InterviewSolution
Saved Bookmarks
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. |
Find out the capacitance of the earth ? (Radius of the earth = 6400 km) |
| Answer» `C=4pi in_(0)R=(6400xx10^(3))/(9xx10^(9))=711mu F` | |
| 102. |
Read the following statements (a) Non polar molecules have uniform charge distribution (b) Polar molecules have non-uniform charge distribution ( c) Polar molecules are already polarized (d) Molecules are not already polarized without electric field in Non-polar molecules.A. only `a` & `b` are correctB. only `c` & `d` are correctC. only `c` is wrongD. all are correct |
| Answer» Correct Answer - D | |
| 103. |
The electric field `(vec( E))` between two parallel plates of a capacitor will be uniform if.A. the plate separation `(d)` is equal to area of the plate `(A)`B. the plate separation `(d)` greater when compared to area of the plate `(A)`C. the plate separation `(d)` is less when compared to area of the plate `(A)`D. `2` (or) `3` |
| Answer» Correct Answer - C | |
| 104. |
The capacitance of a capacitor depends onA. the geometry of the platesB. separation between platesC. the dielectric between the platesD. above all |
| Answer» Correct Answer - D | |
| 105. |
In an `RC` circuit while charging, the graph of `1n i` versis time is as shown by the dotted line in the diagram figure. Where `i` is the current. When the value of the resistance is doubled, which of the solid curve best represents the variation of `1n i` versus time |
|
Answer» Correct Answer - B During charging `Q = Q_(0) (1-e^(-t//RC)) = CV (1-e^(-t//RC))` `I = (-CVe^(-t//RC)) xx -(1)/(R ) e^(-t//RC)` `lnI = ln (V)/(R)-(t)/(RC ) rArr lnI =- (t)/(RC )+ln (V)/(R )` slop and intercept both decreases. |
|
| 106. |
An air cylindrical capacitor eith a `dc` voltage `V = 200 V` applied across it is being submerged vertivally into a vessel filled with water at a velocity `v = 5.0 mm//s`. The electrodes of the capacitors are separated by a distance `d = 2.0 mm`, the mean curvature radius of the electrodes is equal to `r = 50 mm`. Find the current flowing in this case along lead wires, if `d lt lt r`. |
|
Answer» Correct Answer - `0.11muA` |
|
| 107. |
The force between the plates of a parallel plate capacitor of capacitance `C` and distance of separation of the plates `d` with a potential difference `V` between the plates, is.A. `(CV^(2))/(2d)`B. `(C^(2)V^(2))/(2d^(2))`C. `(C^(2)V^(2))/(d^(2))`D. `(V^(2)d)/(C )` |
|
Answer» Correct Answer - A `F=(QE)/(2)=(CV)/(2)[(V)/(d)]`. |
|
| 108. |
A capacitor of capacitance as C is given a charge Q. At `t=0`,it is connected to an ideal battery of emf `(epsilon)`through a resistance R. Find the charge on the capacitor at time t.A. `CEe^(-t//CR)+Qe^(-t//CR)`B. `CE(1-e^(-t//CR))-Qe^(-t//CR)`C. `CE(1-e^(-t//CR))+Qe^(-t//CR)`D. `CE(1-e^(-t//CR))` |
|
Answer» Correct Answer - C Similar to above question (or) substitute `t=0` and verify the options. |
|
| 109. |
A capacitor is made of a flat plate of area A and a second plate having a stair -like structure as shown in figure. The width of each stair is `a` and the height is `b` . Find the capacitance of the assembly. A. `(2Ain_(0))/(2(d+b))`B. `(Ain_(0)(3d^(2)+6bd +2b^(2)))/(3d(b+d)(d+2b))`C. `(Ain_(0)(d^(2)+2bd +b^(2)))/(3d(d+b)(d+2b))`D. `(2A in_(0)(d^(2)+2bd +b^(2)))/(3d(d+b)(d+2b))` |
|
Answer» Correct Answer - B In Equivalent circuit `C_(1),C_(2),C_(3)` are parallel `C_(1)=(A in_(0))/(3d),C_(2)=(A in_(0))/(3(d+b)),C_(3)=(A in_(0))/(3(d +2b))`. |
|
| 110. |
In the adjoining circuit, the capacity between the points A and B will be - A. CB. 2CC. 3CD. 4C |
| Answer» Correct Answer - B | |
| 111. |
In the following figure, the charge on each condenser in the steady state will be - A. `3mu C`B. `6mu C`C. `9mu C`D. `12mu C` |
| Answer» Correct Answer - D | |
| 112. |
The magnitude of charge in steady state on either of the plates of condenser C in the adjoining circuit is A. CEB. `(CER_(2))/((R_(1)+r))`C. `(CER_(2))/((R_(2)+r))`D. `(CER_(1))/((R_(2)+r))` |
| Answer» Correct Answer - C | |
| 113. |
The work done against electric in increasing the potential difference of a condenser from 20 V to 40 V is W. The work done in increasing its potential difference from 40 V to 50 V will beA. 4WB. `(3W)/(4)`C. 2WD. `(W)/(2)` |
| Answer» Correct Answer - B | |
| 114. |
Two metal plates are separated by a distance `d` in a parallel plate condenser. A metal plate of thickness `t` and of the same area is inserted between the condenser plates. The value of capacitance increases by ….times.A. `(d-t)/(d)`B. `(1-(t)/(d))`C. `(t -(t)/(d))`D. `(1)/((1-(t)/(d)))` |
|
Answer» Correct Answer - D `C=(epsilon_(0)A)/(d-t+(t)/(k)),k= infty`. |
|
| 115. |
In a charged capacitor the enery is stored in `( r)` is less than at `B`.A. both in positive and negative chargesB. positive chargesC. the edges of the capacitor platesD. the electric field between the plates |
| Answer» Correct Answer - D | |
| 116. |
You have a parallel plate capacitor, a spherical capacitor and a cylindrical capacitor. Each capacitor is charged and then removed from the same battery. Consider the following situations : I:· Separation between the plates of parallel plate capacitor is reduced II: Radius of the outer spherical shell of the spherical capacitor is increased Ill: Radius of the outer cylinder of cylindrical capacitor is increased. Which of the following is correct?A. In each of these situations I, II and Ill, charge on the given capacitor remains the same and potential difference across it also remains the same.B. In each ofthese situations I, II and Ill, charge on the given capacitor remains the same but potential difference, in situations I and III, decreases, and in situation II, increases.C. In each of these situations I, II and Ill, charge on the given capacitor remains the same but potential difference, in situations I, decreases, iind in situations II and III, increases.D. Charge on the capacitor in each situation changes. It increases in all these situations but potential difference remains the same. |
|
Answer» Correct Answer - C |
|
| 117. |
The plates of a parallel plate capacitor are, not exactly parallel. The surface charge density therefore :A. Is higher at the closer endB. The surface charge densitywill not be uniformC. Each plate will have the same potential at each pointD. The electric field is smallest where the plates are closest |
|
Answer» Correct Answer - A::B::C |
|
| 118. |
Figure shows a capacitor having three layers of equal thickoess and same area as that of its plates. Layer-I is free space, Layer-II is a conductor and Layer-Ill is a dielectric of dielectric constant `k`. Calculate the ratio of energy stored in region III to total energy stored in capacitor when a potential difference is applied across the plates of capacitor. |
|
Answer» Correct Answer - `1/(k+1)` |
|
| 119. |
A parallel plate capacitor has two layers of dielectric as shown in figure. This capacitor is connected across a battery. The graph which shows the variation of electric field `(E)` and distance `(X)` from left plate. B. C. D. |
|
Answer» Correct Answer - A Field in dielectric is `(E)/(K)` when `E` is the filed in air. |
|
| 120. |
Two identical parallel plate capacitors of same dimensions joined in series are connected to a constant voltage source. When one of the plates of one capacitor are brought closer to the other plate :A. The voltage on the capacitor whose plates come closer is greater than the voltage on the capacitor whose plates are not movedB. The voltage on the capacitor whose plate come closer is lesser than the voltage on the capacitor whose plates are not movedC. The voltage on the two capacitor remain equalD. The applied voltage is divided among the two inversely as the capacitance |
|
Answer» Correct Answer - B::D |
|
| 121. |
What is potential at a distance `r (lt R)` in a dielectric sphere of uniform charge density `rho`, radius R and dielectric constant `epsilon_(r)`. |
|
Answer» `V_(A)=V_(B)+(W_(B rarr A))/(q)` `V=(Q)/(4pi in_(0)R)+underset(R)overset(r)int (rho r)/(3 in_(0)in_(r))(-dr)=(Q)/(4 pi in_(0)R)+(rho (R^(2)-r^(2)))/(3 in_(0) in_(r))` `V_("outside")=(KQ)/(r)` |
|
| 122. |
Six plates each of area `A` arranged as shown· in figure. The separation between adjoining plates is `d`. Find the equivalent capacitance between points `A` and `B` : A. `(epsilon_(0)A)/d`B. `(7epsilon_(0)A)/d`C. `(6epsilon_(0)A)/d`D. `(5epsilon_(0)A)/d` |
|
Answer» Correct Answer - A |
|
| 123. |
In an isolated parallel plate capacitor of capacitance C the four surfaces have charges `Q_(1), Q_(2), Q_(3)` and `Q_(4)` as shown in the figure. The potential difference between the plates is : A. `(Q_(1)+Q_(2))/(C)`B. `|(Q_(2))/(C)|`C. `|(Q_(3))/(C)|`D. `(1)/(C) [(Q_(1)+Q_(2))-(Q_(3)-Q_(4))]` |
| Answer» Correct Answer - B::C | |
| 124. |
In an isolated parallel plate capacitor of capacitancee `C`, the four surface have chrges `Q_(1),Q_(2),Q-(3)` and `Q_(4)` as shown. The potential difference between the plates is A. `(Q_(1)+Q_(2)+Q_(3)+Q_(4))/(2C)`B. `(Q_(2)+Q_(3))/(2C)`C. `(Q_(2)-Q_(3))/(2C)`D. `(Q_(1)+Q_(4))/(2C)` |
|
Answer» Correct Answer - C Facing surface have equal and opposite charges `Q_(3)=-Q_(2)` charge on capacitor `Q_(2) or -Q_(3)` `V=(Q-(20))/(c)` `V=(-Q_(3))/(c)`….(ii) `(i)+(ii)` `2V=(Q_(2)-Q_(3))/(c)impliesV=(Q_(2)-Q-930)/(2C)` |
|
| 125. |
A condenser stores.A. potentialB. chargeC. currentD. energy in magnetic field |
| Answer» Correct Answer - B | |
| 126. |
Out of the following statements (A) The capacity of a conductor is affected due to the presence of an uncharged isolated conductor (B) A conductor can hold more charge at the same potential if it is surrounded by dielectric medium.A. Both `A` and `B` are correctB. Both `A` and `B` are wrongC. `A` is correct and `B` is wrongD. `A` is wrong and `B` is correct. |
| Answer» Correct Answer - A | |
| 127. |
A metal slab of thickness, equal to half the distance between the plates is introduced between the plates of a parallel plate capacitor as shown. Find its capacity. . |
|
Answer» When capacitor is partially filled with dielectric capacity `c=(epsilon_(0)A)/([d-t(1-(1)/(k))])` For metal slab of thickness `t = d//2`, `C=(in_(0)A)/(d-t)` (`K = infty` for metal slab) `= (in_(0)A)/(d-(d)/(2))=2(in_(0)A)/(d)`. |
|
| 128. |
The charge flowing through the cell on cloing the key `k` is equal to : A. `(CV)/(4)`B. CVC. `(4)/(3)CV`D. `(3)/(4) CV` |
|
Answer» Correct Answer - A `C_("net") =(3)/(4) C` when key was open `q = (3)/(4) CV` when key was closed `3C` becomes short circuited. Net charge on `C` is now `q = CV` `Delta q = q - q = (CV)/(4)`. |
|
| 129. |
The area of the positive plate is `A_(1)` and the area of the negative plate is `A_(2)(A_(2) lt A_(1))`. They are parallel to each other and are separated by a distance `d`. The capacity of a condenser with air dielectric is.A. `(epsilon_(0)A_(1))/(d)`B. `(epsilon_(0)A_(2))/(d)`C. `(epsilon_(0)A_(1)A_(2))/(d)`D. `(epsilon_(0)A_(1))/(A_(2)d)` |
|
Answer» Correct Answer - B Effective area only `:. A_(2)`. |
|
| 130. |
What is equivalent capacitance of ladder circuit shown in figure between points `A` and `B`? A. `2/3muF`B. `4/3muF`C. InfiniteD. `(1+sqrt(3))muF` |
|
Answer» Correct Answer - B |
|
| 131. |
When two isolated conductors A and B are connected by a conducting wire positive charge will flow from . A. A to BB. B to AC. will not flowD. can not say. |
|
Answer» Correct Answer - B Charge always flows from higher potential body to lower potential body Hence, `V_(A)=(30)/(10)=3V rArr V_(B)=(20)/(5)=4V " " "As" V_(B) gt V_(A) :. (B)` is correct Answer. |
|
| 132. |
The charge across the capacitor in two different `RC` circuit `1` and `2` are potted as shown in figure. Identify the correct statement `(s)` related to the `R_(1),R_(2),C_(1)` and `C_(2)` of the two `RC` circuit-A. `R_(1) gtR_(2)` if `E_(1) =E_(2)`B. `C_(1) lt C_(2)` if `E_(1) = E_(2)`C. `R_(1)C_(1) gt R_(2)C_(2)`D. `(R_(1))/(R_(2)) lt (C_(2))/(C_(1))` |
|
Answer» Correct Answer - D `I = (V)/(R )e^(-t//RC)` Since `tau_(2) gt tau_(1) = R_(2)C_(2) gt R_(1)C_(1) = (R_(1))/(R_(2))lt (C_(2))/(C_(1))` |
|
| 133. |
A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by `Q_0` , `V_0`, `E_0` and `U_0` respectively. A dielectric slab is now introduced to fill the space between the plates with battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous one asA. `Q gt Q_(0)`B. `V gt V_(0)`C. `E gtE_(0)`D. `U lt U_(0)` |
|
Answer» Correct Answer - A (A) `Q_(0) = CV & Q = (KC)V rArr Q gt Q_(0)` (B) `V = V_(0)` (C ) `E = E_(0)` (D) `U_(0) = (1)/(2) CV^(2) &U = (1)/(2)(KC)V^(2)rArr U gt U_(0)` |
|
| 134. |
A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by `Q_0` , `V_0`, `E_0` and `U_0` respectively. A dielectric slab is now introduced to fill the space between the plates with battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous one asA. `Q gt Q_(0)`B. `V gt V_(0)`C. `E gt E_(0)`D. `U gt U_(0)` |
| Answer» Correct Answer - A::D | |
| 135. |
In the shown circuit involving a resistor of resistance `R Omega`, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for `t_(0) = RC` seconds and then key is pushed back to position 1 for `t_(0) = RC`seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice verse to be negligible. The current through the resistance at t = 1.5 RC seconds isA. `(E)/(e^(2)R)(1-(1)/(e))`B. `(E)/(eR)(1-(1)/(e))`C. `(E)/(R)(1-(1)/(e))`D. `(E)/(sqrt(e)R)(1-(1)/(e))` |
| Answer» Correct Answer - D | |
| 136. |
In the shown circuit involving a resistor of resistance `R Omega`, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for `t_(0) = RC` seconds and then key is pushed back to position 1 for `t_(0) = RC`seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice verse to be negligible. The charge on capacitor at t = 2RC second isA. CEB. CE `(1-(1)/(e))`C. CE `((1)/(e)-(1)/(e^(2)))`D. CE `(1-(1)/(e)+(1)/(e^(2)))` |
| Answer» Correct Answer - C | |
| 137. |
In the shown circuit involving a resistor of resistance `R Omega`, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for `t_(0) = RC` seconds and then key is pushed back to position 1 for `t_(0) = RC`seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice verse to be negligible. Then the variation of charge on capacitor with time is best represented byA. B. C. D. |
| Answer» Correct Answer - C | |
| 138. |
In the `R-C` circuit shown in the figure the total enegry of `3.6 xx 10^(-3)J` is dissipated in the `10 Omega` resistor when the switch `S` is cloed. The initially charge on the capacitor is A. `60 muC`B. `120 muC`C. `60 sqrt(2)muC`D. `(60)/(sqrt(2))muC` |
|
Answer» Correct Answer - B `H = U_(i) - U_(f) rArr H = U_(i)` `H= (q^(2))/(2C) rArr 3.6 xx 10^(-3) = (q^(2))/(2xx2xx10^(-6))` `q = 120 muC` |
|
| 139. |
A charged capacitor is allowed to discharge through a resistor by closing the key at the instant `t=0`. At the instant `t=(ln 4)mus`, the reading of the ammeter falls half the initial value. The resistance of the ammeter is equal to A. `1M Omega`B. `1 Omega`C. `2 Omega`D. `2M Omega` |
|
Answer» Correct Answer - C `I = I_(0)e^(-t//tau)` `2=e^(t//tau) rArr ln2 = (t)/(tau) rArr tau = (ln4)/(ln2) rArr tau = 2muS` `RC = 2xx10^(-6) rArr (2+R_(A)) xx 0.5 xx 10^(-6) =2xx10^(-6)` `2+R_(A) = 4 rArr R_(A) = 2 Omega` |
|
| 140. |
A charged capacitor is allowed to discharge through a resistor by closing the key at the instant `t=0`. At the instant `t=(ln 4)mus`, the reading of the ammeter falls half the initial value. The resistance of the ammeter is equal to A. `0`B. `2Omega`C. `oo`D. `2M Omega` |
|
Answer» Correct Answer - A `I = I_(0)e^(-t//tau) rArr 2=e^(t//tau) rArr ln2 = (t)/(tau)` `tau = (ln2)/(ln2) rArr tau = 1muS rArr RC = 1xx 10^(-6)` `(2+R_(A)) xx 0.5 xx 10^(-6) = 1xx 10^(-6)` `2+R_(A) = 2rArr R_(A = 0` |
|
| 141. |
A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles:A. The charge on the capacitor increasesB. The voltage across the plate increasesC. The capacitance increasesD. the electrostatic energy stored in the capacitor increases |
|
Answer» Correct Answer - B::D |
|
| 142. |
A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles:A. the charges on the capacitor increasesB. the voltage across the capacitor increasesC. the capacitance increasesD. the electrostatic energy stored in the capacitor increases. |
|
Answer» Correct Answer - B::D `C` decreases, so `V` increases and `U` also increases. |
|
| 143. |
A capacitor with no dielectric is connected to a battery at t=0.Consider a point A in the connecting wires and a point B in between the plates.A. There is no current through A.B. There is displacement current through B till electric fieldchanges between the plates.C. There is a current through A as long as the charging is notcomplete.D. The current always flows between the plates of capacitor. |
|
Answer» Correct Answer - B::C |
|
| 144. |
The equivalent capacitance between `P` and `Q` is. .A. `10 mu F`B. `20 mu F`C. `5 mu F`D. `15 mu F` |
|
Answer» Correct Answer - C From Left `C=(10 xx10)/(10+10)=5 muF` `C^(1)=5+5 =10 muF ,C^(11)=(10 xx10)/(10+10)=5 muF` and so on finally `C_(eff)=(10xx10)/(10+10)=5 mu F`. |
|
| 145. |
The effective capacitance between the point `P` and `Q` in the given figure is. .A. `4 mu F`B. `16 mu F`C. `26 mu F`D. `10 mu F` |
|
Answer» Correct Answer - A `C^(1)=(C_(1)C_(2))/(C_(1)+C_(2)),C^(11) =(C_(3)C_(4))/(C_(3)+C_(4))C_(eff)=C^(1)+C^(11)`. |
|
| 146. |
Two conducting spheres of radii `r_(1)` and `r_(2)` having charges `Q_(1)` and `Q_(2)` respectively are connected to each other. There isA. no change in the energy of the systemB. an increase in the energy of the systemC. always a decrease in the energy of the systemD. a decrease in the energy of the system unless `Q_(1)r_(2)=Q_(2)r_(1)` |
|
Answer» Correct Answer - D If potentials of two spheres are same i.e., `V_(1)=V_(2)implies(Q_(1))/(4piin_(0)r_(1))=(Q_(2))/(4piin_(0)r_(2))` `Q_(1)r_(2)=Q_(2)r_(1)` There will be no redistribution of charge and hence no loss of energy |
|
| 147. |
There are two conducting spheres of radius a and b `(b gt a)` carrying equal and opposite charges. They are placed at a separation d (`gt gt gt ` a and b). The capacitance of system is A. `(4 pi epsilon_(0))/(a-b-d)`B. `(4pi epsilon_(0))/((1)/(a)-(1)/(b)-(1)/(d))`C. `(4pi epsilon_(0))/((1)/(a)+(1)/(b)-(1)/(d))`D. `(4 pi epsilon_(0))/((1)/(a)+(1)/(b)-(2)/(d))` |
| Answer» Correct Answer - D | |
| 148. |
Two conductors carrying equal and opposite charges produce a non uniform electric field along X- axis given by `E=(Q)/(in_(0)A)(1+Bx^(2))` where `A` and `B` are constant. Separation between the conductors along X-axis is `X`. Find the capacitance of the capacitor formed. |
|
Answer» Potential difference between the conductors is given by `V=V_(+) -V_(-) = underset(0) overset(x) int Edx` `rArr V= underset(0) overset(x) int (Q)/(in_(0)A)(1-Bx^(2))dx` or `V=(Q)/(in_(0) A)(x-(Bx^(3))/(3))_(O)^(X)=(Q)/(in_(0)A) (X+(BX^(3))/(3))` Capacity `C=(Q)/(V) =(in_(0) A)/(X(1+(BX^(2))/(3)))`. |
|
| 149. |
Two parallel plate capacitors A and B have the same separation `d=8.85xx10^-4m` between the plates. The plate area of A and B are `0.04m^2` and `0.02m^2` respectively. A slab of dielectric constant (relative permittivity) `K=9` has dimensions such that it can exactly fill the space between the plates of capacitor B. (i) The dielectric slab is placed inside. A as shown in figure (a). A is then charged to a potential difference of 110V. Calculate the capacitance of A and the energy stored in it. The battery is disconnected and then the dielectric slab is moved from A. Find the work done by the external agency in removing the slab from A. (iii) The same dielectric slab is now placed inside B, filling it completely, The two capacitors A and B are then connected as shown in figure(c). Calculate the energy stored in the system. |
|
Answer» Correct Answer - (a) `C_(A)=2.0 xx 10^(-9)F,U_(A)=1.21 xx 10^(-5)J` (b) `W=4.84 xx 10^(-5) J` ( c) `U=1.1 xx 10^(-5)J`. (a) `C_(A)=2.0 xx 10^(-9)F,U_(A)=1.21 xx 10^(-5)J` (b) `W=4.84 xx10^(-5)J (c) `U=1.1 xx10^(-5)J`. |
|
| 150. |
In the fig. shown the circuit is in steady state. If the battery is disconnected give the capacitor charge as a function of time.A. `Q=200 e^(-3t//10)`B. `Q=200e^(-3t//1000)`C. `Q=200 e^(-3t)`D. `Q = 100 e^(-3000 t)` |
|
Answer» Correct Answer - B `q =q_(0) e^(-t//r) q_() = 200 muc tau = RC` `R` is effective resistance across capacitor. |
|