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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.
| 1551. |
The electrostatic potential `(phi _r)` of a spherical symmetric system , kept at origin is shown in the adhacent figure, and given as `phi _e = q/( 4 pi varepsilon_0 r) (r le R_o)` `phi _e = q/( 4 pi varepsilon_0 R_o) (r le R_0)` Which of the following option `s` in //are correct ?A. For spherical region ` r le R_0` total electrostatic energy stored id zero .B. Within ` r= 2R_0`, total charge is `q`.C. There will be no charge anywhere except at ` r = R_0`.D. Electric field is discontinuous at ` r= R_0`. |
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Answer» Correct Answer - A::B::C::D Potential graph is of uniformly charged spherical sedll , so option ` A, B, C, D` are correct . |
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| 1552. |
A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric filed due to the sphere at a distance r from its centreA. increases as r increases, for `r le R`B. decreases as r increses, for `0 lt r lt oo`.C. decreases as r increases, fo `R lt r lt oo`D. is discontinuous at r = R |
| Answer» Correct Answer - A::C | |
| 1553. |
A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric filed due to the sphere at a distance r from its centreA. increases as `r` increases for `r lt R`B. decreases as `r` increases for `0 lt r lt oo`C. decreases as `r` increases for `R lt r lt oo`D. is discontiues at `r = R` |
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Answer» Correct Answer - A::C for `r lt R, E = (kQ)/(R^(3)) r, i.e., E prop r` for `r gt R, i.e., R lt r lt oo, E = (kQ)/(r^(2)), i.e., E prop (1)/(r^(2))` |
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| 1554. |
A positively charged thin metal ring of radius R is fixed in the xy plane with its centre at the origin O. A negatively charged particle P is released from rest at the point `(0, 0, z_0)` where `z_0gt0`. Then the motion of P isA. (a) periodic, for all values of `z_0` satisfying `0ltz_0ltoo`B. (b) simple harmonic, for all values of `z_0` satisfying `0ltz_0leR`C. (c) approximately simple harmonic, provided `z_0lt ltR`D. (d) such that P crosses O and continues to move along the negative z axis towards `z=-oo` |
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Answer» Correct Answer - A::C Let Q be the charge on the ring, the negative charge `-q` is released from point `P(0, 0, Z_0)`. The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be `E=(1)/(4piepsilon_0)*(QZ_0)/((R^2+Z_0^2)^(3//2)` Therefore, force on charge P will be towards centre as shown, and its magnitude is `F_e=qE=(1)/(4piepsilon_0)*(Qq)/((R^2+Z_0^2)^(3//2)).Z_0` ...(1) Similarly, when it crosses the origin, the force is again towards centre O. Thus the motion of the particle is periodic for all values of `Z_0` lying between 0 and `oo`. Secondly if `Z_0lt lt R, (R^2+Z_0^2)^(3//2)~~R^3` `F_e=(1)/(4piepsilon_0)*(Qq)/(R^3).Z_0` [From equation 1] i.e. the restoring force `F_e prop -Z_0`. Hence the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position). |
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| 1555. |
Two idential capacitors are joined in parallel, charged to a potential `V` and then separated and then connected in series i.e. the positive plate of one is connected to negative of the otherA. charges on the free plates connected together are destroyedB. the charges on the free plates are enhancedC. the energy stored in the system increasesD. the potential difference between the free plates becomes 2V. |
| Answer» Correct Answer - D | |
| 1556. |
An infinite number of capacitors, having capacitances `1muF, 2muF, 4muF and 8muF…..` are connected in series. The equivalent capacitance of the system isA. infiniteB. `0.25muF`C. `0.5muF`D. `2muF` |
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Answer» Correct Answer - C `1/C=1/1+1/2+1/4+1/8+.......` = 1 + 1 (approximately) `therefore" "C=1//2=0.5muF` |
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| 1557. |
A `4 muF` capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged `2 mu F` capacitor. How much electrostatic energy of the first capacitor is disspated in the form of heat and electromagnetic radiation ? |
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Answer» Here, `C_(1) = 4 muF = 4xx10^(-6) F, V_(1) = 200 vol t` Initial electrostatic energy stored in `C_(1) si U_(1) = (1)/(2) C_(1) V_(1)^(2) = (1)/(2) xx 4xx10^(-6)xx200xx200 = 8xx10^(-2) joul e` When `4 mu F` capacitor is connected to uncharged capacitor of `2 mu F`, charge flows and both acquire a common potential `V = ("total charge")/("total capacity") , V = (C_(1) V_(1))/(C_(1) + C_(2)) = (4xx10^(-6)xx200)/((4+2) 10^(-6)) = (800)/(6) vol t` `:.` Final electrostatic energy of both capacitors `U_(2) = (1)/(2) (C_(1) + C_(2)) V^(2) = (1)/(2) xx6xx10^(-6) xx (800)/(6) xx (800)/(6) = 5.33xx10^(-2) joul e` `:.` Energy dissipated in the form of heat and electromagnetic radiation `U_(1) - U_(2) = 8xx10^(-2) - 5.33xx10^(-2) = 2.67xx10^(-2) joul e` |
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| 1558. |
Three uncharged capacitors of capacities `C_(1),C_(2)` and `C_(3)` are connected as shown in the figure to one another and the potentials `V_(1),V_(2)` and `V_(3)` respectively. Then the potential at `O` will be .A. `(V_(A) + V_(B) + V_(D))/(C_(1) + C_(2) + C_(3))`B. `(V_(A) C_(1) + V_(B) C_(2) + V_(D) C_(3))/(C_(1) + C_(2) + C_(3))`C. `(V_(A) V)(B) + V_(B) V_(D) + (V_(D) V_(A))/(C_(1) + C_(2) + C_(3))`D. `(V_(A) V_(B) V_(D))/(C_(1) C_(2) + C_(2) C_(3) + C_(3) C_(1))` |
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Answer» Correct Answer - B Let `q_(1), q_(2), q_(3)` be the charges on `C_(1), C_(2), C_(3)` resectively From Fig `V_(A) - V_(0) = (q_(1))/(C_(1))` or `q_(1) = (V_(A) - V_(0)) C_(1)` `V_(B) - V_(0) = (q_(2))/(C_(2))` or `q_(2) = (V_(B) - V_(0)) C_(2)` `V_(D) - V_(0) = (q_(3))/(C_(3))` or `q_(3) = (V_(D) - V_(0)) C_(3)` As `q_(1) = q_(2) + q_(3)` `:. (V_(A) - V_(0)) C_(1) = (V_(B) - V_(0)) C_(2) + (V_(D) - V_(0)) C_(3)` which gives, `V_(0) = (V_(A) C_(1) + V_(B) C_(2) + V_(D) C_(3))/(C_(1) + C_(2) + C_(3))` |
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| 1559. |
1000 similar electrified rain drops merge together into one drop so that their total charge remains uncharged. How is the electric energy affected ? |
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Answer» Correct Answer - 100 times Let `q` be the charge on each drop of radius `r`. Energy before merger `U_(1) = 1000 xx (q^(2))/(2C) = 10^(3) (q^(2))/(2C)` On merger, `(4)/(3) pi R^(3) = 1000xx (4)/(3) pi r^(3), :. R = 10r`. Capacity becomes 10 times. Energy after merger `= U_(2) = ((1000 q)^(2))/(2 (10 C)) = 10^(5) (q^(2))/(2C)` `:. U_(2)//U_(1) = 100` |
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| 1560. |
Two parallel palate capacitors `X` and `Y` have the same area of plates and same separation between then. `X` has air between the plates and `Y` contains a dielectric medium of `in_(r) = 4,` Calculate (i) capacitance of `X` and `Y` if equivalent capacitance fo combination is `4 mu F`. (ii) pot diff between the plates of `X` and `Y`. (iii) What is the ratio of electrostatic energy stored in `X` and `Y` ? |
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Answer» Correct Answer - `5 mu F, 20 mu F, 9.6 V, 2.4 V ; 4 : 1` `C_(1)` is capacity of `X , C_(2)` is capacity of `Y` fig. `C_(2) = 4 C_(1)`. As `(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) :. (1)/(4) = (1)/(C_(1)) + (1)/(4 C_(1)) = (5)/(4 C_(1))` `C_(1) = 5 mu F , C_(2) = 20 mu F` `q_(1) = q_(2) = C_(s) V = 4xx12 = 48 mu C` `V_(1) = (q_(1))/(C_(1)) = (48)/(5) = 9.6 V, V_(2) = (q_(2))/(C_(2)) = (48)/(20) = 2.4 V` `(U_(1))/(U_(2)) = (C_(2))/(C_(1)) = 4` |
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| 1561. |
Two capacitors of `25 mu F` and `100 mu F` are connected in series to a source of `120 V`. Keeping their charges uncharged, they are separated and connected in parallel to eachother. Find out (i) pot. Diff. between the plates of each capacitor (ii) energy loss in the process. |
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Answer» Correct Answer - `38.4 V, 0.052 J` Here, `C_(1) = 25 muF, C_(2) = 100 muF, C_(s) = 20 muF`, `q = C_(s) V_(s) = 2400 mu C` on each capacitor `C_(p) = 125 mu F, V_(p) = (2400+2400)/(125) = 38.4 V` Loss of energy `= (1)/(2) C_(s) V_(s)^(2) - (1)/(2) C_(p) V_(p)^(2)` |
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| 1562. |
In the circuit shown in fig, the energy stored in both capacitors is `U_(1)`. If swich `S` is opened and a dielectric slab of constant 5 is put in free spaces of the capacitors, the energy stored is found to be `U_(2)`. Calcualte `U_(1)//U_(2). |
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Answer» Correct Answer - `5//13` In Fig, when `S` is closed, `U_(1) = (1)/(2) CV^(2) = CV^(2)` when `S` is open and electric is introduced. Capacity of each condenser `= KC = 5 C`. Condenser to left is connected to battery. Its potential remains `V`. But condenser on right is not connected to battery. So on introducing dielectric its potential becomes `(V)/(5)` `:.` Total energy, `U_(2) = (1)/(2) (5 C) V^(2) + (1)/(2) (5C) ((V)/(5))^(2)` `= (1)/(2) (5 C) V^(2) (1 + (1)/(25))` `U_(2) = (5 CV^(2))/(2) xx (26)/(25) = (13)/(5) CV^(2)` `:. (U_(1))/(U_(2)) = (CV^(2))/((13)/(5) CV^(2)) = (5)/(13)` |
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| 1563. |
Calculate the capacitance of a spherical capacitor consisting of two concentric spheres of radius 0.50m, 0.60m. The material filled in the space between the two spheres has a dielectric constant of 6. |
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Answer» Here, `C = ?, r_(a) = 0.50m`, `r_(b) = 0.60m, K = 6` `C = KC_(0) = K (4pi in_(0) r_(a) r_(b))/(r_(b) - r_(a))` `= (6)/(9xx10^(9)) ((0.5) (0.6))/((0.6 - 0.5)) = 2xx10^(-9)F`. |
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| 1564. |
When two uncharged, conducting spheres of radius `0.1mm` each collide 7 electrons get transferred from one to the ther. The potential difference between the spheres will be aboutA. `4xx10^(-6)V`B. `2xx10^(-6)V`C. `2xx10^(-4)V`D. `10^(-4)V` |
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Answer» Correct Answer - C::D |
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| 1565. |
An isolated conducting sphere of charge Q and radius R is connected to a similar uncharged sphere (kept at a large distance) by using a high resistance wire. After a long time is the amount of heat loss? |
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Answer» When two conducting spheres of equal radii are connected, charge is equally distributed on them (Result VI). So, we can say that heat loss of system : `DeltaH=U_(i)-U_(f)=(Q^(2)/(8pi epsi_(0)R)-0)-((Q^(2)//4)/(8pi epsi_(0)R)+(Q^(2)//4)/(8pi epsi_(0)R))=Q^(2)/(16pi epsi_(0)R)` |
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| 1566. |
A conducting sphere of radius `R` and carrying a charge `Q` is joined to an uncharged conducting sphere of radius `2R`. The charge flowing between them will beA. `Q//4`B. `Q//3`C. `Q//2`D. `2Q//3` |
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Answer» Correct Answer - D |
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| 1567. |
A point charge q is placed at a distance of r from cebtre of an uncharged conducting sphere of rad `R(lt r)`. The potential at any point on the sphere isA. zeroB. `(1)/(4piepsilon_(0))(q)/(r)`C. `(1)/(4pi epsilon_(0))(qR)/(r^(2))`D. `(1)/(4piepsilon_(0))(qr^(2))/(R)` |
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Answer» Correct Answer - B |
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| 1568. |
Two long straight parallel wires carry charges `lambda_(1) and lamba_(2)` per unit length. The distance between them is d. Calculate the magnitude of force externed on the length of one due to charge on the other. |
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Answer» Correct Answer - `(lambda_(1) lambda_(2))/(2pi in_(0) d)` Field intensity at the location of wire 2 due to charge on wire 1 is `E = lambda_(1)//2 pi in_(0) d` Charge on unit length of wire `2 , q = lambda_(2)` `:.` Force on unit length of wire 2, `f = qE = lambda_(2). (lambda_(1))/(2 pi in_(0)d) = (lambda_(1) lambda_(2))/(2pi in_(0) d)` |
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| 1569. |
Find the magnitude of uniform electric field E of which the direction is shown in figure if an electron entering with velocity l00m/s making `30^(@)` comes out making `60^(@)` , after a time numerically equal to m/e ofelectron where m is mass ofelectron and e is electronic charge : |
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Answer» Correct Answer - [100] |
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| 1570. |
Two long straight parallel wires carry charges `lambda_(1) and lamda_(2)` per unit length. The distance between them is d. Calculate the magnitude of force externed on the length of one due to charge on the other. |
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Answer» Correct Answer - `[(lambda^(2))/(2epsi_(0))]` |
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| 1571. |
A charged spherical conductor in a medium of permittivity e basa surface charge density σ. At an outside point, a distance r from the centre of the conductor, the electric field intensity is(A) \(\cfrac σ ɛ\)(B) \(\cfrac σ {ɛr}\)(C) \(\cfrac1{4\piɛ_0}\cfrac σ {r^2}\)(D) \(\cfrac1{4\piɛ}\cfrac σ {r^2}\) |
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Answer» Correct option is (A) \(\cfrac σ ɛ\) |
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| 1572. |
Match the following `{:(,"Table - 1",,"Table - 2",),((A),sigma^(2)//epsi_(0),(P),C^(2)//J - m,),((B),epsi_(0),(Q),"Farad",),((C),("ampere - second")/("Volt"),(R),J//m^(3),),((D),(V)/(E),(S),"metre",):}` |
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Answer» Correct Answer - (A)R,(B)P,(C)Q,(D)S |
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| 1573. |
Assertion: If we see along the axis of a chaged ring, the magnitudeof electric field is minimum at centre and magnitude of electridc potential is maximum. Reason: Electric field is a vector quantity while electric potential is scalar.A. If both Assertion and Reason are true but Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B `|E|=0` minimum at centre and `|V|=1/(4piepsilon_0).q/R` is maximum at centre. |
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| 1574. |
`ABC` is an equilateral triangle. Charges `+q` are placed at each corner. The electric intensity at `O` will be A. `(1)/(4 pi epsi_(0)). (q)/(r^(2))`B. `(3)/(4 pi epsi_(0)). (q)/(r)`C. zeroD. `(1)/(4 pi epsi_(0)). (3q)/(r^(2))` |
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Answer» Correct Answer - C |
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| 1575. |
Three point charges are placed at the corner of an equilateral triangle. Assuming only electrostatic forces are acting.A. The system will be in equilibrium if the charges have the same magnitude but not all have the same sign.B. The system will be in equilibrium if the charges have different magnitudes and not all have the same sign.C. The system will be in equilibrium if the charges rotate about the centre of the triangle.D. The system can never be in equilibrium. |
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Answer» Correct Answer - D |
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| 1576. |
Assertion: When two unlike charges are brought nearer, their electrostatic potentiasl energy decreases. Reason: All conservative forces act in the direction of decreasing potential energy.A. If both Assertion and Reason are true but Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A::B Two unlike charges come together when left freely. |
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| 1577. |
A cable consisting of a wire `3 mm` thick dielectric of relative permitively 10. Calculate the capacitance of `1 km` length of the cable. |
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Answer» Correct Answer - `0.506 muF` Here `C_(1) = 3 muF, C_(2) = 6 muF` `r_(b) = 1.5 + 3 = 4.5 mmn = 4.5xx10^(3) m` `L = 1 km = 10^(3) m , K = 10 , C = ?` `C = (2pi in_(0) L K)/(2.303 log_(10) ((r_(b))/(r_(a)))) = (2pi xx 8.85xx10^(-12)xx10^(3)xx10)/(2.303 log ((4.5xx10^(-3))/(1.5xx10^(-3))))` `= 50.6xx10^(-8) F = 0.506 mu F` |
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| 1578. |
An electric cable of diameter 10 cm runs under water (k for water = 81) for a distance of 5 km. If the cable carries a charge of 0.01 C on its surface. The electric intensity at a distance of 95 cm from the surface of the cable isA. `2.4xx10^(2)N//C`B. `4.4xx10^(2)N//C`C. `4.4xx10^(3)N//C`D. `4.4xx10^(-2)N//C` |
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Answer» Correct Answer - B `E=1/(4piepsi_(0)k)(2q)/r=1/(4piepsi_(0)k)(2Q)/(rL)` `=(9xx10^(9)xx2xx0.01)/(81xx1xx5xx10^(3))=4.4xx10^(2)N//C.` |
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| 1579. |
The surface density of charge on a conductor situated in air, is `2xx10^(-4)C//m^(2)`. The electric intensity at a point very near to its surface isA. `2.259xx10^(7)N//C`B. `2.261xx10^(9)N//C`C. `3.321xx10^(7)N//C`D. `3.321xx10^(9)N//C` |
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Answer» Correct Answer - A `E=sigma/(in_(0)k)=(2xx10^(-4))/(8.85xx10^(-12)xx1)` `= 2.259xx10^(7)N//C` |
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| 1580. |
An electric cable of diameter 10 cm runs under water for a distance of 5 km. If the cable carries a charge of 0.01 C on its surface then its linear charge density isA. `2muC//m`B. `0.5muC//m`C. `4muC//m`D. `1.5muC//m` |
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Answer» Correct Answer - A `lamda=q/L=0.01/(5xx10^(3))=2muC//m` |
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| 1581. |
A spherical balloon of radius R charged uniformly on its surface with surface density `sigma`. Find work done against electric forces in expanding it upto radius 2R. |
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Answer» Correct Answer - `(-pisigma^(2)R^(3))/(epsi_(0))` |
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| 1582. |
Two point charges `q_(1)= 20 muC` and `q_(2)=25 muC` are placed at (-1, 1, 1) m and (3, 1, -2)m, with respect to a coordinate system. Find the magnitude and unit vector along electrostatic force on `q_(2)` ? |
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Answer» Correct Answer - `0.18N.((4hati-3hatk)/(5))` |
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| 1583. |
A point charge `q` is placed at orign Let ` vec E_A, vec E_B ` and `vec E_C ` be the electric field at theree points ` (A) ( 1,23, ), B (A, 1,1-1)` and ` C( 2,2,2) ` due to charge (q). Then- [i] ` vec E_A _|_ vec E_B` [ii] `|vec E_B |=4|vecE_(c)|`selsect the correct alternative .A. only [i] is correctB. only [ii] is correctC. bothe [i] is corrctD. both [i] and [iii] are wrong |
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Answer» Correct Answer - C ` vec E_A = (Kq)/(14) . ( hat I + 2 hat j + 3 hat k)/( sqrt (14) rArr vec E_A = (Kq)/(14)) . Ohat A` ` vec E_B = (Kq)/3 ( hat I + hat j - hat k)/(sqrt 14) rArr vec E_A = (Kq)/ 3 . O hat B` ` vec E_C = (Kq) /(12) ( 2 hat I + 2 hat j + 2 hat k)/(sqrt 12) rArr vec E_(C) = (Kq)/(12). O hat C` since ` hat E_A. vec E_B = 0` so ` vec E_a _|_ vec E_A and |vec E_B | = vec E_C|`. |
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| 1584. |
The differeence in the effective capacity of two similar capacitor when joined in series and then in paralllel is 6 `mu` F the capacity of each capacitor isA. `2muF`B. `4muF`C. `8muF`D. `16muF` |
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Answer» Correct Answer - B When two similar capacitor are joined in series, then effective capacity is `(1)/(C_(s))=(1)/(C)+(1)/(C)=(2)/(C)` `therefore" "C_(s)=(C)/(2)` When joined in parallel, `C_(p)=C+C=2C` Since, `" "C_(p)-C_(s)=6muF` `therefore" "2C-(C)/(2)=6` `rArr" "4C-C=12` `"or "C=4muF` |
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| 1585. |
A body gains `25xx 10^(18)` electrons, by rubbing, it with another body. Find the charge on the body.A. chargeB. massC. resistanceD. inductance |
| Answer» Correct Answer - A | |
| 1586. |
Why is earth considered as zero of potential in practice ? Justify. |
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Answer» If we consider earth to be a conducting sphere surrouded by air its capacitance will be equal to `4 pi in_(0)R`. Since redius R of earth is very large. So capacitance is also very large. If a charge q is given to earth, it will increase its potential by `V = (q)/(C )` , As C is very large, `:. V rarr 0` for all finite charges. Hence earth is considered as zero of potential. |
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| 1587. |
How many `6muF, 200V` condensers are needed to make a condenser of `18muF, 600?`A. 9B. 18C. 3D. 27 |
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Answer» Correct Answer - D In series arrangement, potential difference is the sum of the individual potential difference of each capacitor. `"i.e. "V=V_(1)+V_(2)+V_(3)+…` `therefore" "600=x xx 200` x = 3 So, there should be 3 capacitors in series to obtain the required potential difference. The equivalent capacitance of the 3 capacitors in series is `(1)/(C_("eq"))=(1)/(6)+(1)/(6)+(1)/(6) rArr C_("eq")=2` |
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| 1588. |
Two condensers of capacities `8muF, and 4muF` are connected in parallel across a potential difference of 120 V. The charge and potential difference across `4muF` capacitior is respectivelyA. 1 mC, 10 VB. 0.2 mC, 20 VC. 0.4 mC, 60 VD. 0.48 mC, 120 V |
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Answer» Correct Answer - D `Q_(1)=C_(1)V=4xx10^(-6)xx120=0.48mC` In parallel combination potential is same. Therefore, V = 120 V |
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| 1589. |
A condenser is connected across another charged condenser. The energy in two condensers will beA. equal to the energy in the initial condenserB. less than that in the initial condenserC. more than that in the initial condenserD. more or less depending upon the relative capacities of the two condensers |
| Answer» Correct Answer - B | |
| 1590. |
The minimum number of condensers each capacitance of `2 muF`, in order to obtain result capacitance of `5 muF` will beA. 4B. 5C. 6D. 3 |
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Answer» Correct Answer - A |
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| 1591. |
Condensers of capacities 5F and 10F are connected in parallel in a circuit with a cell of e.m.f. 2V. The capacity of the condenser to be connected in series with parallel combination of the condensers to get 1C charge to flow in the circuit isA. 0.62 FB. 0.52 FC. 1.62 FD. 0.42 F |
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Answer» Correct Answer - B `C_(p)=C_(1)+C_(2)=5+10=15F` `C_(S)=(C*C_(p))/(C+C_(p))` Now," "`Q=C_(S)*V` `Q=((C*C_(p))/(C+C_(p)))V` `1=((Cxx15)/(C+15))xx2` `therefore" "C=0.52F.` |
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| 1592. |
Two parallel plate condensers of capacity `20 mF` and `30 mF` are charged to the potentials of 30 V and respectively. If likely charged plates are conneted together then the common potential difference beA. 100 VB. 50 VC. 24 VD. 10 V |
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Answer» Correct Answer - C |
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| 1593. |
The equivalent capacity of n identical condensers each of capacity C connented in series is given byA. `n^(2)C`B. nCC. `n/C`D. `C/n` |
| Answer» Correct Answer - D | |
| 1594. |
Four condensers having capacities 2pF, 3pF, 4pF and 6pF are connected in series. The equivalent capacity of the combination isA. 8.0 pFB. 0.8 pFC. 1.8 pFD. 0.4 pF |
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Answer» Correct Answer - B `1/C_(S)=1/C_(1)+1/C_(2)+1/C_(3)+1/C_(4)` `=1/2+1/3+1/4+1/6` `therefore" "C_(S)=0.8pF.` |
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| 1595. |
The resultant capacity of number of condensers connected in parallel is equal toA. sum of reciprocals of their individual capacityB. sum of their individual capacityC. difference of the capacities of each condenserD. none of these |
| Answer» Correct Answer - B | |
| 1596. |
The equivalent capacity of number of condensers can be increased if they are connected inA. seriesB. parallelC. both series and parallelD. neither series nor parallel |
| Answer» Correct Answer - B | |
| 1597. |
What is a capacitors? |
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Answer» Capacitor is a device used to store electric charge and electrical energy. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology. |
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| 1598. |
Two identical point charges of magnitude -q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the / displacement?(a) A1 and A2 (b) B1 and B2 (c) both directions (d) No stable |
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Answer» correct answer is (b) B1 and B2 |
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| 1599. |
Which charge configuration produces a uniform electric field? (a) point charge (b) infinite uniform line charge(c) uniformly charged infinite plane (d) uniformly charged spherical shell |
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Answer» (c) uniformly charged infinite plane |
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| 1600. |
Define surface density of charge and potential of a charged and potential of a charged spherical conductor. Establish a relation between them. |
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Answer» Surface density of charge is defined as the charge per unit area of a charged conductor . If r is the radius of a spherical conductor having charge q, then its surface density of charge, `sigma = (q)/(4pi r^(2))` …(i) Electric potentail of the spherical charged condutor, `V = (q)/(4pi in_(0) r) or q = 4pi in_(0) r xx V` Putting this value in (i), we get `sigma = (4pi in_(0) rV)/(4pi r^(2)) = (in_(0) V)/(r )` |
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