1 Crore+ students have signed up on EduRev. Have you? 
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 T_{C}(0). An interesting property of superconductors is that their critical temperature becomes smaller than T_{C} (0) if they are placed in a magnetic field, i.e., the critical temperature T_{C} (B) is a function of the magnetic field strength B. The dependence of T_{C} (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 B_{1} (solid line) and B_{2} (dashed line). If B_{2} is larger than B_{1} which of the following graphs shows the correct variation of R with T in these fields?
From the given graph it is clear that with increase of the magnitude of magnetic field (B), the critical temperature T_{C} (B) decreases.
Given B_{2} > B_{1}. Therefore for B_{2}, the temperature at which the resistance becomes zero should be less. The above statement is true for graph (a).
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 T_{C}(0). An interesting property of superconductors is that their critical temperature becomes smaller than T_{C} (0) if they are placed in a magnetic field, i.e., the critical temperature T_{C} (B) is a function of the magnetic field strength B. The dependence of T_{C} (B) on B is shown in the figure.
Q.2.A superconductor has T_{C} (0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its T_{C} decreases to 75 K. For this material one can definitely say that when
We know that as B increases, T_{C }decreases but the exact dependence is not known.
Given at B = 0, T_{C} = 100 K and at B = 7.5T, T_{C} = 75 K ∴ At B = 5T, T_{C} should be between 75 K and 100 K.
STATEMENT1 : 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.
STATEMENT2 : Resistance of a metal increases with increase in temperature.
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.
If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a
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.
A wire when connected to 220 V mains supply has power dissipation P_{1}. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is P_{2}. Then P_{2} : P_{1} is
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
If a current is passed through a spring then the spring will
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.
If in the circuit, power dissipation is 150 W, then R is
The equivalent resistance is R_{eq } =
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).
According to Faraday's first law of electrolysis m = ZIt ⇒ m∝ It
If θ_{i} , is the inversion temperature, θ_{n} is the neutral temperature, θ_{c} is the temperature of the cold junction, then
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.
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.
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
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.
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
According to Faraday’s first law of electrolysis m = Z × q For same q,
m ∝ Z
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
i_{g} × G = (i – i_{g}) S
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
A 220 volt, 1000 watt bulb is connected across a 110 volt mains supply . The power consumed will be
When this bulb is connected to 110 volt mains supply we get
The total current supplied to the circuit by the battery is
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
Arithmetic mean > Geometric mean Minimum value of n is 4
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
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
In the first case
In the second case
The termistors are usually made of
Thermistors are usually made of metaloxides with high temperature coefficient of resistivity.
Time taken by a 836 W heater to heat one litre of water from 10°C to 40°C is
= 1 × 4180 × (40 – 10) = 4180 × 30
(∴ ΔQ = heat supplied in time t for heating 1L water from 10°C to 40°C)
The thermo emf of a thermocouple varies with the temperature q of the hot junction as E = aθ+bθ^{2} in volts where the ratio a/b is 700°C. If the cold junction is kept at 0°C, then the neutral temperature is
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.
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
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
Two thin, long, parallel wires, separated by a distance ‘d’ carry a current of ‘i’ A in the same direction. They will
(attractive as current is in the same direction)
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
Resistance of half the coil = R/2
∴ As R reduces to half, ‘H’ will be doubled.
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 
∴R = 100Ω
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10divisions 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 
KEY CONCEPT : Resistance of Galvanometer,
Here i_{g} = Full scale deflection current = 150/10 = 15 mA
V = voltage to be measured = 150 volts (such that each division reads 1 volt)
Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are R_{1}and R_{2} (R_{1} > R_{1}). If the potential difference across the source having internal resistance R_{2} is zero, then
Potential difference across second cell = V = ε – IR_{2} = 0
R + R_{1} + R_{ 2}  2R _{2} = 0
R + R_{1}  R _{2} = 0 ∴ R = R _{2}  R_{1}
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 Z_{1} and Z_{2} respectively the charge which flows through the silver voltameter is
Mass deposited
From equations (i) and (iii),
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
The internal resistance of the cell,
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 ?
An energy source will supply a constant current into the load if its internal resistance is
Internal resistance (r) is
constant.
The Kirchhoff's first law (∑i = 0) and second law (∑iR = ∑E), where the symbols have their usual meanings, are respectively based on
NOTE : Kirchhoff's first law is based on conservation of charge and Kirchhoff's second law is based on conservation of energy.
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 l_{B}/l_{A} of their respective lengths must be
ρB = 2ρ_{A}
d_{B} = 2d_{A}
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
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).
The current I drawn from the 5 volt source will be
The network of resistors is a balanced wh eatstone bridge. The equivalent circuit is
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
R^{1} = R_{0 }[1 + α × 100] = 100 ....(1)
R_{2 }= R_{0} [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 = R_{0}e^{αΔT}
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 S_{1 }and S_{2} connected in parallel. The condition for the bridge to be balanced will be
An electric bulb is rated 220 volt  100 watt. The power consumed by it when operated on 110 volt will be
The resistance of the bulb is
The power consumed when operated at 110 V is
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
Required ratio
where C = Capacitance of capacitor V = Potential difference,e = emf of battery
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
KEY CONCEPT : We know that R_{t }= R_{0} (1 + αt ), where R_{t} is the resistance of the wire at t ºC, R_{0} is the resistance of the wire at 0ºC and α is the temperature coefficient of resistance.
Dividing (iii) by (iv), we get
or, 6 – R_{0} = 10 – 2 R_{0} or, R_{0} = 4Ω .
Shown in the figure below is a meterbridge set up with null deflection in the galvanometer.
The value of the unknown resistor R is
According to the condition of balancing
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
Let j be the current density.
On applying superposition as mentioned we get
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
As shown above E=
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
Applying kirchoff's loop law in AB P_{2}P_{1}A we get 2i + 5  10i_{1}=0 ....(i)
Again applying kirchoff's loop law in P_{2 }CDP_{1}P_{2} we get, 10 i_{1} + 2 – i + i_{1}= 0.....(ii)
Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t_{1} is the time taken for the energy stored in the capacitor to reduce to half its initial value and t_{2} is the time taken for the charge to reduce to onefourth its initial value. Then the ratio t_{1}/ t_{2 }will be
Initial energy of capacitor,
Final energy of capacitor,
∴t_{1} = time for the charge to reduce to 1/√2 of its initial value and t_{2} = time for the charge to reduce to 1/4 of its initial value
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
In Series, R = R_{1}+R_{2}
If a wire is stretched to make it 0.1% longer, its resistance will :
Resistance of wire
∴Fractional change in resistance
∴ Resistance will increase by 0.2%
Two electric bulbs marked 25W – 220V and 100W – 220V are connected in series to a 440V supply. Which of the bulbs will fuse?
The current upto which bulb of marked 25W 220V, will not fuse
The current flowing through the circuit
Thus the bulb marked 25W220 will fuse.
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?
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 V_{1} – V_{2} = 117.073 – 106.66 = 10.04 Volt (approximately)
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.
StatementI : Higher the range, greater is the resistance of ammeter.
StatementII : To increase the range of ammeter, additional shunt needs to be used across it.
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.
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:
Total power consumed by electrical appliances in thebuilding, P_{total} = 2500W
Watt = Volt × ampere
⇒ 2500 = V × I
⇒2500 = 220 I
(Minimum capacity of main fuse)
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 :
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
In the circuit shown, the current in the 1Ω resistor is :
From KVL – 6 + 3I_{1} + 1 (I_{i} – I_{2}) = 0
6 = 3 I_{1} + I_{1} – I_{2}
4I_{1} – I_{2} = 6 ...(1)
– 9 + 2I_{2} – (I_{1} – I_{2}) + 3I_{2} = 0
– I_{1} + 6I_{2} = 9 ...(2)
On solving (1) and (2)
I_{1} = 0.13A
Direction Q to P, since I_{1} > I_{2}.
Considering potential at P as 0V and at Q as x volt, then
The temperature dependence of resistances of Cu and undoped Si in the temperature range 300400 K, is best described by :
258 videos633 docs256 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
258 videos633 docs256 tests








