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# Test: 35 Year JEE Previous Year Questions: Current Electricity

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## 55 Questions MCQ Test Physics For JEE | Test: 35 Year JEE Previous Year Questions: Current Electricity

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Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 1

### Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value to zero as their temperature is lowered below a critical temperature TC(0). An interesting property of superconductors is that their critical temperature becomes smaller than TC (0) if they are placed in a magnetic field, i.e., the critical temperature TC (B) is a function of the magnetic field strength B. The dependence of TC (B) on B is shown in the figure. Q.1. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields B1 (solid line) and B2 (dashed line). If B2 is larger than B1 which of the following graphs shows the correct variation of R with T in these fields?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 1

From the given graph it is clear that with increase of the magnitude of magnetic field (B), the critical temperature TC (B) decreases.
Given B2 > B1. Therefore for B2, the temperature at which the resistance becomes zero should be less. The above statement is true for graph (a).

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 2

### Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value to zero as their temperature is lowered below a critical temperature TC(0). An interesting property of superconductors is that their critical temperature becomes smaller than TC (0) if they are placed in a magnetic field, i.e., the critical temperature TC (B) is a function of the magnetic field strength B. The dependence of TC (B) on B is shown in the figure. Q.2.A superconductor has TC (0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its TC decreases to 75 K. For this material one can definitely say that when

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 2

We know that as B increases, TC decreases but the exact dependence is not known.
Given at B = 0, TC = 100 K and at B = 7.5T, TC  = 75 K ∴ At B = 5T, TC  should be between 75 K and 100 K.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 3

### STATEMENT-1 :  In a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance. STATEMENT-2 : Resistance of a metal increases with increase in temperature.

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 3

When the temperature of metal increases; its resistance increases.
Therefore statement - 2 is correct.

For a meter bridge when null point N is obtained we get

When the unknown resistance is put inside an enclosure, maintained at a high temperature, then X increases. To maintain the ratio of null point l should also increase. But if we want to keep the null point at the initial position (i.e., if we want no change in the value of l) there to maintain the ratio, S should be increased.
Therefore statement - 1 is false.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 4

If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 4

KEY CONCEPT : To convert a galvanometer into a voltmeter we connect a high resistance in series with the galvanometer.
The same procedure needs to be done if ammeter is to be used as a voltmeter.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 5

A wire when connected to 220 V mains supply has power dissipation P1. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is P2. Then P2 : P1 is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 5

Case 2  : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is R/2. These are connected in parallel

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 6

If a current is passed through a spring then the spring will

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 6

When current is passed through a spring then current flows parallel in the adjacent turns.
NOTE : When two wires are placed parallel to each other and current flows in the same direction, the wires attract each other.
Similarly here the various turns attract each other and the spring will compress.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 7

If in the circuit, power dissipation is 150 W, then R is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 7

The equivalent resistance is Req  =

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 8

The mass of product liberated on anode in an electrochemical cell depends on (where t is the time period for which the current is passed).

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 8

According to Faraday's first law of electrolysis m = ZIt ⇒ m∝ It

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 9

If θi , is the inversion temperature, θn is the neutral temperature, θc is the temperature of the cold junction, then

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 9

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 10

The length of a wire  of a potentiometer is 100 cm, and the  e. m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal  resistance is 0.5Ω. If the balance point is obtained at l = 30 cm from the positive end, the e.m.f. of the battery is

where i is the current in the potentiometer wire.

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 10

rom the principle of potentiometer, V ∝ l

V = emf of battery, E = emf of standard cell.

L = length of potentiometer wire

In this arrangement, the internal resistance of the battery E does not play any role as current is not passing through the battery.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 11

The thermo e.m.f. of a thermo -couple is 25 μV / o C at room temperature. A galvanometer of  40 ohm resistance, capable of detecting current  as low as 10 -5 A , is connected with the thermo couple. The smallest temperature difference that can be detected by this system is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 11

Let θ be the smallest temperature difference that can be detected by the thermocouple, then

I × R = (25 × 10–6) θ

where I is the smallest current which can be detected by the galvanometer of resistance R.

∴ 10–5 × 40 = 25 × 10 – 6 × θ

∴ θ = 16°C.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 12

The negative Zn pole of a Daniell cell, sending a constant current through a circuit, decreases in mass by 0.13g  in 30 minutes. If the electeochemical equivalent of  Zn and Cu are 32.5 and 31.5 respectively, the increase in the mass of the positive  Cu pole in this time is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 12

According to Faraday’s first law of electrolysis m =  Z × q For same q,

m ∝ Z

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 13

An ammeter reads upto 1 ampere. Its internal resistance is 0.81ohm. To increase the range to 10 A the value of the required shunt is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 13

ig × G = (i – ig) S

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 14

A 3 volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The  current I, in the circuit will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 14

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 15

A 220 volt, 1000 watt bulb  is connected across a 110 volt mains supply . The power consumed will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 15

When this bulb is connected to 110 volt mains supply we get

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 16

The total current supplied to the circuit by the battery is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 16

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 17

The resistance of the series combination of two resistances is S. when they are joined in parallel the total resistance is P.
If S = nP then the Minimum possible value of n is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 17

Arithmetic mean > Geometric mean Minimum value of n is 4

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 18

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii arein the  ratio of 4/3 and 2/3, then the ratio of the current passing through the wires will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 18

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 19

In a meter bridge experiment null point is obtained at 20 cm. from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4 X against Y

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 19

In the first case

In the second case

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 20

The termistors are usually made of

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 20

Thermistors are usually made of metaloxides with high temperature coefficient of resistivity.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 21

Time taken by a 836 W  heater to heat one litre of water from 10°C to 40°C is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 21

= 1 × 4180 × (40 – 10) = 4180 × 30

(∴ ΔQ = heat supplied in time t for heating  1L water from 10°C to 40°C)

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 22

The thermo emf of a thermocouple varies with the temperature q of the hot junction as E = aθ+bθ2 in volts where the ratio a/b is 700°C. If the cold junction is kept at 0°C, then the neutral temperature is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 22

Neutral temperature is the temperature of  a  hot junction at which E is maximum.

Neutral temperature can never be negative hence no θ is possible.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 23

The electrochemical equivalent of a metal is 3.35109-7 kg per Coulomb. The mass of the metal liberated at the cathode when a 3A current is passed for 2 seconds will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 23

The mass liberated m, electrochemical equivalent of a metal Z,  are related as m = Zit

⇒ m = 3.3*10-7*3*2 = 19.8*10-7 kg

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 24

Two thin, long, parallel wires, separated by a distance ‘d’ carry a current of ‘i’ A in the same direction. They will

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 24

(attractive as current is in the same direction)

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 25

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 25

Resistance of half the coil = R/2

∴ As R reduces to half, ‘H’ will be doubled.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 26

In the circuit , the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be -

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 26

∴R = 100Ω

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 27

A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10-divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be -

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 27

KEY CONCEPT : Resistance of Galvanometer,

Here ig = Full scale deflection current = 150/10 = 15 mA

V = voltage to be measured = 150 volts (such that each division reads 1 volt)

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 28

Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are R1and R2 (R1  > R1). If the potential difference across the source having internal resistance R2 is zero, then

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 28

Potential difference across second cell  = V = ε – IR2 = 0

R + R1 + R 2 - 2R 2 = 0
R + R1 - R 2 = 0   ∴ R = R 2 - R1

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 29

Two voltameters, one of copper and another of silver, are joined in parallel. When a total charge q flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are Z1 and Z2 respectively the charge which flows through the silver voltameter is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 29

Mass deposited

From equations (i) and (iii),

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 30

In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2W, the balancing length becomes 120 cm. The internal resistance of the cell is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 30

The internal resistance of the cell,

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 31

The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100 W and 200 V lamp when not in use ?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 31

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 32

An energy source will supply a constant current into the load if its internal resistance is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 32

Internal resistance (r) is

constant.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 33

The Kirchhoff's first law (∑i = 0) and second law  (∑iR = ∑E), where the symbols have their usual meanings, are respectively based on

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 33

NOTE : Kirchhoff's first law is based on conservation of charge and Kirchhoff's second law is based on conservation of energy.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 34

A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. then for the two wires to have the same resistance, the ratio lB/lA of their respective lengths must be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 34

ρB = 2ρA

dB = 2dA

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 35

A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold, then, an electric current will

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 35

At cold junction, current flows fr om Antimony to Bismuth (because current flows from metal occurring later in the series to metal occurring earlier in the thermoelectric series).

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 36

The current I drawn from the 5 volt source will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 36

The network of resistors is a balanced wh eatstone bridge. The equivalent circuit is

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 37

The resistance of a bulb filmanet is 100Ω at a temperature of 100°C. If its temperature coefficient of resistance be 0.005 per °C, its resistance will become 200 Ω at a temperature of

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 37

R1 = R0 [1 + α × 100] = 100 ....(1)

R2 = R0 [1 +α × T] = 200 ....(2)

On dividing we get

NOTE : We may use this expression as an approximation because the difference in the answers is appreciable.
For accurate results one should use R = R0eαΔT

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 38

In a Wheatstone's bridge, three resistances P, Q an d R connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for the bridge to be balanced will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 38

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 39

An electric bulb is rated 220 volt - 100 watt. The power consumed by it when operated on 110 volt will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 39

The resistance of the bulb is

The power consumed when operated at 110 V is

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 40

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 40

Required ratio

where C = Capacitance of capacitor  V = Potential difference,e = emf of battery

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 41

The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C.The resistance of the wire at 0°C will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 41

KEY CONCEPT : We know that Rt = R0 (1 + αt ), where Rt is the resistance of the wire at t ºC,  R0 is the resistance of the wire at 0ºC and α is the temperature coefficient of resistance.

Dividing (iii) by (iv), we get

or, 6 – R0 = 10 – 2 R0  or,  R0 = 4Ω .

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 42

Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer.

The value of the unknown resistor R is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 42

According to the condition of balancing

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 43

Question No. 40 and 41 are based on the following paragraph.
Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:

(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.

(ii) Calculate field E(r) at distance ‘r’ from A by using  Ohm’s law E =ρj, where j is the current per unit area at ‘r’.

(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.

(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.

Q.40. ΔV measured between B and C is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 43

Let j be the current density.

On applying superposition as mentioned we get

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 44

Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:

(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.

(ii) Calculate field E(r) at distance ‘r’ from A by using  Ohm’s law E =ρj, where j is the current per unit area at ‘r’.

(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.

(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.

Q.41. For current entering at A, the electric field at a distance ‘r’ from A is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 44

As shown above E=

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 45

A 5V battery with internal resistance 2 Ω and a 2V battery with internal resistance 1Ω are  connected to a 10W resistor as shown in the figure.

The current in the  10Ω resistor is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 45

Applying kirchoff's loop law in AB P2P1A we get -2i + 5 - 10i1=0 ....(i)

Again applying kirchoff's loop law in P2 CDP1P2 we get,      10 i1 + 2  – i + i1= 0.....(ii)

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 46

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t1/ t2 will be

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 46

Initial energy of capacitor,

Final energy of capacitor,

∴t1 = time for the charge to reduce to 1/√2 of  its initial value and t2 = time for the charge to reduce to 1/4 of  its initial value

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 47

Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are α1 and α2 . The respective temperature coefficients of their series and parallel combinations are nearly

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 47

In Series, R = R1+R2

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 48

If a wire is stretched to make it 0.1% longer, its resistance will :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 48

Resistance of wire

∴Fractional change in resistance

∴ Resistance will increase by 0.2%

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 49

Two electric bulbs marked 25W – 220V and 100W – 220V are connected in series to a 440V supply. Which of the bulbs will fuse?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 49

The current upto which bulb of marked 25W -220V, will not fuse

The current flowing through the circuit

Thus the bulb marked 25W-220 will fuse.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 50

The supply voltage to room is 120V. The resistance of the lead wires is 6Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 50

Power of bulb = 60 W (given)

Power of heater = 240W (given)

Voltage across bulb before heater is switched on,

Voltage across bulb after heater is switched on,

Hence decrease in voltage V1 – V2 = 117.073 – 106.66 = 10.04 Volt (approximately)

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 51

This questions has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes into two Statements.

Statement-I : Higher the range, greater is the resistance of ammeter.
Statement-II : To increase the range of ammeter, additional shunt needs to be used across it.

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 51

Statements I is false and Statement II is true For ammeter, shunt resistance

Therefore for I to increase, S should decrease, So additional S can be connected across it.

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 52

In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of electric mains is 220 V. The minimum capacity of the main fuse of the building will be:

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 52

Total power consumed by electrical appliances in thebuilding, Ptotal = 2500W

Watt = Volt × ampere

⇒ 2500 = V × I
⇒2500 = 220 I

(Minimum capacity of main fuse)

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 53

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close to :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 53

Here V = potential difference

l = length of wire

n = no. of electrons per unit volume of conductor.

e = no. of electrons

Placing the value of above parameters we get resistivity

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 54

In the circuit shown, the current in the 1Ω resistor is :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 54

From KVL – 6 + 3I1 + 1 (Ii – I2) = 0

6 = 3 I1 + I1 – I2
4I1 – I2 = 6 ...(1)
– 9 + 2I2 – (I1 – I2) + 3I2 = 0
– I1 + 6I2 = 9 ...(2)
On solving (1) and (2)

I1 = 0.13A
Direction Q to P, since I1 > I2.

Considering potential at P as 0V and at Q as x volt, then

Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 55

The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K, is best described by :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Current Electricity - Question 55

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