InterviewSolution
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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. |
The equivalent capacitance of the circuit shown, between points A and B will be- A. `2/3 muF`B. `5/3 muF`C. `8/3 muF`D. `7/3 muF` |
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Answer» Correct Answer - C `C_(eq)=(2xx1)/(2+1)+2=8/3 muF` |
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| 2. |
What fraction of the energy drawn from the charging battery is stored in as capacitorA. `100%`B. `75%`C. `50%`D. `25%` |
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Answer» Correct Answer - C Energy down from the charge battery =QE Energy stored in capacitor `=1/(2) QE` So % of energy stored =`(1//2 QE)/(QE) xx100=50 %` |
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| 3. |
After charging a capacitor the battery is removed. Now by placing a dielectric slab between the platesA. the potential difference between the plates and the energy stored will decreases but the charge on plates will remains sameB. the charge on the plates will decreases and the potential difference between the plates will increasesC. the potential difference between the plates will increases and energy stored will decreases but the charge on the plates will remains sameD. the potential difference, energy stored and the charge will remain unchanged |
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Answer» Correct Answer - A Battery is removed, charge on plate is constant. By placing a dielectric slab between the plates, capacitances increases so that potential difference between the plates and energy stored will decreases |
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| 4. |
If a thin metal foil of same area is placed between the two plates of a parallel plate capacitor of capacitance C, then new capacitance will beA. CB. 2CC. 3CD. 4C |
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Answer» Correct Answer - A Metal foil placed between the plates of capacitor do not affect the capacitance |
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| 5. |
Two capacitors `A` and `B` are connected in series with a battery as shown in the figure. When the switch `S` is closed and the two capacitors get charged fully, then A. The potential difference across the plates of A is 4V and across the plates of B is 6VB. The potential difference across the plates of A is 6V and across the plates of B is 4VC. The ratio of electric energies stored in A and B is `2:3`D. The ratio of charges on A and B is 3:2 |
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Answer» Correct Answer - B In serues `q_(A)=q_(B)=q=10 xx(2xx3)/5xx10^(-6) =12 mu C` `V_(A)=q/(C_(V))=6V, V_(B)=q/(C_(B))=4V` in series `U prop 1/C rArr (U_(A))/(U_(B)) =(C_(B))/(C_(A))=3/2` |
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| 6. |
The capacitance of the capacitors of plate areas `A_1` and `A_2(A_1 lt A_2)` at a distance d is A. `(epsilon_(0)A_(1))/d`B. `(epsilon_(0)A_(2))/d`C. `(epsilon_(0)(A_(1)+A_(2)))/(2d)`D. `(epsilon_(0)sqrt(A_(1)A_(2)))/d` |
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Answer» Correct Answer - A `C=(epsilon_(0)A_(1))/d` `A_(1) to ` overlap area. |
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| 7. |
You have been given a parallel plate air capacitor having capacitance C, a battery of emf `epsilon` and three dielectric blocks having dielectric constants `K_(1), K_(2) and K_(3)` such that `K_(1) gt K_(2) gt K_(3)`. Describe a sequence of steps such as connecting or disconnecting the capacitor to the battery, inserting or taking out of one of the dielectrics etc -so that the capacitor ends up having maximum possible energy stored in it. [Each dielectric block fills completely the space between the plates]. Write this maximum energy. |
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Answer» Correct Answer - Insert dielectric with constant `k_(1)`, connected the battery, disconnected the battery and remove the dielectric. `U_(max)=(1)/(2)K_(1)^(2)V^(2)C` |
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| 8. |
An air capacitor is charged upto a potential `V_1`. It is connected in parallel to an identical uncharged capacitor filled with a dielectric medium. After redistribution of charge if the potential difference of this combination is V, then the dielectric constant of the substance will beA. `V^(2)(V_(1)-V)`B. `V(V_(1)-V)`C. `V/((V_(1)-V))`D. `((V_(1)-V))/V` |
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Answer» Correct Answer - D `V=(CV_(1))/(C+KC)rArr K=(V_(1)-V)/V` |
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| 9. |
A neutral conducting ball of radius R is connected to one plate of a capacitor (Capacitance = C), the other plate of which is grounded. The capacitor is at a large distance from the ball. Two point charges, q each, begin to approach the ball from infinite distance. The two point charges move in mutually perpendicular directions. Calculate the charge on the capacitor when the two point charges are at distance x and y form the centre of the sphere. |
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Answer» Correct Answer - `Q=(qRC)/((4piin_(0)R+C))((1)/(x)+(1)/(y))` |
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| 10. |
In the circuit shown in the figure all the capacitors have capacitance C. (a) Find the charge on capacitors marked as 1 and 2 when a battery of emf V is connected across points A and B. (b) Find the equivalent capacitance across points C and D marked in the Figure. (c) Find the equivalent capacitance across points E and G. |
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Answer» Correct Answer - (a) Zero on both (b) 2C (c) 2C |
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| 11. |
Four large identical metallic plates are placed as shown in the Figure. Plate 2 is given a charge Q. All other plates are neutral. Now plates 1 and 4 are earthed. Area of each plate is A. (a) Find charge appearing on right side of plate 3. (b) Find potential difference between plates 1 and 2. |
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Answer» Correct Answer - (a) `(Q)/(4)` (b) `(3)/(4) (Qd)/(in_(0)A)` |
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| 12. |
Three identical large metal plates each of area S are at distance `d and 2d` from each other as shown. Metal plate A is uncharged, while metal plates `B and C` have charges `+ Q and - Q` respectively. Metal plates A and C are connected by a conducting wire through a switch K. How much electrostatic energy is lost when the switch is closed ? |
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Answer» Correct Answer - `(Q^(2)d)/(6 in_(0)S)` |
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| 13. |
Lower plate of a parallel plate capacitor is fixed on a horizontal insulating surface. The upper plate is suspended above it using on elastic cord of force constant K. The upper plate has negligible mass and area of each plate is A. When there is no charge on the plates the equilibrium separation between them is d. When a potential difference = V is applied between the plates the equilibrium separation changes to x. (a) Calculate V as a function of x. (b) Find the value of x for which V is maximum. Calculate the maximum value of `V (= V_(max))` (c) What will happen if `V gt V_(max)` ? (d) Plot a rough graph showing variation of equilibrium separation (x) with V |
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Answer» Correct Answer - (a) `V=sqrt((2K(d-x)x^(2))/(in_(0)A))` (b) `V_(max) = sqrt((8Kd^(3))/(27 in_(0)A)) "at" x=(2d)/(3)` (c) System cannot remain in equilibrium. |
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| 14. |
Two identical metal plates with area A and mass m are kept separated by help of three insulating springs as shown in the Figure. The equilibrium separation between the plates is `d_(0) (gt gt sqrt(A))` and force constant of each spring is K. When a constant voltage source having emf V is connected to the plates, the equilibrium separation changes to d. Assume that the lower plate is fixed and the upper plate is free to move. (a) Find V in terms of given parameters. (b) If the upper disc is slightly displaced from its equilibrium position and released, calculate the time period of its oscillation. |
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Answer» Correct Answer - (a) `V=sqrt((6Kd^(2)(d_(0)-d))/(in_(0)A))` (b) `T=2pi sqrt((m.d)/(3K(3d-2d_(0))))` |
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| 15. |
One plate of a parallel plate capacitor is tilted by a small angle about its central line as shown in the Figure. The tilt angle `theta` is small. Both the plates are square in shape with side length a and separation between their centers is d. Find the capacitance of the capacitor. Given : `l n(1+x)=x-(x^(2))/(2)+(x^(3))/(3)+....` `l n(1-x)=-(x+(x^(2))/(2)+(x^(3))/(3)+....)` |
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Answer» Correct Answer - `(in_(0)a^(2))/(d)[1+(a^(2)theta^(2))/(12d^(2))]` |
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| 16. |
Two conducting sphere of radii a and b are placed at separation d. It is given that `d gt gt a and d gt gt b` so that charge distribution on both the sphere remains spherically symmetric. Assume that a charge `+ q` is given to the sphere of radius a and `-q` is given to the sphere of radius b. (i) Write the electrostatic energy (U) of the system and calculate the capacitance of the system using the expression of U. (ii) Prove that the capacitance of the system is given by `(1)/(C)=((1)/(4 pi in_(0)a)+(1)/(4 pi in_(0)b))"if" d rarr infty` |
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Answer» Correct Answer - (i) `U=(q^(2))/(8 pi in_(0))((1)/(a)+(1)/(b)-(2)/(d));C=(4 pi in_(0))/((1)/(a)+(1)/(b)-(2)/(d))` |
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| 17. |
Two concentric spherical shells have radii a and `b (gt a)`. Write the capacitance of the system in following cases. (a) Positive terminal of the battery is connected to the outer shell and its other terminal and the inner shell are grounded. (b) Positive terminal of the battery is connected to the inner shell and its negative terminal is grounded. (c) A terminal of the battery is connected to the inner shell and the other terminal along with the outer shell is grounded. (d) A terminal of the battery is connected to the outer shell and the other terminal is grounded. |
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Answer» Correct Answer - (a) `4 pi in_(0)((b^(2))/(b-a))` (b) `4 pi in_(0)a` (c) `4 pi in_(0)((ab)/(b-a))` (d) `4 pi in_(0) b` |
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| 18. |
A hollow spherical conductor of radius R has a charge Q on it. A small dent on the surface decreases the volume of the spherical conductor by 2%. Assume that the charge density on the surface does not change due to the dent and the electric field in the dent region remains same as other points on the surface. (a) `Delta E` is the electrostatic energy stored in the electric field in the shallow dent region and E is the total electrostatic energy of the spherical shell. Find the ratio `(Delta E)/(E)` (b) Using the ratio obtained in part (a) calculate the percentage change in capacitance of the sphere due to the dent. |
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Answer» Correct Answer - (a) `(1)/(150)` (b) 0.67 % |
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| 19. |
The space between the conductors of a spherical capacitor is half filled with a dielectric as shown is Figure. The dielectric constant is K. (a) If a charge is given to the capacitor write the ratio of free charge density on the inner sphere at point A and B. (b) Write the ratio of capacitance with dielectric and without dielectric. |
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Answer» Correct Answer - (a) `(sigma_(A))/(sigma_(B))=K` (b) `(1+K)/(2)` |
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| 20. |
Two students A and B were taught that electric field near a uniformly charged large surface is normal to the surface and is equal to `(sigma)/(2 in_(0))`. They were also told that field near the surface of a conductor is `(sigma)/(in_(0))` normal to the conductor where s is charge density on the conductor surface. Now both of them were asked to write field between the plates of an ideal parallel plate capacitor having charge density `sigma` and `-sigma` on its plates. Student A said that field can be seen as superposition of field due to two large charged surfaces. He wrote the answer as `E=(sigma)/(2in_(0))+(sigma)/(2in_(0))=(sigma)/(in_(0))` Student B thought that a capacitor has conducting plates and therefore field due to each plate Must be `(sigma)/(in_(0))` . He wrote his answer as `E=(sigma)/(in_(0))+(sigma)/(in_(0))=(2sigma)/(in_(0))` Who is wrong and where is the flaw in thinking ? |
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Answer» Correct Answer - B is wrong |
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| 21. |
Electrical susceptibility `(chi)` of a dielectric material is defined as `chi = in_(r) -1` where `in_(r)` is its relative permittivity. An isolated parallel plate capacitor carries some charge and the field in the dielectric present between its plates is E. Express the electric field due to induced charge on dielectric surface in terms of `chi` and E. |
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Answer» Correct Answer - `chi E` |
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| 22. |
Metallic sphere of radius R is charged to potential V. Then charge q is proportional toA. VB. RC. bothD. None |
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Answer» Correct Answer - C `q=CV=4piepsilon_(0) RV` |
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| 23. |
In the circuit shwon in fig. switch S is closed at time t = 0. Let `I_1 and I_2` be the currents at any finite time t, then the ratio `I_1// I_2` A. is constantB. increases with timeC. decreases with timeD. first increases and then decreases |
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Answer» Correct Answer - B `I_(1)=V/R e^(t/(2RC)), i_(2)=V/R e^(t/(RC))` `(i_(1))/(i_(2)) e^(t/(2RC))`, Increase with time |
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| 24. |
Two capacitors, `3mu F` and `4mu F`, are individually charged across a 6V battery. After being disconnected from the battery, they are connected together with the negative place of one attached to the positive plate of the other. What is the common potential?A. 6VB. `(6/7)V`C. 2VD. `(3/2)V` |
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Answer» Correct Answer - B `V=(Q_("total"))/(C_(eq)) =(4xx6-3xx6)/(4+3)=6/7V` |
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| 25. |
Between the plates of a parallel plate condenser, a plate of thickness `t_(1)` and dielectric constant `k_(1)` is placed. In the rest of the space, there is another plate of thickness `t_(2)` and dielectric constant `k_(2)`. The potential difference across the condenser will beA. `Q/(Aepsilon_(0))((t_(1))/(k_(1))+(t_(2))/(k_(2)))`B. `(epsilon_(0)Q)/A((t_(1))/(k_(1))+(t_(2))/(k_(2)))`C. `Q/(Aepsilon_(0))((k_(1))/(t_(1))+(k_(2))/(t_(2)))`D. `(epsilon_(0)Q)/A(k_(1)t_(1)+k_(2)t_(2))` |
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Answer» Correct Answer - A `1/(C_(eq))=(t_(1))/(k_(1)epsilon_(0)A)+(t_(2))/(k_(2)epsilon_(0)A)` `V=Q/(C_(eq))=Q/(epsilon_(0)A)((t_(1))/(K_(1))+(t_(2))/(K_(2)))` |
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| 26. |
In a parallel plate capacitor the separation between the two plates is maintained by a dielectric of dielectric constant K and thickness `d`. The dielectric material is not rigid and has a young’s modulus of Y. Capacitance of the capacitor is `C_(0)` if applied potential difference is nearly zero. At higher potentials the attractive force between the plates compresses the dielectric (by a small amount) and reduces the gap between the plates. Change in K can be neglected due to compression in the dielectric. Find the change in capacitance when a battery of V volt is connected across it. |
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Answer» Correct Answer - `Delta C ~= (C_(0)K^(2)V^(2)in_(0))/(2d^(2)Y)` |
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| 27. |
A parallel plate capacitor has square plates of side length L kept at a separation d. The space between them is filled with a dielectric whose dielectric constant changes as `K = e^(betax)` where `x` is distance measured from the left plate towards the right plate, and b is a positive constant. A poten- tial difference of V volt is applied with left plate positive. (i) What happens to capacitance of the capacitor if d is increased ? What is the smallest possible capacitance that can be obtained by changing d ? (ii) Write the expression of electric field between the plates as a function of `x`. |
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Answer» Correct Answer - (i) Capacitance decreases, `C_(min) = in_(0) L^(2) beta` (iii) `E=(betaVe^(-betax))/(1-e^(=-betad))` |
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| 28. |
Two large conducting plates, identical in size, are placed parallel to each other at a separation d. Each plate has area A. One of the plates is cut into two equal parts and then a battery of emf V is connected across these two pieces. Find the work done by the battery in supplying charge to the plates. |
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Answer» Correct Answer - `(in_(0)A)/(4d)V^(2)` |
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| 29. |
Two capacitors A and B with capacitors `3 muF and 2 muF` are charged to a potential difference of 100 V and 180 V respectively. One plate of two capacitors are connected as shown. Now switch S is closed so as to connect a cell of 100 V to the free plates of two capacitors. (a) Find charge on the two capacitors after the switch is closed. (b) Calculate heat generated in the circuit |
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Answer» Correct Answer - (a) `q_(A)=84 muC ;q_(B)=144muC` (b) `19.44 mJ` |
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| 30. |
In the circuit shown in figure, each capacitor has a capacity of `3 muF`. The equivalent capacity between `A` and `B` is A. `3/5 muF`B. `3 muF`C. `6 muF`D. `5 muF` |
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Answer» Correct Answer - D `C_(eq)=(6xx3)/(6+3) +3=5 muC` |
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| 31. |
In the given circuit it is known that the capacitor A has a capacitance of `2 muF` and carries a charge of `40 muC`. Capacitor C has a capacitance of `6 muF` and carries a charge of `180 muC`. The positive plate of both capacitors has been indicated in the Figure. Capacitance of capacitor B is `3 muF`. Calculate charge on B after the switch S is closed. |
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Answer» Correct Answer - `(240)/(11) muC` |
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| 32. |
Find heat dissipated in the circuit after switch S in closed. `C = 2 muF`. |
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Answer» Correct Answer - 1.8 mJ |
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