m) due to some charges situated on the x-axis is given by `V(x)=20//(x^2-4) vol t`A. `(5)/(3) (V)/(mum)` and in the positive x-direction.B. `(10)/(9) (V)/(mum)` and in the negative x-direction.C. `(10)/(9) (V)/(mum)` and in the positive x-direction.D. `(5)/(3) (V)/(mum)` and in the negative x-direction. |
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Answer» Correct Answer - 3 Common potential |
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| 83. |
An electric and proton are sent into an electric a fileld .The ratio of forcr experienced by them isA. `1:1`B. `1:1840`C. `1840:1`D. `1:9.11` |
| Answer» Correct Answer - 1 | |
| 84. |
Two very small particles `1` and `2` each of mass `0.5kg` and of charge `q_(1)=1mc` and `q_(2)=1muC` resepectivley are connected by a mass less spring of spring constant `200Nm^(-1)` and placed on a horizontal rough surface.The particle `1` is fixed and `2` is free to move ,Initially the spring is in its natural length `(2m)` where `q_(2)` is released.If coefficient of static friction between `2` and horizontal surface is `0.5` the separation between the charges when they are in equilibrium will be (Neglect gravitational interactions between `1` and `2g=10ms^(-2)` |
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Answer» Correct Answer - 2 Based on friction and Coulombus Law `F=(9xx10^(9)xx10^(-9))/(4)=2.25N` which is less `mumg` acting on `2` rArr final separation =2m only. |
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| 85. |
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the center of the cube.A. zeroB. `(32q)/(b^(2))`C. `(q)/(2b^(2))`D. `(q)/(b^(2))` |
| Answer» Correct Answer - 1 | |
| 86. |
A positive charge `q_(0)` placed at a point `P` near a charged of metal sphere experience a force of repulsion of magnaitude `F`, the electric field `E` of the charge metal, charge at `P` isA. `(F)/(q_(0))`B. `lt (F)/(q_(0))`C. `gt(F)/(q_(0))`D. `F` |
| Answer» Correct Answer - 2 | |
| 87. |
A charge is moved against repulsion.Then there isA. decreasing its kinetic energyB. increasing its potential energyC. increasing both the energiesD. decreasing both the energies |
| Answer» Correct Answer - 3 | |
| 88. |
The coulomb electrostatic force is defined forA. two spherical charges at restB. two spherical charges in motionC. two point charge in motionD. two point charges at rest |
| Answer» Correct Answer - 4 | |
| 89. |
The p.d `(V_(B)-V_(C)` between two point from `C` to `B`A. does not depend on the pathB. depends on the pathC. depends on test chargeD. independent of electric field |
| Answer» Correct Answer - 1 | |
| 90. |
A positive point charge `+Q` is fixed in space .A negative point charge `-q` of mass `m` revolves around a fixed charge in elliptical orbits .The fixed charge `+Q `is at one focus of the ellipse.The only force acting on negative charge is the electrostatic force due to positive charge is the electrostatic force due ot positive charge.Then which of the following statement is true A. Linear momentum of negative point charge is conservedB. Angular momentum of negative point charge about fixed positve charge is conservedC. Total Kinetic energy of negative point charge is conserved.D. Electrostatic potential energy of system of both point charge is conserved. |
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Answer» Correct Answer - B Only angular momentum `x` & total energy are conserved but not kinetic energy. |
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| 91. |
an electron and proton are placed in an electric field.The forces acting on them are `bar(F_(1))` and `bar(F_(2))` and their acclelerations are `bar(a_(1))` and `bar(a_(2))` respectively thenA. `bar(F_(1))=bar(F_(2))`B. `bar(F_(1)) = -bar(F_(2))`C. `|bar(a_(1))|=|bar(a_(2))|`D. `|bar(a_(1))|lt|bar(a_(2))|` |
| Answer» Correct Answer - 2 | |
| 92. |
Drawings `I` and `II` show two samples of electric field lines A. The electric fields in both I and II are produced .by negative charge located somewhere on the left and positive charges located somewhere on the rightB. In both I and II the electric field is same every whereC. In both cases the field becomes stronger on moving form left to rightD. The electric field in I is the same everywhere ,but in II the electric field becomes stronger on moving from left and right |
| Answer» Correct Answer - 4 | |
| 93. |
The force of attraction between two charges separated by certain distance in air is `F_(1)` .If the space between the charges is completely filled with dielectric of constant `4` the force becomes `F_(2)` .If half of the distance between the charges is filled with same dielectric the force between the charges is `F_(3)` Then `F_(1):F_(2):F_(3)` isA. `16:9:4`B. `9:36:4`C. `4:1:2`D. `36:9:16` |
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Answer» Correct Answer - 4 `F=(1)/(4piepsilon_(0)epsilon_(r))(q_(1)q_(2))/(d^(2))` |
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| 94. |
Potential at the tip of a pointed conductor isA. maximumB. minimumC. zeroD. same as at any other point |
| Answer» Correct Answer - 4 | |
| 95. |
A block of mass `m` is attached to a spring of force constant k .Charges on the block is `q`. A horizontal electric field E is acting in the directions as shown.Block is released with the spring in unstretched position block will execute `SHM` (b) time period of osciallation is `2pisqrt(m)/(k)` (c) amplitude of oscillation is `(qE)/(k)` Block will oscillate but not simple harmonically choose the correct answerA. `a` and `b` are trueB. `d` is trueC. `a,b,c` are trueD. `a,b,c,d` are true |
| Answer» Correct Answer - 3 | |
| 96. |
A hollow closed conductor of irregular shape is given some charge . Which of the following statements are correct ?A. The entire charge will appear on its outer surfaceB. All points on the conductor will have the same potentialC. All points on its surface will have the same charge densityD. All points near its surface and outside will have the same electric intensity. |
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Answer» Correct Answer - A,B Electric field intensity of a hollow conductor. |
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| 97. |
A thin walled spherical conducting shells `S` of radius `R` is given charge `Q`,the same amount of charge is also placed at its centre `C`.Which of the following statements are correct.A. on the outer surface of `S`, the charge density is `(Q)/(2piR^(2))`B. The electric field is zero all points inside `S`.C. At a point just outside `S`, the electric fiels is double the field at a point just inside `S`.D. At any point inside `S` the electric field is inversely proportional to the square of its distance from `C`. |
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Answer» Correct Answer - A,C,D Charge density at the outer surface is `(2Q)/(4piR^(2))=(Q)/(2piR^(2))` `E_("Inside")=(Q)/(4piepsilon_(0)R^(2)),E_("outside")=(2Q)/(4piepsilon_(0)R^(2))E_("outside")=2E_("inside")` |
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| 98. |
Four equipotnetial curves in an electric filed are shown in the figure `A,B,C` are `E_(A),E_(B),E_(C)` then A. `E_(A)=E_(B)=E_(C)`B. `E_(A)gt E_(B)gtE_(C)`C. `E_(A)lt E_(B)ltE_(C)`D. `E_(A)gt E_(B)ltE_(C)` |
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Answer» Correct Answer - 3 `E=-(dV)/(dx)` |
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| 99. |
A partical of mass `1kg` and carrying `0.01C` is at rest on an inclined plane of angle `30^(@)` with horizontal when an electric field of `(490)/(sqrt(3))NC^(-1)` applied parallel to horizontal .The cofficient of friction isA. `0.5`B. `(1)/(sqrt(3))`C. `(sqrt(3))/(2)`D. `(sqrt(3))/(7)` |
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Answer» Correct Answer - 4 `N=mg sin theta+qE sin theta `mg sin theta=muN+qEcos theta` |
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Electric field on the axis of a small electric dipole at a distance r is `vec(E)_(1)` and `vec(E)_(2)` at a distance of `2r` on a line of perpendicular bisector isA. `vec(E)_(2)=-vec(E)_(1)//8`B. `vec(E_(2))=-vec(E)_(1)//16`C. `vec(E_(2))=-vec(E)_(1)//14`D. `vec(E_(2))=-vec(E)_(1)//8` |
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Answer» Correct Answer - 2 `E_("axis")=(2kp)/(r^(3))` and `E_("biesector")=(kp)/(2r^(3)` |
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