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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.
| 101. |
Select the correct statement :(Only force on a particle is due to electric field)A. A charged particle always movies along the electric line of forceB. A charged particle may move along the line of forceC. A charged particle never move along the line of forceD. A charged particle moves along the line of force only if released from rest. |
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Answer» Correct Answer - B If field is straight then the charge moves along the straight line .if the fielss is a curve then the electrostatic force is not sufficient to change the direction along the curved field line . |
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| 102. |
Electrical force between two point charge is `200N`, if we increase `10%` charge on the other then elecrtrical force between them for the same direction becomes.A. `198N`B. `100N`C. `200N`D. `99N` |
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Answer» Correct Answer - 1 `F=(1)/(4piepsilon_(0))(q_(1)q_(2))/(r^(2)),q_(1)^(1)=(110)/(100)q_(1)`and `q_(2)^(1)=(90)/(100)q_(2)` |
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| 103. |
If the flux of the electric field through a closed surface is zero,A. The electric field must be zero every where on the surfaceB. The electric field must not be zero everywhere on the surfaceC. the change inside the surface must be zeroD. The charge in the vicnity of the surface must be zero |
| Answer» Correct Answer - 3 | |
| 104. |
Mark the correct option:A. Gauss law is valid only for unsymmetraically charge distributionsB. Gauss law is valid only for charge placed in vacuumC. The electric field calculated byGauss law is the field due to the charge outside the Gaussian surface.D. The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface. |
| Answer» Correct Answer - 4 | |
| 105. |
Find the electric field potentail and strength at the centre of a hemisphere fo raidus `R` ahcged uniformly with the the surface density `sigma`.A. `(sigmaR)/(epsilon_(0)`B. `(2sigmaR)/(epsilon_(0))`C. `(sigmaR)/(2epsilon_(0))`D. `(sigmaR)/(3epsilon_(0))` |
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Answer» Correct Answer - A The area of shaded element is `dA=2pirRd theta` `=2piR^(2)sintheta d theta rArr dV=(1)/(4piepsilon_(0))(dq)/(R)` `=(1)/(4piepsilon_(0))(sigma2piR^(2)sinthetad theta)/(R)=(sigma R)/(2epsilon_(0))sinthetad theta` `V=(sigmaR)/(2epsilon_(0))int_(0)^(pi//2)sintheta d theta =(sigmaR)/(2epsilon_(0))` |
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| 106. |
As one penetrates uniformly charged conducting sphere,what happens to the electric field strengthA. decreases inverserly as the square of the distanceB. decreases inversely as the distanceC. becomes zeroD. increases inversely as the square of distance |
| Answer» Correct Answer - 3 | |
| 107. |
In space of horizontal `EF (E = (mg)//q)` exist as shown in figure and a mass m is released at the end of a light rod. If mass m is releases from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position .A. `sqrt((g)/(l))`B. `sqrt((2g)/(l))`C. `sqrt((3g)/(l))`D. `sqrt((5g)/(l))` |
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Answer» Correct Answer - B `W_(E)+W_(g)=DeltaK.E` Use work energy theorem `(Eq)l sintheta+mgl(1-costheta)=(1)/(2)IomegaQ^(2)` |
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| 108. |
A bullet of mass m and charge `q` is fired towards a solid uniformly charged sphere of radius `R ` and total charge ` + q`. If it strikes the surface of sphere with speed `u`, find the minimum speed `u` so that it can penetrate through the sphere , (Neglect all resistance forces or friction acting on bullet except electrostatic forces ) .A. `(q)/(sqrt(2piepsilon_(0)mR))`B. `(q)/(sqrt(4piepsilon_(0)mR))`C. `(q)/(sqrt(8piepsilon_(0)mR))`D. `(sqrt(3)q)/(sqrt(4piepsilon_(0)mR))` |
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Answer» Correct Answer - B `(TE)_("surface")=(PE)_("centre")` Velocity of centre is nearly zero then it can across the sphere `(1)/(4piepsilon_(0))(q.q)/(R)+(1)/(2)mv^(2)=(3)/(2)((1)/(4piepsilon_(0))(q)/(R))` |
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| 109. |
A particle of mass m and charge q is fastened to one end of a string of length. The other end of the string is fixed to the point O. The whole sytem liles on as frictionless horizontal plane. Initially, the mass is at rest at A. A uniform electric field in the direction shown in then switfched on. Then A. the speed of the particle when it reaches `B` is `sqrt((2qEl)/(m))`B. the speed of the particle when it reaches `B` is `sqrt((qEl)/(m))`C. the tension in the string when particle reaches at `B` is `(Eq)/(2)`D. the tension in the string when the particle reaches at `B` is `qE`. |
| Answer» Correct Answer - 2 | |
| 110. |
Conside r two concentric spherical surface `S_(1)` with radius `a` and `S_(2)` with radius `2a` ,both centred on the origin. There is a charge `+q` at the origin, and no other charges.Compare the flux `phi` through `S_(1) with the flux `phi_(2)` through `S_(2)`.A. `phi_(1)=4phi_(2)`B. `phi_(1)=2phi_(2)`C. `phi_(1)=phi_(2)`D. `phi_(1)=phi_(2)//2` |
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Answer» Correct Answer - 3 Flux through both will be same as net charge enclosed by both is same |
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| 111. |
An infinite number of charges each equal to q, are placed along the X-axis at `x = 1, x = 2, x = 4, x = 8`,…….. and so on. (i) find the electric field at a point `x = 0` due to this set up of charges. (ii) What will be the electric field if the above setup, the consecutive charges have opposite signs.A. `(q)/(3piepsilon_(0))`B. `(q)/(6piepsilon_(0))`C. `(q)/(2piepsilon_(0))`D. `(q)/(4piepsilon_(0))` |
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Answer» Correct Answer - 1 `E=(q)/(4piepsilon_(0))[(1)/(1^(2))+(1)/(2^(2))+(1)/(4^(2))+..........]` |
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| 112. |
A protons of mass `m` charge `c` is released form rest in a uniform electric field of strength `E` .The time taken by it to travel a distance `d` in the field isA. `sqrt((2de)/(mE))`B. `sqrt((2dm)/(Ee))`C. `sqrt((2dE)/(me))`D. `sqrt((2dEe)/(dm))` |
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Answer» Correct Answer - 2 `s=(1)/(2)(qE)/(m)t^(2)` |
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| 113. |
Two cahrges of `50muC` and `100muC` are separated by a distance of `0.6m` .The intensity of electric filed at a point midway between them isA. `50xx10^(6)v//M`B. `5xx10^(6)M//V`C. `10xx10^(6)V//m`D. `10XX10^(-6)V//M` |
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Answer» Correct Answer - 2 ` E=(1)/(4piepsilon_(0))(q_(1))/(x_(1)^(2))-(1)/(4piepsilon_(0))(q_(2))/(x_(2)^(2))` |
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| 114. |
A mass m carrying a charge `q` si suspended form a string and placed in a uniform horizontal electric field of intensity `E`.The angel made by the string with the vertical in the equilibrium position isA. `theta=tan^(-1)""(mg)/(Eq)`B. `theta=tan^(-1)""(m)/(Eq)`C. `theta=tan^(-1)""(Eq)/(m)`D. `theta=tan^(-1)""(Eq)/(mg)` |
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Answer» Correct Answer - 4 `qE=mg tantheta` |
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| 115. |
Two point charges `Q` and `-3Q` are placed at some distance apart. If the electric field at the location of `Q` is `E` then at the locality of `-3Q`, it isA. `vec(E)`B. `-vec(E)`C. `+(vec(E))/(3)`D. `-(vec(E))/(3)` |
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Answer» Correct Answer - 3 `vec(E)=(1)/(4piepsilon_(0))(Q)/(r^(3))vec(r)` |
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| 116. |
An electric dipole is placed at the centre of a sphere. Mark the correct options:A. the flux of the electric field through the sphere is zeroB. the electric field is zero at every point of the sphereC. the electric potential is zero everywhere on the sphereD. the electric potential is zero on a circle on the surface. |
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Answer» Correct Answer - A,D Conceptual based on electric dipole and electric field intersity at the centre of a sphere. |
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| 117. |
An electric field converges at the origin whose magnitude is given by the expression `E=100r N//C`, where r is the distance measured from the origin.A. total charge contained in any spherical volume with its centre at the origin is negativeB. total charge contained in any spherical volume irrespective of the location of its centre is negativeC. total energy contained in a spherical volume of radius `3cm` with its centre at the originD. total charge contained in a spherical volume of radius `3cm` with its centre at the origin has magnitude `3xx10^(-9)C` |
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Answer» Correct Answer - A,B,C `E=-(dv)/(dr)` |
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| 118. |
Two identical metal spheres posses `+60C` and `-20C` of charges .They are brought in contact and then separated by `10cm` .the force between them isA. `36xx10^(13)N`B. `36xx10^(14)N`C. `36xx10^(12)N`D. `3.6xx10^(12)N` |
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Answer» Correct Answer - 1 `F=(1)/(4piepsilon_(0))(q_(1)+q_(2))^(2)/(4d^(2))` |
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| 119. |
Statement 1: Any charge will move from electric potential `v_(1)` and `V_(2)` by its own only when `v_(1) lt V_(2)` Statement2: Electron moves form `V_(1)=2V` towards `V_(2)=4V` |
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Answer» Correct Answer - D Since field is zero for a charged conductor on the surface and inside it also.so the surface of a charged conductor is always equipotential .Also, for equipotential surface,Hence `int(vec(E).vec(dl))=0` `rArr vec(E)_|_ vec(dl)` .Hence both are true and statement -`2` is the correct explanation to statement -`1`. |
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| 120. |
A body has a charge of `9.6xx10^(-20)` coulomb.It isA. PossibleB. not possibleC. may (or) may not possibleD. Data not sufficient |
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Answer» Correct Answer - 2 `Q=+-` "ne n is integer" |
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| 121. |
A point positve charge is brought near an isolated conducting sphere as shown in figure the electric field is best given by A. Figure(i)B. Figure(ii)C. figure(iii)D. Figure(iv) |
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Answer» Correct Answer - A When a point positive charge is brought near an isolated conducting sphere,the nearer side of it induces negative charge and further side positive ,Now electric field lines start form the positive charge and end at the negative charge.Also the field lines are normal to the conductor.Figure (i) is the best representation. |
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| 122. |
A point charge `+q` is placed at a distance `d` from an isolated conducting plane. The field at a point `P` on the other side of plane isA. directed perpendicular to the plane and away from the planeB. directed perpendicular to the plane but towards the planeC. directed radilally away form the point hcarageD. directed radially towards the point charge |
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Answer» Correct Answer - A when a point charge `+q` is placed at a distance `d` from and isolated conducting plane ,the surface of the plane towards the charge induces negative charge and an equal positive charge and an equal positive charge develops on the opposite side of the plane .So the field at a point `P` on the other side of the plane is away from the plane and directed perpendicular to the surface. |
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| 123. |
The electric flux through the surface A. in figure (iv) is the largestB. in figure(iii) is the leastC. in figure (ii) is same as in figure (iii) but in smaller then figure (iv)D. is the same for all the figures |
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Answer» Correct Answer - D The electric flux through the all four surface is same i.e, `q//e_(0)` .It does not depend on size and shape of the surface but depends on the amount of charge enclosed by the surface only. |
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| 124. |
Figures shown electric field lines in which as electric dipole `vec(p)` is placed as shown .Which of the following statement is correct? A. The dipole will not experience any forceB. The dipole will experience a force towards rightC. The dipole will experience a force towards leftD. the dipole will experience a force upward |
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Answer» Correct Answer - C The electic field lines represent a non-uniform electric field since the spacing between line is greater on right so the electric field is more strong at left than that at right. Hence the force on negative charge wil be more and towards left.So the dipole will experience a force towards left. |
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| 125. |
A force of `4N` is acting between two charges in air.If the space between them is completelly filled with glass `(epsilon_(r)=8)` then the new force will beA. `2N`B. `5N`C. `0.2N`D. `0.5N` |
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Answer» Correct Answer - 4 `F^(|)=(F)/(K)` |
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| 126. |
A charged particle is free to move in an electric fields is iniitally at rest ,ThenA. It will always move perpendicular to the line of forceB. it will always move along the line of force in the direction of the fieldC. it will always move along the line of force opposite to the direction of the fieldD. it will always move along the line of force in the direction of the fields or opposite to the direction of the field depending on the nature of the charge |
| Answer» Correct Answer - 4 | |
| 127. |
Dimension of `epsilon_(0)` areA. `[M^(-1)L^(-3)T^(4)A^(2)]`B. `[M^(0)L^(3)T^(3)A^(2)]`C. `[M^(-1)L^(-3)T^(3)A]`D. `[M^(-1)L^(-3)TA^(2)]` |
| Answer» Correct Answer - 1 | |
| 128. |
An electron moves with a velocity `vec(v)` in are electric field `vec(E)` if the angle between `vec(V)` and `Vec(E)` is neither `0` nor `pi` ,then path followed by the electron isA. straight lineB. circleC. ellipseD. parabola |
| Answer» Correct Answer - 4 | |
| 129. |
`E=-(dV)/(dr)` here negative sign signified thatA. `E` is positiveB. `E` is negativeC. `E` is increase when `V` decreaseD. `E` is directed in the direction of decreasing `V` |
| Answer» Correct Answer - 4 | |
| 130. |
An electron of mass `m` and charge `e` is accelerated from rest through a potential difference `V` in vacuum. The final speed of the electron will be |
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Answer» Correct Answer - 3 `(1)/(2)mv^(2)=eV` |
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| 131. |
Two charges`+q_(1)` and `-q_(2)` are placed at A and B respectively. A line of force emanates from `q_(1)` at an angle `alpha` with the line AB. At what angle will it terminate at `-q_(2)` ? |
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Answer» we know that number of lines of forces emerge is proportional to magnitude of the charge.The field lines emanating form `Q_(1)` ,spread out equally in all directions.The number of fields lines or flux through cone of half angle `theta` is `(Q_(1))/(4pi)2pi(1-costheta)`.Similarly the number of lines of force terminating on `-Q_(2)` at an angle `phi` is `(Q_(1))/(4pi)2pi(1-costheta)`,The total lines of force emanating from `Q_(1)` is equal to the total lines fo force terminating on `Q_(2)` `rArr=(Q_(1))/(4pi)2pi(1-costheta)=(Q_(2))/(4pi)2pi(1-cosphi)` `(Q_(1))/(2)(1-costheta)=(Q_(2))/(2)(1-cosphi),Q_(1)sin^(2)theta//2=Q_(2)sin^(2)phi//2` `sin phi//2=sqrt((Q_(1))/(Q_(1)))sintheta//2 rArr 2sin^(-1){sqrt((Q_(1))/(Q_(2)))sintheta//2}` |
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| 132. |
Consider a region inside which there are various types of charges but the total charge is zero ,.At points outside the regionA. the electric field is necessarily zeroB. the electric field is due to the dipole moment of the charge distribution onlyC. the dominant electric field is `prop(1)/(r^(3))` for large `r`, where `r` is the distance from a origin in this regionD. the work done to move a charged particle along a closed path ,away form the region will be zero |
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Answer» Correct Answer - C,D The electric field at points outside the origin is due to all the lndividual charges and their distribution over the region.As the total charge is zero ,the region can be supposed to contain a number of dipoles .For a point outside the region ,the dominant electric field is due to dipoles.ie.e `prop (1)/(r^(3)` for large `r`. Also as the electric field is conservation ,the work done to move a charged particle along a closed path is zero.options (c ) and (d) are correct. |
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| 133. |
IF an electorn enters into a space between the plates of a parallel plate capacitor at an an angle `alpha` with the plates an leaves at an angle `beta` to the plates find the ratio of its kinetic energy while entering the capacitor of that while leaving. |
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Answer» Let `u` be the velocity of electron while entering the filed and `v` be the velocity when it leaves the plates.Compound of velocity parallel to the plates will remain unchanged. Hence `u cos alpha=u cos beta:.(u)/(v)=(cos beta)/(cos alpha)` `therefore (((1)/(2)m u^(2)))/(((1)/(2)mv^(2)))=((u)/(v))^(2)=((cosbeta)/(cosalpha))^(2)` |
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| 134. |
Two electric dipoles each of dipolemoment `p=6.2xx10^(-30)C-m` are placed with their axis along the same line and their centres at a distanced =`10^(-8)cm`.The force of attraction between dipole isA. `2.1xx10^(-16)N`B. `2.1xx10^(-12)N`C. `2.1xx10^(-10)N`D. `2.1xx10^(-8)N` |
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Answer» Correct Answer - 4 `u_(i)=0, u_(f)=2xx(1)/(4piepsilon_(0)).(e^(2))/((d//2))` `DeltaPE=KE=(1)/(2) mv^(2)` calcualte `v` |
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| 135. |
A small ball if mass `2 xx 10^(-3) Kg` having a charges of ` 1 mu C` is suspended by a string of length ` 0. 8 m` Another identical ball having the same charge is kept at the point of suspension . Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make complete revolution . |
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Answer» To complete the circle at top most point `T_(2)=0` `Mg-(q^(2)/(4piepsilon_(0)l^(2)))=(MV^(2))/(l)` `rArr V^(2)-g^(l)=-(q^(2))/(4piepsilon_(0)Ml)....(1)` from law of conservation of law `(1)/(2)m u^(2)=(1)/(2)mv^(2)+mg2l....(2)` from(1) and (2) `u=sqrt(4gl-(q^(2))/(4piepsilon_(0)ml))=5.86m//s` |
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| 136. |
the angle between of the electric dipolemoment `p` and the electric field `E` when the dipole is in stable equilibriumA. `0`B. `pi//4`C. `pi//2`D. `pi` |
| Answer» Correct Answer - 1 | |
| 137. |
Electric field intensity (E) due to an electric dipole varies with distance (r ) from the point of the center of dipole as :A. `(1)/(r )`B. `(1)/(r^(2))`C. `(1)/(r^(3))`D. `r^(2)` |
| Answer» Correct Answer - 3 | |
| 138. |
An electric dipole is placed in a non uniform electric field increasing along the `+ve` direction of " X-axis " .In which direction does the dipole A. move along `+ve` direction of "X-axis", rotate clockwiseB. move along `-ve` direction of "X-axis" ,rotate clockwiseC. move along `+ve` direction of "X-axis" ,rotate anti clockwiseD. move along `-ve` direction of "X-axis" ,rotate anti clockwise |
| Answer» Correct Answer - 1 | |
| 139. |
An electric dipole placed withits axis in the direction of a unifrom electric field experienceA. a force but not torqueB. a torque but no forceC. a force as well as a torqueD. neither a force nor a torque |
| Answer» Correct Answer - 4 | |
| 140. |
A dipole is placed in `x-y` plane parallel to the line `y=2x` there exist a unifrom electric field among `z-axis` .Net force acting on the dipole will be zero.But it can experience some torque,we can show that the direction of this torque will be parallel to the lineA. `y=2x+1`B. `y=-2x`C. `y=-(1)/(2)x`D. `y=-(1)/(2)x+2` |
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Answer» Correct Answer - C,D Torque will be perpendicular to the line `y=2x` and it should be in `xy` plane ,because electric field in z-direction .the line in option (C ) and (d) both are perpendicular to `y=2x` |
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| 141. |
The acceleration of a charged particle at a unifrom electric field isA. proportonal to its charge onlyB. inversely proportional to its mass onlyC. proportional to its specific chargeD. inversely proportional to specific charge |
| Answer» Correct Answer - 3 | |
| 142. |
An electric dipole is along a unifrom electric field.If it is deflected by `60^(@)` ,work done by an agent is `2xx10^(-19)J`.then the work done by an agent if it is deflected by `30^(@)` futherA. `2.5xx10^(-19)J`B. `2xx10^(-19)J`C. `4xx10^(-19)J`D. `2xx10^(-16)J` |
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Answer» Correct Answer - 2 `W_(1)=pE(1-costheta)and W_(2)=pE(costheta_(1)-costheta_(2))` |
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| 143. |
An electric dipole when placed in a uniform electric field `E` will have minimum potential energy, if the positive direction of dipole moment makes the following angle with `E`A. zeroB. `pi//2`C. `pi`D. `3pi//2` |
| Answer» Correct Answer - 1 | |
| 144. |
the dipole moment of the given system is A. `sqrt(3ql)` along perpendicular biesector of `q-q` line .B. `2ql` along perpendicular bisector of `q-q` lineC. `qlsqrt(2)` along perpendicular bisector of `q-q` lineD. `0` |
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Answer» Correct Answer - 1 `p_(1)=lq=p_(2)`and `P_(R)=sqrt(3)ql` |
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| 145. |
Two charge when kept at a distance of `1m` apart vacuum have a some force of repulsion .if the force of repulsion between these two charges be same ,whenplaced in an oil of dielectric constant `4`the distance of separation isA. `0.25m`B. `0.4m`C. `0.5m`D. `0.6m` |
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Answer» Correct Answer - 3 `t^(1)=(t)/(sqrt(k))` |
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| 146. |
the electric field is given by `vec(E)=(vec(F))/(q_(0))` here the test charge `q_(0)` should be (a) Infinitesimally small and positive (b) Infintestimally small and negativeA. only aB. only bC. a (or) bD. neither a and b |
| Answer» Correct Answer - 4 | |
| 147. |
Two point sized identiacal sphere carrying charge `q_(1)` and `q_(2)` on them are separated by acertain distance the mutal force between them is `F` these two are brought in contant and kept at the same separation Now the force between them is `F^(1)` then .`(F^(1))/(F)=((q_(1)+q_(2))^(2))/(4q_(1)q_(2))` |
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Answer» when charges separated by certain distance the force is given by Then `F=(1)/(4piepsilon_(0))(q_(1)q_(2))/(r^(2))……..(1)` when charges brought in contant and kept at the same distance the force is given by `F^(1)=(1)/(4piepsilon_(0))(q_(1)+q_(2))^(2)/(4r^(2))……..(2)` " From " (1) and (2), `:.(F^(1))/(F)=((q_(1)+q_(2))^(2))/(4q_(1)q_(2))` |
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| 148. |
Two points charge `q_(1)` and `q_(2)(=q_(1)//2)` are placed at points `A(0,1)` and `B(1,0)` as shown in the figure.The electric field vector at point `P(1,1)` makes an angle `q` with the x-axis ,then the angle `q` is A. `tan^(-1)""((1)/(2))`B. `tan^(-1)""((1)/(4))`C. `tan^(-1)""(1)`D. `tan^(-1)""(0)` |
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Answer» Correct Answer - 1 `w=Fs,W=qvec(E).vec(S)` |
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| 149. |
Figures shows three spherical and equipotential surfaces `1,2` and `3` round a point charge `q`.The potential difference `v_(1)-v_(2)=V_(2)-V_(3)` if `t_(1)` and `t_(2)` be the distance between them ,then A. `t_(1)=t_(2)`B. `t_(1) gtt_(2)`C. `t_(1) ltt_(2)`D. `t_(1) let_(2)` |
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Answer» Correct Answer - 3 `E=(-dv)/(dt)` |
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| 150. |
The sketch below show cross section so equipotential surfaces between two charged.Conductors that are shown in solid black.Some points on the equipotenital 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. `2xx10^(-5)J`B. `-2xx10^(-5)J`C. `4xx10^(-5)J`D. `-4xx10^(-5)J` |
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Answer» Correct Answer - D `W=q.DeltaV` `=-1xx10^(-6)[20-(-20)]` `=-4xx10^(-5)J` |
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