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Solution : 

Electric potential at a point is 1 volt means 1 joule of work is done in moving 1 unit positive charge from infinity to that point.

The SI unit of resistance is ohm.
One ohm is the resistance offered by a conductor when the current passing through it is one amPere and the potential difference across its ends is one volt.
By Ohm's law, V = IR
R = V/I = 0.8 V/ 0.2 A =4 ohm

Potential difference of 1 volt means the amount of work done when a unit charge moves from one point to the other point in an electric field.

The resistance of the rheostat will not change at all as it does not depend on the voltage applied across it. 

EARTHING OF THE APPLIANCE : To protect from the electric shock electric appliance is earthed. Sometimes due to break of insulation of wires, live wire comes in contact with the body of the appliance and we get a fatal shock when the appliance is touched. If the appliance is earthed, the current will pass to the earth and we remain protected from the electric shock.

Correct Answer is: (c) Q/2ε₀

Correct Answer is: (b) V (1 - a/b)

Let initial charge on B be Q. Then, V = kQ/b. There is no initial charge on A, as A and B are at the same potential. Let q charge be now given to A. Potential of A now is

k q/a + k Q/b = 0 or q = - Qa/b.

Potential of B now is

k(Q + q/b) = k/b [Q - Qa/b] = k Q/b (1 - a/b) = V (1 - a/b)

Correct Answer is: (c) remain stationary

The field inside a conductor is zero. Hence, the point charge experiences no force. 

(A) Resistance of copper strip decreases because it has positive temperature coefficient of resistance. 

Resistors across AC and DE are in parallel i.e. R_AC and R_DE are in parallel.

So

\(\frac1{R'_p}=\frac1{R_{AC}}+\frac1{R_{DE}}\)

⇒ \(\frac1{R'_p}=\frac1{30}+\frac1{30}\)

⇒ \(\frac1{R'_p}=\frac{1+1}{30}+\frac2{30}\)

⇒ R'_p = \(\frac{30}2\) = 15Ω

Now R’_p and R_BC are I series. 

R’_s = R’_p + R_BC 

⇒ R’_s = 15 + 15 = 30Ω. 

Again, R_AB and R’_s are in parallel.

\(\frac{1}{R^"_p}=\frac{1}{R_{AB}}+\frac{1}{R'_s}\)

\(\frac{1}{R^"_p}=\frac{1}{15}+\frac{1}{30}\)

⇒ \(\frac{1}{R^"_p}=\frac{1+2}{30}+\frac{3}{30}\)

⇒ R"_p = \(\frac{30}3\) = 10Ω

Equivalent resistance of the above circuit is 10Ω. 

Now, current flowing through the circuit is

l = \(\frac{V}{R} =\frac3{10}=0.3\)

The value of current is 0.3A

Shushant told his teacher that ammeter was broke by him he did not hide from his teacher this shows that Shushant is very honest and respectful towards his teacher. Also, in spite of his group mates telling him not to tell to the teacher he developed enough courage to tell everything to his teacher this show that he is courageous and confident student who accept his mistake.

Correct Answer is: (a, c, & d)

(i) 10^-8 ohm × m (ii) 10^-3 ohm × m (iii) 10¹⁰ ohm × m.

Solution : 

One coulomb of charge is that quantity of charge which exerts a force of 9 × 109 Newton on an equal charge is placed at a distance of 1 m from it.

Solution : 

Coulomb is the unit of electric charge

Correct  Answer is: (a) T

However, the time period of a spring–block system does not depend on g.

False A fuse wire has a high melting point.

Correct Answer is: (d) 30 V

Let Q amount of charge flow through the MN branch.

V = 60 V = Q/2C + Q/C + Q/2C = 2 Q/C

or Q = 30C/C V = 30 V.

Potential difference between M and N = Q/C = 30C/C V = 30 V.

Correct Answer is: (c) 4 V 

Charges on the two capacitors are 36 μC and 72 μC. When they are connected in opposition, 

the total charge on the system = (72 - 36) μC = 36 μC.

This is shared between the capacitors so that they acquire the same potential difference. Let this be V.

(3 μF)V + (6 μF)V = 36 μC or V = 4 V.

Correct Answer is: (a, b, & d)

Correct Answer is: (a) cell is ℰ

Correct Answer is: (a, b, c, & d)

Correct Answer is: (a, d)

Correct Answer is: (a, b, c, & d)

Correct Answer is: (a) F ∝ Q², (b) F ∝ 1/A

Correct Answer is: (a, c)

The electric bulbs are filled with chemically inactive as Nitrogen or Argon. 

As when the current is passed through the bulb, the buld filament made of tungsten glow due to its high resistivity if exposed to air will cause the decomposition of tungsten metal in air which can leds 

Tungsten element to melt thus causing circuit to break. 

Thus for the long life of the bulb the electric bulb is filled with chemically inactive as nitrogen and argon.

True

Due to attraction of the atoms, the free electrons cannot leave the metal.

V = 220 V, P = 100W 

R=? 

We know that P = V₂/R 

Thus 

R = V₂/P = 2202/100 = 484 ohm 

The commercial unit of electric energy is kilowatt-hour (kWh).

Kilowatt-hour is the commercial unit of electric energy.

For maximum effective resistance, the resistors must be connected in series combination. If there are n resistors each of resistance R, then the maximum effective resistance = nR.

For minimum effective resistance, the resistors must be connected in parallel combination. So the minimum effective resistance =R/n

Ratio of the maximum to minimum resistance is (nR)/(R/n)=n²:1.

The resistance of the bulb is defined as R =V²/P So the resistance of 50W bulb is double than the resistance of 100W bulb. When they are connected in series the current through both bulbs is same. Hence 50-watt bulb will be brighter because P=I²R. In parallel, the voltage will be same in both bulbs. So, the 100-watt bulb will be brighter because of P=V²/R.

Charge (Q) = 150 coulomb

Time (t) = 1 minute = 60 sec.

Current (I) = ?

I=Q/t

=150/60=2.5 Amp.