415 + Interview Questions in ELECTROMAGNETISM IN GENERAL AWARENESS Page 8 InterviewSolution

351.	A wire is bent to form the double loop shown in figure. There is a uniform magnetic field directed into the plane of the loop. If the magnitude of this field is decreasing current will flow from: A. `A` to `B` and `C` to `D`B. `B` to `A` and `D` to `C`C. `A` to `B` and `D` to `C`D. `B` to `A` and `C` to `D`
Answer» Correct Answer - C

Discussion

352.	Two coils of inductances L1 and L2 are linked such that their mutual inductance is M.(a) M = L1 + L2 (b) M = 1/2 (L1 + L2)(c) The maximum value of M is (L1 + L2).(d) The maximum value of M is √(L1L2) .
Answer» Correct Answer is:(d) The maximum value of M is √(L₁L₂) .

Discussion

353.	In which way does the wire move which is placed perpendicular to the magnetic field and the direction of current is Shown in figure? [The direction of the magnetic field is into the page.i.e x]
Answer» Correct option is (B

Discussion

354.	A current-carrying ring is placed in a magnetic field. The direction of the field is perpendicular to the plane of the ring (i) There is no net force on the ring (ii) The ring will tend to expand (iii) The ring will tend to contract (iv) Either (ii) or (iii) depending on the directions of the current in the ring and the magnetic fieldA. There is no net forces on the rigB. The ring will tend to expand.C. The ring will tend to contractD. Either (b) or (c) depending on direction of the current in the ring and the magnetic field
Answer» Correct Answer - A::D

Discussion

355.	Why does the needle get deflected by the magnet?
Answer» Because of strength of the magnetic field of the magnet, the needle gets deflected since it is in the field

Discussion

356.	A current carrying wire of length L is placed perpendicular to a uniform magnetic field B. Then the force acting on the wire with current is ……………………..A) 0 B) ILB C) 2ILB D) ILB/2
Answer» Correct option is B) ILB

Discussion

357.	The magnetic force on a current carrying wire which is placed along a magnetic field is ………….. A) F = BILB) F = BIL sinθ C) F = BIL cosθD) zero
Answer» Correct option is D) zero

Discussion

358.	The magnetic force on a current carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field, is A) 0 B) ILB C) 2ILB D) ILB/2
Answer» Correct option is: B) ILB

Discussion

359.	Can you determine the magnetic force on a current carrying wire which is placed along a magnetic field?
Answer» F = BIl sin θ. If the current carrying wire is placed along direction field θ = 0. ∴ F = 0

359.

Can you determine the magnetic force on a current carrying wire which is placed along a magnetic field?

Answer»

F = BIl sin θ. If the current carrying wire is placed along direction field θ = 0.

∴ F = 0

Discussion

360.	Statement X : The picture on the TV screen is distorted when you bring a bar magnet near the TV screen as the magnetic field exerts a force on the moving charges.Statement Y : When you bring a bar magnet away from the TV screen you will get a clear picture on the screen because the force exerted by the magnet on the charges disapp-ear. A) Both statements are true B) Both statements are false C) X is true, Y is false D) X is false, Y is true
Answer» A) Both statements are true

Discussion

361.	Is the motion of electrons reaching the screen affected by the magnetic field of the bar magnet?
Answer» Yes. The motion of electrons reaching the screen is affected by the magnetic field of the bar magnet.

Discussion

362.	Rajkumar said to you that the magnetic field lines are open and they start at north pole of bar magnet and end at south pole. What questions do you ask Rajkumar to correct him by saying “field lines are closed”?
Answer» If the magnetic field lines start at north pole and end at south pole, where do the lines go from south pole? What is happening within the bar magnet? Are the magnetic field lines passing through bar magnet? What is the direction of magnetic field lines inside the bar magnet? (Recall the solenoid activity). Can you say now, that the magnetic field lines are open?

Discussion

363.	In the following diagram an arrow shows the motion of the bar magnet. (1) State in which direction the current flows A to B or B to A? (2) Name the law used to come to the conclusion
Answer» 1. Current flows from A and B as N-pole is formed at end B and S-pole is is formed at end A 2. Name of the law is Lenz’s law i.e such polarity is formed at coil which opposes the cause which produces

363.

In the following diagram an arrow shows the motion of the bar magnet. (1) State in which direction the current flows A to B or B to A? (2) Name the law used to come to the conclusion

Answer»

1. Current flows from A and B as N-pole is formed at end B and S-pole is is formed at end A

2. Name of the law is Lenz’s law i.e such polarity is formed at coil which opposes the cause which produces

Discussion

364.	A coil ABCD mounted on an axle is placed between the poles N and S of permanent magnet as shown in figure.a) In which direction will the coil begin to rotate when the current is passed through the coil in direction ABCD by connecting a battery at the ends A and D of the coil. b) Why is commutator necessary for continuous rotation of the coil?
Answer» a) Anti-clockwise direction. b) Commutator is needed to change the direction of current in the coil after each half rotation of coil.

364.

A coil ABCD mounted on an axle is placed between the poles N and S of permanent magnet as shown in figure.a) In which direction will the coil begin to rotate when the current is passed through the coil in direction ABCD by connecting a battery at the ends A and D of the coil. b) Why is commutator necessary for continuous rotation of the coil?

Answer»

a) Anti-clockwise direction.

b) Commutator is needed to change the direction of current in the coil after each half rotation of coil.

Discussion

365.	What are the use of cartridge fuses?
Answer» To protect electric appliances like television sets, computers, which are highly expensive, cartridge fuses are used.

Discussion

366.	Three capacitors A , B and C are connected to a battery of 25volt as shown in the figure. The ratio of charges on capacitors A A. `5 : 2 : 3`B. `5 : 3 : 2`C. `2 : 5: 3`D. `2 : 3 : 5`
Answer» Correct Answer - A

Discussion

367.	A circuit is shown in the figure below. Find out the charge of the condenser having capacity `5muF` A. 4.5 micro coulombB. 6.0 micro coulombC. 9.0 micro coulombD. 30 micro coulomb
Answer» Correct Answer - C

Discussion

368.	Statement I : A capacitor blocks d.c. Statement II : This is because capacitative reactance of condenser is `X_(C) = (1)/(omegaC) = 1/(2pifC)` and d.c. `f = 0`.A. If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement– I.B. If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement – I.C. If Statement - I is true but Statement - II is false.D. If Statement - I is false but Statement - II is true.
Answer» Correct Answer - A

Discussion

369.	A conductor PQ, with Vector PQ = Vector r, moves with a velocity Vector v in a uniform magnetic field of induction Vector B. The emf induced in the rod is(a) (Vector v X Vector B). Vector r(b) Vector v.(Vector r X Vector B)(c) Vector B. (Vector r X Vector v)(d) \|Vector r X (Vector v X Vector V)\|
Answer» Correct Answer is: (a) (Vector v X Vector B). Vector r Consider a unit charge inside the conductor. As the conductor moves with a velocity Vector v, the charge inside it will experience the force Vector F = Vector v X Vector B. If this unit charge is now moved from one end of the conductor to the other, its displacement is Vector r. Hence, the work done on it is Vector F. Vector r = (Vector v X Vector B). Vector r. This, by definition, is the emf across the conductor.

369.

A conductor PQ, with Vector PQ = Vector r, moves with a velocity Vector v in a uniform magnetic field of induction Vector B. The emf induced in the rod is(a) (Vector v X Vector B). Vector r(b) Vector v.(Vector r X Vector B)(c) Vector B. (Vector r X Vector v)(d) |Vector r X (Vector v X Vector V)|

Answer»

Correct Answer is: (a) (Vector v X Vector B). Vector r

Consider a unit charge inside the conductor. As the conductor moves with a velocity Vector v, the charge inside it will experience the force Vector F = Vector v X Vector B. If this unit charge is now moved from one end of the conductor to the other, its displacement is Vector r. Hence, the work done on it is Vector F. Vector r = (Vector v X Vector B). Vector r. This, by definition, is the emf across the conductor.

Discussion

370.	An electric dipole is placed along the x-axis at the origin `O.A` point `P` is at a distance of `20 cm` from this origin such that `OP` makes an angle `(pi)/(3)` with the `x`-axis. If the electric field at `P` makes an angle `theta` with x-axis, the value of `theta` would beA. `(pi)/(3)`B. `(pi)/(3) + tan^(-1)((sqrt(3))/2)`C. `(2pi)/(3)`D. `tan^(-1)((sqrt(3))/(2))`
Answer» Correct Answer - B

Discussion

371.	The sides of rectangular block are 2 cm, 3 cm ad 4 cm. The ratio of the maximum to minimum resistance between its parallel faces isA. `4`B. `3`C. `2`D. `1`
Answer» Correct Answer - B

Discussion

372.	A hydrogen atom is a paramagnetic. A hydrogen molecule is –A. DiamagneticB. ParamagneticC. FerromagneticD. None of these
Answer» Correct Answer - A

Discussion

373.	A line joining places of zero declination is calledA. isogonic linesB. isoclinic linesC. agonic linesD. aclinic lines
Answer» Correct Answer - C

Discussion

374.	What is an a.c. generator or dynamo used for?
Answer» An a.c. generator or dynamo is used to produce electric city by mutual induction i.e. rotating a coil in magnetic field.

Discussion

375.	What is Ampere’s swimming rule?
Answer» Ampere’s swimming rule: Imagine a swimmer along the length of the conductor in which current flows. Then the left-hand motion of the person gives the direction of deflection of the north pole of the magnetic needle placed near to it.

Discussion

376.	What is Electromagnet?
Answer» Electromagnet: An electromagnet is a temporary strong magnet and is just a solenoid with its winding on a soft iron core.

Discussion

377.	Explain the Flemmings left-hand rule.
Answer» Flemmings left-hand rule: When a current carrying wire is placed in a magnetic field, a force is exerted on the wire. According to Flemming’s rule, hold the forefinger, middle finger, and thumb of the left hand at right angles to one another, then the forefinger points in the direction of force due to the magnetic field. Middle finger points in the direction of current and the thumb points in the direction of motion of the conductor.

377.

Explain the Flemmings left-hand rule.

Answer»

Flemmings left-hand rule:

When a current carrying wire is placed in a magnetic field, a force is exerted on the wire. According to Flemming’s rule, hold the forefinger, middle finger, and thumb of the left hand at right angles to one another, then the forefinger points in the direction of force due to the magnetic field. Middle finger points in the direction of current and the thumb points in the direction of motion of the conductor.

Discussion

378.	Define the Permanent magnet.
Answer» Permanent magnet: A permanent magnet is made from steel. As steel has more retentivity than iron, it does not lose its magnetism easily.

Discussion

379.	Magnetic field produced by electrons in aotms and molecules is due to their -A. spin motion onlyB. orbital motion onlyC. spin and orbital motion bothD. neither spin nor orbital motion.
Answer» Correct Answer - C

Discussion

380.	Torques `tau_(1)` and `tau_(2)` are required for a magnetic needle to remain perpendicular to the magnetic fields at two different places. The magnetic field at those places are B1 and B2 respectively, then `(B_(1))/(B_(2))` isA. `(tau_(1))/(tau_(2))`B. `(tau_(2))/(tau_(1))`C. `(tau_(1) + tau_(2))/(tau_(1) - tau_(2))`D. `(tau_(1) - tau_(2))/(tau_(1) + tau_(2))`
Answer» Correct Answer - A

Discussion

381.	Two identical parallel plate air capacitors are connected in series to a battery of emf V. If one of the capacitor is completely filled with dielectric material of constant K, then potential difference of the other capacitor will becomeA. `(K + 1)/(K)V_(0)`B. `(K)/(K + 1)V_(0)`C. `(K + 1)/(2K)V_(0)`D. `(2K)/(K + 1)V_(0)`
Answer» Correct Answer - B

Discussion

382.	A battery charges a parallel plate capacitor of thickness `(d)` so that an energy `[U_(0)]` is stored in the system. A slab of dielectric constant `(K)` and thickness `(d)` is then introduced between the plates of the capacitor. The new energy of the system is given byA. `KU_(0)`B. `K^(2)U_(0)`C. `(U_(0))/(K)`D. `U_(0)//K^(2)`
Answer» Correct Answer - A

Discussion

383.	Four condenser are joined as shown in the adjoining figure. The capacity of each is `8 muF`. The equivalent capacity between the points `A` and `B` will be A. `32 muf`B. `2muf`C. `8muf`D. `16muf`
Answer» Correct Answer - A

Discussion

384.	A capacitor of `2 muF` is charged to its maximum emf of `2V` and is discharged through a resistance of `10^(4)omega` . Current in the circuit after `0.02 s` will be-A. `10^(–4)A`B. `1.4 xx 10^(5) A`C. `7.4 xx 10^(-5)A`D. `3.7 xx 10^(-5)A`
Answer» Correct Answer - C

Discussion

385.	In the circuit shown here `C_(1) = 6muF, C2 = 3muF` and battery `B = 20V`. The Switch `S_(1)` is first closed. It is then opened and afterwards `S_(2)` is closed. What is the charge finally on `C_(2)` A. `120 muC`B. `80muC`C. `40muC`D. `20muC`
Answer» Correct Answer - C

Discussion

386.	Name two safety measures commonly used in electric circuits and appliances.
Answer» Electric fuse and Miniature Circuit Breaker (MCB).

Discussion

387.	A metal wire of mass m slides without friction on two horizontal rails spaced distance d-apart as shown in figure. The rails are situated in a uniform magnetic field B, directed vertically upwards, and a battery is sending a current I through them. Find the velocity of the wire as a function of time, assuming it to be at rest initially. A. `(Bid)/(m) t`B. `0`C. `Bidmat`D. none
Answer» Correct Answer - A

Discussion

388.	An electrical heater and a capacitor are joined in series across a `220V, 50Hz, AC` supply. The potential difference across the heater is `90V`. The potential difference across the capacitor will be about:A. `200 V`B. `130 V`C. `110V`D. `90V`
Answer» Correct Answer - A

Discussion

389.	An electric lamp designed for operation on `10V AC` is connected to a `220V AC` supply, through a choke coil of inductance `2H` for proper operation. The angular frequency of the `AC` is `100sqrt(10)rad//s`. If a capacitor is to be used in place of the choke coil its cpacitance must beA. `1 muF`B. `2 muF`C. `5 muF`D. `10 muF`
Answer» Correct Answer - C

Discussion

390.	A charged capacitor discharges through a resistance `R` with time constant `tau`. The two are now placed in series across an `AC` source of angular frequency `omega =1/(tau)` . The impedance of the circuit will beA. `R/(sqrt(2))`B. `R`C. `sqrt(2R)`D. `2R`
Answer» Correct Answer - C

Discussion

391.	Two coils, A and B, are linked such that emf ℰ is induced in B when the current in A is changing at the rate I. If i current is now made to flow in B, the flux linked with A will be (a) (ℰ/I)i (b) ℰi I (c) (ℰ I)i (d) iI/ℰ
Answer» Correct Answer is: (a) (ℰ/I)i Let M = mutual inductance between A and B. e_B = M (di_A/dt) or ℰ = MI or M = ℰ / I ϕ_A = Mi_B= (ℰ / I) i.

391.

Two coils, A and B, are linked such that emf ℰ is induced in B when the current in A is changing at the rate I. If i current is now made to flow in B, the flux linked with A will be (a) (ℰ/I)i (b) ℰi I (c) (ℰ I)i (d) iI/ℰ

Answer»

Correct Answer is: (a) (ℰ/I)i

Let M = mutual inductance between A and B.

e_B = M (di_A/dt)

or ℰ = MI

or M = ℰ / I

ϕ_A = Mi_B= (ℰ / I) i.

Discussion

392.	A charged capacitor discharges through a resistance R with time constant τ. The two are now placed in series across an AC source of angular frequency ω = 1/τ The impedance of the circuit will be(a) R/√2(b) R(c) √2R(d) 2R
Answer» Correct Answer is: (c) √2R τ = RC ∴ ω = 1/RC Z = √(R² + (1/ωC)² = √(R² + R²) = √2R

392.

A charged capacitor discharges through a resistance R with time constant τ. The two are now placed in series across an AC source of angular frequency ω = 1/τ The impedance of the circuit will be(a) R/√2(b) R(c) √2R(d) 2R

Answer»

Correct Answer is: (c) √2R

τ = RC

∴ ω = 1/RC

Z = √(R² + (1/ωC)² = √(R² + R²) = √2R

Discussion

393.	In the circuit shown, the symbols have their usual meanings. The cell has emf `varepsilon.X` is initially joined to `Y` for a long time. Then `X` is joined to `Z`. The maximum charge on `C` at any later time will beA. `(varepsilon)/(Rsqrt(LC))`B. `(varepsilonR)/(2sqrt(LC))`C. `(varepsilonsqrt(LC))/(2R)`D. `(varepsilonsqrt(LC))/(R)`
Answer» Correct Answer - D

Discussion

394.	At t = 0, an inductor of zero resistance is joined to a cell of emf ℰ through a resistance. The current increases with a time constant τ. The emf across the coil after time t is(a) ℰ t/τ.(b) ℰ (1 - e -t/τ)(c) ℰ e -t/τ(d) ℰ e 2-t/τ
Answer» Correct Answer is: (c) ℰ e ^-t/τ i = i₀(1 - e ^-t/τ) di/dt = -i₀ (-1/τ) e ^-t/τ = ℰ /R. R/L. e^-t/τ or L di/dt = ℰ e ^{-t/τ =} emf across the coil.

394.

At t = 0, an inductor of zero resistance is joined to a cell of emf ℰ through a resistance. The current increases with a time constant τ. The emf across the coil after time t is(a) ℰ t/τ.(b) ℰ (1 - e -t/τ)(c) ℰ e -t/τ(d) ℰ e 2-t/τ

Answer»

Correct Answer is: (c) ℰ e ^-t/τ

i = i₀(1 - e ^-t/τ)

di/dt = -i₀ (-1/τ) e ^-t/τ = ℰ /R. R/L. e^-t/τ

or L di/dt = ℰ e ^{-t/τ =} emf across the coil.

Discussion

395.	Which of the following statements is, are correct? A) A long wire wound in a closely packed helix is called solenoid.B) The magnetic field lines set up by solenoid resemble those of a bar magnet. C) One end of the solenoid behaves like a north pole and other behaves like a south pole when electric current flows through it.D) All of the above
Answer» D) All of the above

Discussion

396.	Which of the following is true regarding solenoid ? (A) strength of the magnetic field is independent of the material on which the coil is wound. (B) when the coil is wound around a soft iron core, it behaves like a temporary magnet.A. Only AB. Only BC. Both A and BD. None of the above
Answer» Correct Answer - B

Discussion

397.	In the circuit shown, X is joined to Y for a long time, and then X is joined to Z. The total heat produced in R2 is(a) Lℰ2 / 2R12(b) Lℰ2 / 2R22(c) Lℰ2 / 2R1R2(d) Lℰ2R2 / 2R13
Answer» Correct Answer is: (a) Lℰ²/2R₁² Steady-state current in L = i₀ = ℰ/R₁. Energy stored in L = 1/2 Li₀² = 1/2 L(ℰ²/R₁²) = heat produced in R₂ during discharge.

397.

In the circuit shown, X is joined to Y for a long time, and then X is joined to Z. The total heat produced in R2 is(a) Lℰ2 / 2R12(b) Lℰ2 / 2R22(c) Lℰ2 / 2R1R2(d) Lℰ2R2 / 2R13

Answer»

Correct Answer is: (a) Lℰ²/2R₁²

Steady-state current in L = i₀ = ℰ/R₁.

Energy stored in L = 1/2 Li₀² = 1/2 L(ℰ²/R₁²) = heat produced in R₂ during discharge.

Discussion

398.	The wire loop `PQRSP` formed by joining two semicircular wires of radii ` R_(1)` and `R_(2)` carries a current `I` as shown . The magnitude of the magnetic induction at the center `C` is ……….. A. `(mu_(0)I)/2(1/(R_(1))+1/(R_(2)))`B. `(mu_(0)I)/4(1/(R_(1))+1/(R_(2)))`C. `(mu_(0)I)/2(1/(R_(1))-1/(R_(2)))`D. `(mu_(0)I)/4(1/(R_(1))-1/(R_(2)))`
Answer» Correct Answer - D

Discussion

399.	When a coil is joined to a cell, current in it grows with a time constant `tau`. The coil is disconnected from the cell before the current has reached its steady-state value, and, it is then short-circuited. The current will now decrease with a time constantA. `tau`B. `gt tau`C. `lt tau`D. either (b) or (c) depending on the instant at which it was disconnected from the cell
Answer» Correct Answer - A

Discussion

400.	At `t=0`, an inductor of zero resistance is joined to a cell of emf `varepsilon` through a resistance. The current increases with a time constant `tau`. The emf across the coil after time `t` isA. `varepsilont//tau`B. `varepsilon(1-e^(-t/tau))`C. `varepsilone^(-t//tau)`D. `varepsilone^(-2//tau)`
Answer» Correct Answer - C

Discussion

Explore topic-wise InterviewSolutions in .

Two coils of inductances L1 and L2 are linked such that their mutual inductance is M.(a) M = L1 + L2 (b) M = 1/2 (L1 + L2)(c) The maximum value of M is (L1 + L2).(d) The maximum value of M is √(L1L2) .

In which way does the wire move which is placed perpendicular to the magnetic field and the direction of current is Shown in figure? [The direction of the magnetic field is into the page.i.e x]

Why does the needle get deflected by the magnet?

A current carrying wire of length L is placed perpendicular to a uniform magnetic field B. Then the force acting on the wire with current is ……………………..A) 0 B) ILB C) 2ILB D) ILB/2

The magnetic force on a current carrying wire which is placed along a magnetic field is ………….. A) F = BILB) F = BIL sinθ C) F = BIL cosθD) zero

The magnetic force on a current carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field, is A) 0 B) ILB C) 2ILB D) ILB/2

Can you determine the magnetic force on a current carrying wire which is placed along a magnetic field?

Is the motion of electrons reaching the screen affected by the magnetic field of the bar magnet?

Rajkumar said to you that the magnetic field lines are open and they start at north pole of bar magnet and end at south pole. What questions do you ask Rajkumar to correct him by saying “field lines are closed”?

In the following diagram an arrow shows the motion of the bar magnet. (1) State in which direction the current flows A to B or B to A? (2) Name the law used to come to the conclusion

What are the use of cartridge fuses?

Three capacitors A , B and C are connected to a battery of 25volt as shown in the figure. The ratio of charges on capacitors A A. `5 : 2 : 3`B. `5 : 3 : 2`C. `2 : 5: 3`D. `2 : 3 : 5`

A circuit is shown in the figure below. Find out the charge of the condenser having capacity `5muF` A. 4.5 micro coulombB. 6.0 micro coulombC. 9.0 micro coulombD. 30 micro coulomb

A conductor PQ, with Vector PQ = Vector r, moves with a velocity Vector v in a uniform magnetic field of induction Vector B. The emf induced in the rod is(a) (Vector v X Vector B). Vector r(b) Vector v.(Vector r X Vector B)(c) Vector B. (Vector r X Vector v)(d) |Vector r X (Vector v X Vector V)|

The sides of rectangular block are 2 cm, 3 cm ad 4 cm. The ratio of the maximum to minimum resistance between its parallel faces isA. `4`B. `3`C. `2`D. `1`

A hydrogen atom is a paramagnetic. A hydrogen molecule is –A. DiamagneticB. ParamagneticC. FerromagneticD. None of these

A line joining places of zero declination is calledA. isogonic linesB. isoclinic linesC. agonic linesD. aclinic lines

What is an a.c. generator or dynamo used for?

What is Ampere’s swimming rule?

What is Electromagnet?

Explain the Flemmings left-hand rule.

Define the Permanent magnet.

Magnetic field produced by electrons in aotms and molecules is due to their -A. spin motion onlyB. orbital motion onlyC. spin and orbital motion bothD. neither spin nor orbital motion.

Four condenser are joined as shown in the adjoining figure. The capacity of each is `8 muF`. The equivalent capacity between the points `A` and `B` will be A. `32 muf`B. `2muf`C. `8muf`D. `16muf`

A capacitor of `2 muF` is charged to its maximum emf of `2V` and is discharged through a resistance of `10^(4)omega` . Current in the circuit after `0.02 s` will be-A. `10^(–4)A`B. `1.4 xx 10^(5) A`C. `7.4 xx 10^(-5)A`D. `3.7 xx 10^(-5)A`

In the circuit shown here `C_(1) = 6muF, C2 = 3muF` and battery `B = 20V`. The Switch `S_(1)` is first closed. It is then opened and afterwards `S_(2)` is closed. What is the charge finally on `C_(2)` A. `120 muC`B. `80muC`C. `40muC`D. `20muC`

Name two safety measures commonly used in electric circuits and appliances.

An electrical heater and a capacitor are joined in series across a `220V, 50Hz, AC` supply. The potential difference across the heater is `90V`. The potential difference across the capacitor will be about:A. `200 V`B. `130 V`C. `110V`D. `90V`

A charged capacitor discharges through a resistance `R` with time constant `tau`. The two are now placed in series across an `AC` source of angular frequency `omega =1/(tau)` . The impedance of the circuit will beA. `R/(sqrt(2))`B. `R`C. `sqrt(2R)`D. `2R`

Two coils, A and B, are linked such that emf ℰ is induced in B when the current in A is changing at the rate I. If i current is now made to flow in B, the flux linked with A will be (a) (ℰ/I)i (b) ℰi I (c) (ℰ I)i (d) iI/ℰ

A charged capacitor discharges through a resistance R with time constant τ. The two are now placed in series across an AC source of angular frequency ω = 1/τ The impedance of the circuit will be(a) R/√2(b) R(c) √2R(d) 2R

At t = 0, an inductor of zero resistance is joined to a cell of emf ℰ through a resistance. The current increases with a time constant τ. The emf across the coil after time t is(a) ℰ t/τ.(b) ℰ (1 - e -t/τ)(c) ℰ e -t/τ(d) ℰ e 2-t/τ

Which of the following is true regarding solenoid ? (A) strength of the magnetic field is independent of the material on which the coil is wound. (B) when the coil is wound around a soft iron core, it behaves like a temporary magnet.A. Only AB. Only BC. Both A and BD. None of the above

In the circuit shown, X is joined to Y for a long time, and then X is joined to Z. The total heat produced in R2 is(a) Lℰ2 / 2R12(b) Lℰ2 / 2R22(c) Lℰ2 / 2R1R2(d) Lℰ2R2 / 2R13

At `t=0`, an inductor of zero resistance is joined to a cell of emf `varepsilon` through a resistance. The current increases with a time constant `tau`. The emf across the coil after time `t` isA. `varepsilont//tau`B. `varepsilon(1-e^(-t/tau))`C. `varepsilone^(-t//tau)`D. `varepsilone^(-2//tau)`