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#### Question 1:

As a constant potential difference is applied across a bulb, due to Joule's heating effect, the temperature of the bulb increases. As the temperature of the bulb filament increases, its resistance also increases, as resistance R is the function of temperature T. It is given by R = R0(1+αT). With an increase in the value of resistance, the value of current decreases as $i=\frac{V}{R}$. Now, the heat generated by the resistance is constantly radiated to the surroundings. Thus, the value of its temperature is maintained and hence its resistance. As a result, current through the bulb filament becomes constant.

#### Question 2:

As a constant potential difference is applied across a bulb, due to Joule's heating effect, the temperature of the bulb increases. As the temperature of the bulb filament increases, its resistance also increases, as resistance R is the function of temperature T. It is given by R = R0(1+αT). With an increase in the value of resistance, the value of current decreases as $i=\frac{V}{R}$. Now, the heat generated by the resistance is constantly radiated to the surroundings. Thus, the value of its temperature is maintained and hence its resistance. As a result, current through the bulb filament becomes constant.

For the given time t, let the currents passing through the resistance R1and R2 be i1 and i2 , respectively.
Applying Kirchoff's Voltage Law to circuit-1, we get:

$\epsilon -{i}_{1}r-{i}_{1}{R}_{1}=0\phantom{\rule{0ex}{0ex}}⇒{i}_{1}=\frac{\epsilon }{r+{R}_{1}}$
Similarly, the current in the other circuit,
${i}_{2}=\frac{\epsilon }{r+{R}_{2}}$
The thermal energies through the resistances are given by

${i}_{1}^{2}{R}_{1}t={i}_{2}^{2}{R}_{2}t\phantom{\rule{0ex}{0ex}}{\left(\frac{\epsilon }{r+{R}_{1}}\right)}^{2}{R}_{1}t={\left(\frac{\epsilon }{r+{R}_{2}}\right)}^{2}{R}_{2}t\phantom{\rule{0ex}{0ex}}\frac{{R}_{1}}{{\left(r+{R}_{1}\right)}^{2}}=\frac{{R}_{2}}{{\left(r+{R}_{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{\left({r}^{2}+{{R}_{1}}^{2}+2r{R}_{1}\right)}{{R}_{1}}=\frac{\left({r}^{2}+{{R}_{2}}^{2}+2r{R}_{2}\right)}{{R}_{2}}\phantom{\rule{0ex}{0ex}}\frac{{r}^{2}}{{R}_{1}}+{R}_{1}=\frac{{r}^{2}}{{R}_{2}}+{R}_{2}\phantom{\rule{0ex}{0ex}}{r}^{2}\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)={R}_{2}-{R}_{1}\phantom{\rule{0ex}{0ex}}{r}^{2}×\frac{{R}_{2}-{R}_{1}}{{R}_{1}{R}_{2}}={R}_{2}-{R}_{1}\phantom{\rule{0ex}{0ex}}{r}^{2}={R}_{1}{R}_{2}\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{{R}_{1}{R}_{2}}$

#### Question 3:

For the given time t, let the currents passing through the resistance R1and R2 be i1 and i2 , respectively.
Applying Kirchoff's Voltage Law to circuit-1, we get:

$\epsilon -{i}_{1}r-{i}_{1}{R}_{1}=0\phantom{\rule{0ex}{0ex}}⇒{i}_{1}=\frac{\epsilon }{r+{R}_{1}}$
Similarly, the current in the other circuit,
${i}_{2}=\frac{\epsilon }{r+{R}_{2}}$
The thermal energies through the resistances are given by

${i}_{1}^{2}{R}_{1}t={i}_{2}^{2}{R}_{2}t\phantom{\rule{0ex}{0ex}}{\left(\frac{\epsilon }{r+{R}_{1}}\right)}^{2}{R}_{1}t={\left(\frac{\epsilon }{r+{R}_{2}}\right)}^{2}{R}_{2}t\phantom{\rule{0ex}{0ex}}\frac{{R}_{1}}{{\left(r+{R}_{1}\right)}^{2}}=\frac{{R}_{2}}{{\left(r+{R}_{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{\left({r}^{2}+{{R}_{1}}^{2}+2r{R}_{1}\right)}{{R}_{1}}=\frac{\left({r}^{2}+{{R}_{2}}^{2}+2r{R}_{2}\right)}{{R}_{2}}\phantom{\rule{0ex}{0ex}}\frac{{r}^{2}}{{R}_{1}}+{R}_{1}=\frac{{r}^{2}}{{R}_{2}}+{R}_{2}\phantom{\rule{0ex}{0ex}}{r}^{2}\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)={R}_{2}-{R}_{1}\phantom{\rule{0ex}{0ex}}{r}^{2}×\frac{{R}_{2}-{R}_{1}}{{R}_{1}{R}_{2}}={R}_{2}-{R}_{1}\phantom{\rule{0ex}{0ex}}{r}^{2}={R}_{1}{R}_{2}\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{{R}_{1}{R}_{2}}$

No, the rise in the temperature of a resistor on passing  current through it is not an adiabatic process. In an adiabatic process, there is no heat exchange between the system and the surroundings. Here, some part of Joule's heat developed inside the resistor increases the temperature of the resistor and the remaining part is dissipated in the surroundings. Thus, the given process cannot be adiabatic.

#### Question 4:

No, the rise in the temperature of a resistor on passing  current through it is not an adiabatic process. In an adiabatic process, there is no heat exchange between the system and the surroundings. Here, some part of Joule's heat developed inside the resistor increases the temperature of the resistor and the remaining part is dissipated in the surroundings. Thus, the given process cannot be adiabatic.

The battery is doing positive work on a resistor carrying current i. Thus, âˆ†W is positive. The work done on the resistor is used to increase its thermal energy; thus âˆ†Q is positive. As the temperature of the resistor rises, âˆ†U is positive.

#### Question 5:

The battery is doing positive work on a resistor carrying current i. Thus, âˆ†W is positive. The work done on the resistor is used to increase its thermal energy; thus âˆ†Q is positive. As the temperature of the resistor rises, âˆ†U is positive.

The temperature of the hot junction at which the thermo-emf in a thermocouple becomes maximum is called neutral temperature for that thermocouple. For a thermocouple in which the constants a and b have the same sign, the neutral temperature will be less than the temperature of the cold junction of the thermocouple (as ${\theta }_{n}=-\frac{a}{b}$).
There will be no neutral or inversion temperature, as the temperature of the hot junction cannot be less than the temperature of the cold junction.

#### Question 6:

The temperature of the hot junction at which the thermo-emf in a thermocouple becomes maximum is called neutral temperature for that thermocouple. For a thermocouple in which the constants a and b have the same sign, the neutral temperature will be less than the temperature of the cold junction of the thermocouple (as ${\theta }_{n}=-\frac{a}{b}$).
There will be no neutral or inversion temperature, as the temperature of the hot junction cannot be less than the temperature of the cold junction.

If the inversion temperature and neutral temperature are measured in degree Celsius, then it is correct to say that "inversion temperature is always double the neutral temperature." When temperature is measured in other units, such as Kelvin, then inversion temperature is not the double of neutral temperature.

#### Question 7:

If the inversion temperature and neutral temperature are measured in degree Celsius, then it is correct to say that "inversion temperature is always double the neutral temperature." When temperature is measured in other units, such as Kelvin, then inversion temperature is not the double of neutral temperature.

No, the neutral temperature is not always the arithmetic mean of the inversion temperature and the temperature of the cold junction. That is valid only when the unit of temperature is degree Celsius.

#### Question 8:

No, the neutral temperature is not always the arithmetic mean of the inversion temperature and the temperature of the cold junction. That is valid only when the unit of temperature is degree Celsius.

No, the electrodes in an electrolytic cell do not have fixed polarity like that of a battery. If we take an electrolytic cell consisting of the Ag electrodes and the AgNO3 as electrolyte. When the battery is connected to it, the end to which the positive terminal of the battery is connected is the anode and the end to which the negative terminal is connected is the cathode. ${\mathrm{NO}}_{3}^{-}$ ions are deposited at the anode and Ag+ ions are deposited at the cathode. When the connection of the electrolytic cell is reversed, the polarities of the electrodes are also reversed.

#### Question 9:

No, the electrodes in an electrolytic cell do not have fixed polarity like that of a battery. If we take an electrolytic cell consisting of the Ag electrodes and the AgNO3 as electrolyte. When the battery is connected to it, the end to which the positive terminal of the battery is connected is the anode and the end to which the negative terminal is connected is the cathode. ${\mathrm{NO}}_{3}^{-}$ ions are deposited at the anode and Ag+ ions are deposited at the cathode. When the connection of the electrolytic cell is reversed, the polarities of the electrodes are also reversed.

Yes, the resistance of the electrolyte will decrease with an increase in temperature. This is because when the temperature of an electrolytic solution increases, its viscosity decreases and mobility of the ions in the solution increases.

#### Question 1:

Yes, the resistance of the electrolyte will decrease with an increase in temperature. This is because when the temperature of an electrolytic solution increases, its viscosity decreases and mobility of the ions in the solution increases.

Plot (a) is the correct option.

When current passes through a resistor, the heat produced,
H = I2Rt,
where I = current
R = resistance of the resistor
t = time for which current is flowing
This relation shows that the heat produced for a given time in a resistor varies with the square of current flowing through it. Hence, the plot between H vs I should be a parabola symmetric along the H axis, which is represented by curve a.

#### Question 2:

Plot (a) is the correct option.

When current passes through a resistor, the heat produced,
H = I2Rt,
where I = current
R = resistance of the resistor
t = time for which current is flowing
This relation shows that the heat produced for a given time in a resistor varies with the square of current flowing through it. Hence, the plot between H vs I should be a parabola symmetric along the H axis, which is represented by curve a.

Plot (d) is correct.

When current passes through a resistor, the temperature of the resistor increases due to the heat produced in it.
H = i2Rt,
where i = current flowing through the resistor
R = resistance of the resistor
t = time for which the current is flowing

With the increase in the temperature of the resistor, its resistance is also increased. The rate of production of thermal energy in the resistor of the circuit is given by the following relation:
$\frac{dU}{dt}=\frac{d}{dt}\left({i}^{2}Rt\right)={i}^{2}R$,
where i = current flowing through the resistor
R = resistance of the resistor

Rate of production of thermal energy in the resistor is directly proportional to the resistance. So, due to increase in resistance, $\frac{dU}{dt}$ also increases linearly, which is best represented by plot d.

#### Question 3:

Plot (d) is correct.

When current passes through a resistor, the temperature of the resistor increases due to the heat produced in it.
H = i2Rt,
where i = current flowing through the resistor
R = resistance of the resistor
t = time for which the current is flowing

With the increase in the temperature of the resistor, its resistance is also increased. The rate of production of thermal energy in the resistor of the circuit is given by the following relation:
$\frac{dU}{dt}=\frac{d}{dt}\left({i}^{2}Rt\right)={i}^{2}R$,
where i = current flowing through the resistor
R = resistance of the resistor

Rate of production of thermal energy in the resistor is directly proportional to the resistance. So, due to increase in resistance, $\frac{dU}{dt}$ also increases linearly, which is best represented by plot d.

(b) A is correct but B is wrong.

The value of neutral temperature is constant for a thermocouple. It depends on the nature of materials and is independent of the temperature of the cold junction. Inversion temperature depends on the temperature of the cold junction, as well as the nature of the material.

#### Question 4:

(b) A is correct but B is wrong.

The value of neutral temperature is constant for a thermocouple. It depends on the nature of materials and is independent of the temperature of the cold junction. Inversion temperature depends on the temperature of the cold junction, as well as the nature of the material.

(c) It cannot be Joule heat.

Joule heat is directly proportional to the square of the current passing through the resistor. Peltier heat is directly proportional to the current passing through the junction.Thomson heat is also directly proportional to the current passing through the section of the wire. Thus, the heat developed can be either Thomson heat or Peltier heat. But it cannot be Joule heat.

#### Question 5:

(c) It cannot be Joule heat.

Joule heat is directly proportional to the square of the current passing through the resistor. Peltier heat is directly proportional to the current passing through the junction.Thomson heat is also directly proportional to the current passing through the section of the wire. Thus, the heat developed can be either Thomson heat or Peltier heat. But it cannot be Joule heat.

(a) due to both A and B

In Seebeck Effect, a temperature difference between two dissimilar electrical conductors produces a potential difference across the junctions of the two different metals. The cause of this potential difference is the diffusion of free electrons from a high electron-density region to a low electron-density region. The free electron-density of the electrons is different in different metals and changes with change in temperature. Hence, both the statements are the causes of Seebeck Effect.

#### Question 6:

(a) due to both A and B

In Seebeck Effect, a temperature difference between two dissimilar electrical conductors produces a potential difference across the junctions of the two different metals. The cause of this potential difference is the diffusion of free electrons from a high electron-density region to a low electron-density region. The free electron-density of the electrons is different in different metals and changes with change in temperature. Hence, both the statements are the causes of Seebeck Effect.

(b) due to A but not due to B

In Peltier Effect, one of the junctions gets heated up and the other cools down when electric current is maintained in a circuit of material consisting of two dissimilar conductors.
This is caused due to the difference in density of free electrons in different metals. When two different metals are joined to form a junction, the electrons tend to diffuse from the side with higher concentration to the side with lower concentration. If current is forced through the junction, positive or negative work is done on the charge carriers, depending on the direction of the current. Accordingly, thermal energy is either produced or absorbed. Thus, Peltier Effect is caused due to A but not due to B.

#### Question 7:

(b) due to A but not due to B

In Peltier Effect, one of the junctions gets heated up and the other cools down when electric current is maintained in a circuit of material consisting of two dissimilar conductors.
This is caused due to the difference in density of free electrons in different metals. When two different metals are joined to form a junction, the electrons tend to diffuse from the side with higher concentration to the side with lower concentration. If current is forced through the junction, positive or negative work is done on the charge carriers, depending on the direction of the current. Accordingly, thermal energy is either produced or absorbed. Thus, Peltier Effect is caused due to A but not due to B.

(c) due to B but not due to A

If a metallic conductor has non-uniform temperature distribution along its length, the density of the free electrons is different for different sections. The electrons diffuse from the sections with higher concentration to those with lower concentration of free electrons. Thus, there is an emf inside the metal that is known as Thomson emf. If a current is forced through the given conductor, positive and negative work is done on the charge carriers, depending on the direction of current. Thus, thermal energy is either produced or absorbed. Thus, the correct cause of the given effect is given by statement B alone.

#### Question 8:

(c) due to B but not due to A

If a metallic conductor has non-uniform temperature distribution along its length, the density of the free electrons is different for different sections. The electrons diffuse from the sections with higher concentration to those with lower concentration of free electrons. Thus, there is an emf inside the metal that is known as Thomson emf. If a current is forced through the given conductor, positive and negative work is done on the charge carriers, depending on the direction of current. Thus, thermal energy is either produced or absorbed. Thus, the correct cause of the given effect is given by statement B alone.

(c) is a universal constant

Faraday,s constant is a universal constant. Its value is 9.6845×107 C/kg. It does not depend on the amount of the electrolyte, current in the electrolyte and on the amount of charge passed through the electrolyte.

#### Question 1:

(c) is a universal constant

Faraday,s constant is a universal constant. Its value is 9.6845×107 C/kg. It does not depend on the amount of the electrolyte, current in the electrolyte and on the amount of charge passed through the electrolyte.

(a) equal amounts of thermal energy must be produced in the resistors
(d) the temperature may rise equally in the resistors

In a resistor of resistance R, current i is passed for time t then the thermal energy produced in the resistor will be given by
H = i2Rt.
As the resistors are in series, the current through them will be same. Thus, the amount of thermal energy produced in the resistors is same. The rise in the temperature of the resistance will depend on the shape and size of the resistor. Thus, the rise in the temperature of the two resistances may be equal.

#### Question 2:

(a) equal amounts of thermal energy must be produced in the resistors
(d) the temperature may rise equally in the resistors

In a resistor of resistance R, current i is passed for time t then the thermal energy produced in the resistor will be given by
H = i2Rt.
As the resistors are in series, the current through them will be same. Thus, the amount of thermal energy produced in the resistors is same. The rise in the temperature of the resistance will depend on the shape and size of the resistor. Thus, the rise in the temperature of the two resistances may be equal.

(a) the two ends of the copper strip
(b) the copper end and the iron end at the junction
(c) the two ends of the iron strip
(d) the free ends B and C

The copper strip AB and an iron strip AC are joined at A and the junction A is maintained at 0°C and the free ends B and C are maintained at 100°C. In this case, there will be generation of thermo-emf between the points that are at different temperatures. Here, the two ends of the copper, the copper end and the iron end at the junction, the two ends of the iron strip and the free ends B and C are at different temperatures. Hence, there will be potential difference among them.

#### Question 3:

(a) the two ends of the copper strip
(b) the copper end and the iron end at the junction
(c) the two ends of the iron strip
(d) the free ends B and C

The copper strip AB and an iron strip AC are joined at A and the junction A is maintained at 0°C and the free ends B and C are maintained at 100°C. In this case, there will be generation of thermo-emf between the points that are at different temperatures. Here, the two ends of the copper, the copper end and the iron end at the junction, the two ends of the iron strip and the free ends B and C are at different temperatures. Hence, there will be potential difference among them.

(a) there will be no neutral temperature
(b) there will be no inversion temperature

The temperature of the hot junction at which the thermo-emf in the thermocouple becomes maximum is called neutral temperature for that thermocouple. The signs of the constants a and b are same. Therefore from the relation, ${\theta }_{n}=-\frac{a}{b},$ the neutral temperature will be less than the temperature of the cold junction of thermocouple.
Hence, there will be no neutral or inversion temperature, as the temperature of the hot junction cannot be less than the temperature of the cold junction.

#### Question 4:

(a) there will be no neutral temperature
(b) there will be no inversion temperature

The temperature of the hot junction at which the thermo-emf in the thermocouple becomes maximum is called neutral temperature for that thermocouple. The signs of the constants a and b are same. Therefore from the relation, ${\theta }_{n}=-\frac{a}{b},$ the neutral temperature will be less than the temperature of the cold junction of thermocouple.
Hence, there will be no neutral or inversion temperature, as the temperature of the hot junction cannot be less than the temperature of the cold junction.

(c) The rate of liberation of material will remain the same.

In an electrolytic cell, both the electrodes are made of the same material. Thus, on reversing the  terminals of the battery, the direction of the flow of charges will be reversed, but the rate of the electrolysis will remain the same.

#### Question 5:

(c) The rate of liberation of material will remain the same.

In an electrolytic cell, both the electrodes are made of the same material. Thus, on reversing the  terminals of the battery, the direction of the flow of charges will be reversed, but the rate of the electrolysis will remain the same.

(a) the nature of the material

The electrochemical equivalent of a substance is the ratio of the relative atomic mass of the substance to its valency. Thus, it is only dependent on the nature of the material.

#### Question 1:

(a) the nature of the material

The electrochemical equivalent of a substance is the ratio of the relative atomic mass of the substance to its valency. Thus, it is only dependent on the nature of the material.

Given:
Current through the wire, i = 2 A
Resistance of the wire, R = 25 Ω
Time taken, t = 1 min = 60 s
Heat developed across the wire,
H = i2Rt
= 2 × 2 × 25 × 60
= 100 × 60 J = 6000 J

#### Question 2:

Given:
Current through the wire, i = 2 A
Resistance of the wire, R = 25 Ω
Time taken, t = 1 min = 60 s
Heat developed across the wire,
H = i2Rt
= 2 × 2 × 25 × 60
= 100 × 60 J = 6000 J

Given:
Resistance of the coil, R = 100 Ω,
Emf of the battery, V = 6 V,
Change in temperature, âˆ†T = 15°C
Heat produced across the coil,
$H=\frac{{V}^{2}}{R}t$
This heat produced is used to increase the temperature of the coil.

#### Question 3:

Given:
Resistance of the coil, R = 100 Ω,
Emf of the battery, V = 6 V,
Change in temperature, âˆ†T = 15°C
Heat produced across the coil,
$H=\frac{{V}^{2}}{R}t$
This heat produced is used to increase the temperature of the coil.

Let R be the resistance of the coil.
The power P consumed by a coil of resistance R when connected across a supply V is given by

Now, P = 1000 W

#### Question 4:

Let R be the resistance of the coil.
The power P consumed by a coil of resistance R when connected across a supply V is given by

Now, P = 1000 W

(a) Let R be the resistance of the coil.
The power P consumed by a coil of resistance R when connected across a supply V is given by

(b) We know:

(c) Let n be the number of turns in the coil. Then,

#### Question 5:

(a) Let R be the resistance of the coil.
The power P consumed by a coil of resistance R when connected across a supply V is given by

(b) We know:

(c) Let n be the number of turns in the coil. Then,

Let R be the resistance of the bulb. If P is the power consumed by the bulb when operated at voltage V, then

Resistance of the copper wire,

The effective resistance,

The current supplied by the power station,

The power supplied to one side of the connecting wire,

The total power supplied on both sides,

#### Question 6:

Let R be the resistance of the bulb. If P is the power consumed by the bulb when operated at voltage V, then

Resistance of the copper wire,

The effective resistance,

The current supplied by the power station,

The power supplied to one side of the connecting wire,

The total power supplied on both sides,

The resistance of a bulb that consumes power P and is operated at voltage V is given by

(a) Now the supply drops to V' = 180 V.
So, the power consumed,

(b) Now the supply increases to V" = 240 V. Therefore,

#### Question 7:

The resistance of a bulb that consumes power P and is operated at voltage V is given by

(a) Now the supply drops to V' = 180 V.
So, the power consumed,

(b) Now the supply increases to V" = 240 V. Therefore,

Output voltage, V = 220 V ± 1% = 220 V ± 2.2 V
The resistance of a bulb that is operated at voltage V and consumes power P is given by

(a) For minimum power to be consumed, output voltage should be minimum. The minimum output voltage,
V' = (220 − 2.2) V
= 217.8 V
The current through the bulb,

Power consumed by the bulb, P' = i' × V'
= 0.45 × 217.8 = 98.0 W

(b) For maximum power to be consumed, output voltage should be maximum. The maximum output voltage,
V" = (220 + 2.2) V
= 222.2 V
The current through the bulb,

Power consumed by the bulb,
P" = i" × V"
= 0.459 × 222.2 = 102 W

#### Question 8:

Output voltage, V = 220 V ± 1% = 220 V ± 2.2 V
The resistance of a bulb that is operated at voltage V and consumes power P is given by

(a) For minimum power to be consumed, output voltage should be minimum. The minimum output voltage,
V' = (220 − 2.2) V
= 217.8 V
The current through the bulb,

Power consumed by the bulb, P' = i' × V'
= 0.45 × 217.8 = 98.0 W

(b) For maximum power to be consumed, output voltage should be maximum. The maximum output voltage,
V" = (220 + 2.2) V
= 222.2 V
The current through the bulb,

Power consumed by the bulb,
P" = i" × V"
= 0.459 × 222.2 = 102 W

Given that the operating voltage is V and power consumed is P.
Therefore, the resistance of the bulb,

The power fluctuation, p = 150 W. So, the voltage fluctuation that the bulb can withstand,

The bulb will withstand up to 270 V.

#### Question 9:

Given that the operating voltage is V and power consumed is P.
Therefore, the resistance of the bulb,

The power fluctuation, p = 150 W. So, the voltage fluctuation that the bulb can withstand,

The bulb will withstand up to 270 V.

Given the operating voltage V and power consumed P, the resistance of the immersion heater,

Mass of water, m = $\frac{1}{100}$ × 1000 = 10 Kg
Specific heat of water, s = 4200 Jkg$-$1 K$-1$
Rise in temperature, θ = 25°C
Heat required to raise the temperature of the given mass of water,
Q = msθ = 10 × 4200 × 25 = 1050000 J
Let t be the time taken to increase the temperature of water. The heat liberated is only 60%. So,
$\left(\frac{{V}^{2}}{R}\right)$ × t × 60% = 1050000 J
$⇒\frac{\left(220{\right)}^{2}}{48.4}×t×\frac{60}{100}=1050000$
t = 29.17 minutes

#### Question 10:

Given the operating voltage V and power consumed P, the resistance of the immersion heater,

Mass of water, m = $\frac{1}{100}$ × 1000 = 10 Kg
Specific heat of water, s = 4200 Jkg$-$1 K$-1$
Rise in temperature, θ = 25°C
Heat required to raise the temperature of the given mass of water,
Q = msθ = 10 × 4200 × 25 = 1050000 J
Let t be the time taken to increase the temperature of water. The heat liberated is only 60%. So,
$\left(\frac{{V}^{2}}{R}\right)$ × t × 60% = 1050000 J
$⇒\frac{\left(220{\right)}^{2}}{48.4}×t×\frac{60}{100}=1050000$
t = 29.17 minutes

Time taken to boil 4 cups of water, t = 2 minutes
Volume of water boiled = 4 × 200 cc = 800 cc
Initial temperature, θ1 = 25°C
Final temperature, θ2 = 100°C
Change in temperature, θθ2θ1 = 75°C
Mass of water to be boiled, m = 800 × 1 = 800 gm = 0.8 Kg
Heat required for boiling water,
Q = msθ = 0.8 × 4200 × 75 = 252000 J

We know:
1000 watt - hour = 1000 × 3600 watt sec.
∴ Cost of boiling 4 cups of water$=\frac{1}{1000×3600}×252000$
= Rs. 0.7

(b) Initial temperature, θ1 = 5°C
Final temperature, θ2 = 100°C
Change in temperature, θθ2θ1 = 95°C
Q = msθ = 0.8 × 4200 × 95 = 319200
∴ Cost of boiling 4 cups of water$=\frac{1}{1000×3600}×319200$
= Rs 0.09

#### Question 11:

Time taken to boil 4 cups of water, t = 2 minutes
Volume of water boiled = 4 × 200 cc = 800 cc
Initial temperature, θ1 = 25°C
Final temperature, θ2 = 100°C
Change in temperature, θθ2θ1 = 75°C
Mass of water to be boiled, m = 800 × 1 = 800 gm = 0.8 Kg
Heat required for boiling water,
Q = msθ = 0.8 × 4200 × 75 = 252000 J

We know:
1000 watt - hour = 1000 × 3600 watt sec.
∴ Cost of boiling 4 cups of water$=\frac{1}{1000×3600}×252000$
= Rs. 0.7

(b) Initial temperature, θ1 = 5°C
Final temperature, θ2 = 100°C
Change in temperature, θθ2θ1 = 95°C
Q = msθ = 0.8 × 4200 × 95 = 319200
∴ Cost of boiling 4 cups of water$=\frac{1}{1000×3600}×319200$
= Rs 0.09

Case-I : When the supply voltage is 220 V.
Power consumed by the bulb = 100 W
Excess power = 100 − 40 = 60 W
Power converted to light = 60% of 60 W = 36 W

Case-II : When the supply voltage is 200 V.
Power consumed = $\frac{200}{220}×100$ = 82.64 W
Excess power = 82.64 − 40 = 42.64 W
Power converted to light = 60% of 42.64 W = 25.584 W

Percentage drop in light intensity,
$p=\frac{36-25.584}{36}×100\phantom{\rule{0ex}{0ex}}⇒p=28.93\approx 29%$

#### Question 12:

Case-I : When the supply voltage is 220 V.
Power consumed by the bulb = 100 W
Excess power = 100 − 40 = 60 W
Power converted to light = 60% of 60 W = 36 W

Case-II : When the supply voltage is 200 V.
Power consumed = $\frac{200}{220}×100$ = 82.64 W
Excess power = 82.64 − 40 = 42.64 W
Power converted to light = 60% of 42.64 W = 25.584 W

Percentage drop in light intensity,
$p=\frac{36-25.584}{36}×100\phantom{\rule{0ex}{0ex}}⇒p=28.93\approx 29%$

The effective resistance of the circuit,

Current i through the circuit,

Let i' be the current through the 6 Ω resistor. Then,
i' × 6 = (ii') × 2

(a) Heat generated in the 2 Ω resistor,
H = (i - i')2Rt

The heat capacity of the calorimeter together with water is 2000 J K−1. Thus, 2000 J of heat raise the temp by 1 K.
∴ 5832 J of heat raises the temperature by $\frac{5832}{2000}$ = 2.916 K
(b) When the 6 Ω resistor gets burnt, the effective resistance of the circuit,
Reff = 1 + 2 = 3 Ω
Current through the circuit,
i = $\frac{6}{3}$ = 2 A
Heat generated in the 2 Ω resistor =  (2)2× 2 × 15 × 60 = 7200 J
2000 J raise the temperature by 1 K.
∴ 7200 J raise the temperature by $\frac{7200}{2000}$ = 3.6 K .

#### Question 13:

The effective resistance of the circuit,

Current i through the circuit,

Let i' be the current through the 6 Ω resistor. Then,
i' × 6 = (ii') × 2

(a) Heat generated in the 2 Ω resistor,
H = (i - i')2Rt

The heat capacity of the calorimeter together with water is 2000 J K−1. Thus, 2000 J of heat raise the temp by 1 K.
∴ 5832 J of heat raises the temperature by $\frac{5832}{2000}$ = 2.916 K
(b) When the 6 Ω resistor gets burnt, the effective resistance of the circuit,
Reff = 1 + 2 = 3 Ω
Current through the circuit,
i = $\frac{6}{3}$ = 2 A
Heat generated in the 2 Ω resistor =  (2)2× 2 × 15 × 60 = 7200 J
2000 J raise the temperature by 1 K.
∴ 7200 J raise the temperature by $\frac{7200}{2000}$ = 3.6 K .

Given:
Difference in temperature, θ = 0.001°C,
a = − 46 × 10−6 V °C−1
b = − 0.48 × 10−5 V °C−2.
Emf, E = aθ + $\frac{1}{2}$2
E = (− 46 × 10−6) × (0.001) $-\frac{1}{2}$ × (0.48 × 10−6 )× (0.001)2
= − 46 × 10−9 − 0.24 × 10−12
= − 46.0024 × 10−9
= − 4.6 × 10−8 V

#### Question 14:

Given:
Difference in temperature, θ = 0.001°C,
a = − 46 × 10−6 V °C−1
b = − 0.48 × 10−5 V °C−2.
Emf, E = aθ + $\frac{1}{2}$2
E = (− 46 × 10−6) × (0.001) $-\frac{1}{2}$ × (0.48 × 10−6 )× (0.001)2
= − 46 × 10−9 − 0.24 × 10−12
= − 46.0024 × 10−9
= − 4.6 × 10−8 V

Difference in temperature, θ = 40°C
Emf, Ecs = acsθ + $\frac{1}{2}$bcsθ2    ...(1)
acs = [2.76 − (−43.7) μV
= 46.46 μV/°C
bcs = [0.012 − (−0.47) μV/°C
= 0.482 μV/°C2
Putting this value in eq. (1), we get:
Ecs = 46.46 × 10−6 × 40 + $\frac{1}{2}$ × 0.482 × 10−6 × (40)2
= 1.04 × 10−5 V

#### Question 15:

Difference in temperature, θ = 40°C
Emf, Ecs = acsθ + $\frac{1}{2}$bcsθ2    ...(1)
acs = [2.76 − (−43.7) μV
= 46.46 μV/°C
bcs = [0.012 − (−0.47) μV/°C
= 0.482 μV/°C2
Putting this value in eq. (1), we get:
Ecs = 46.46 × 10−6 × 40 + $\frac{1}{2}$ × 0.482 × 10−6 × (40)2
= 1.04 × 10−5 V

Neutral temperature,
${\theta }_{n}=-\frac{a}{b}$

Thus, the neutral temperature,

The inversion temperature is double the neutral temperature, i.e. 659 $°$C.

#### Question 16:

Neutral temperature,
${\theta }_{n}=-\frac{a}{b}$

Thus, the neutral temperature,

The inversion temperature is double the neutral temperature, i.e. 659 $°$C.

(a) Amount of charge required by 1 equivalent mass of the substance = 96500 C
For a monovalent material,
equivalent mass = molecular mass
⇒ Amount of charge required by 6.023 × 1023 atoms = 96500 C
∴ Amount of charge required by 1 atom =

(b) For a divalent material,
equivalent mass =$\frac{1}{2}$molecular mass
⇒ Amount of charge required by $\frac{1}{2}$ × 6.023 × 1023 = 96500 C
∴ Amount of charge required by 1 atom = 1.6 × 2 × 10−19 = 3.2 × 10−19 C

#### Question 17:

(a) Amount of charge required by 1 equivalent mass of the substance = 96500 C
For a monovalent material,
equivalent mass = molecular mass
⇒ Amount of charge required by 6.023 × 1023 atoms = 96500 C
∴ Amount of charge required by 1 atom =

(b) For a divalent material,
equivalent mass =$\frac{1}{2}$molecular mass
⇒ Amount of charge required by $\frac{1}{2}$ × 6.023 × 1023 = 96500 C
∴ Amount of charge required by 1 atom = 1.6 × 2 × 10−19 = 3.2 × 10−19 C

Equivalent mass of silver, EAg = 107.9 g   (âˆµ Ag is monoatomic)
The ECE of silver,
${Z}_{\mathit{A}\mathit{g}}=\frac{{E}_{Ag}}{f}=\frac{107.9}{96500}=0.001118$
Using the formula, m = Zit, we get:
m = 0.00118 × 0.500 × 3600
= 2.01 g
So, 2.01 g of silver is liberated.

#### Question 18:

Equivalent mass of silver, EAg = 107.9 g   (âˆµ Ag is monoatomic)
The ECE of silver,
${Z}_{\mathit{A}\mathit{g}}=\frac{{E}_{Ag}}{f}=\frac{107.9}{96500}=0.001118$
Using the formula, m = Zit, we get:
m = 0.00118 × 0.500 × 3600
= 2.01 g
So, 2.01 g of silver is liberated.

Given:
Mass of silver deposited, m = 3 g
Time taken, t = 3 min. = 180 s
E.C.E. of silver, Z = 1.12 × 10−6kg C−1

Using the formula, m = Zit, we get:
$3×{10}^{-3}=1.12×{10}^{-6}×i×180\phantom{\rule{0ex}{0ex}}$

#### Question 19:

Given:
Mass of silver deposited, m = 3 g
Time taken, t = 3 min. = 180 s
E.C.E. of silver, Z = 1.12 × 10−6kg C−1

Using the formula, m = Zit, we get:
$3×{10}^{-3}=1.12×{10}^{-6}×i×180\phantom{\rule{0ex}{0ex}}$

Let the required time be t.
Mass of 1 litre hydrogen,

Using the formula, m = Zit, we get:

#### Question 20:

Let the required time be t.
Mass of 1 litre hydrogen,

Using the formula, m = Zit, we get:

Given:
Mass of salt deposited, m = 1 g
Current, i = 2 A
Time, t = 1.5 hours = 5400 s

For the trivalent metal salt:
Equivalent mass = $\frac{1}{3}$Atomic weight
The E.C.E of the salt,

(a) Using the formula, m = Zit, we get:

(b) Using the relation between equivalent mass and mass deposited on plates, we get:

#### Question 21:

Given:
Mass of salt deposited, m = 1 g
Current, i = 2 A
Time, t = 1.5 hours = 5400 s

For the trivalent metal salt:
Equivalent mass = $\frac{1}{3}$Atomic weight
The E.C.E of the salt,

(a) Using the formula, m = Zit, we get:

(b) Using the relation between equivalent mass and mass deposited on plates, we get:

Given:
Current, i = 15 A
Surface area of the plate = 200 cm2,
Thickness of silver deposited= 0.1 mm = 0.01 cm
Volume of Ag deposited on one side = 200 × 0.01 cm3 = 2 cm3
∴ Volume of Ag deposited on both side = 4 cm3
Mass of silver deposited,
m = Volume × Specific gravity × 1000 = 4 × 10-3 × 10.5 ×1000 = 42 kg
Using the formula, m = Zit, we get:
42 = ZAg × 15 × t

#### Question 22:

Given:
Current, i = 15 A
Surface area of the plate = 200 cm2,
Thickness of silver deposited= 0.1 mm = 0.01 cm
Volume of Ag deposited on one side = 200 × 0.01 cm3 = 2 cm3
∴ Volume of Ag deposited on both side = 4 cm3
Mass of silver deposited,
m = Volume × Specific gravity × 1000 = 4 × 10-3 × 10.5 ×1000 = 42 kg
Using the formula, m = Zit, we get:
42 = ZAg × 15 × t

Given:
Mass of silver deposited, m = 2.68 g
Time, t = 10 minutes = 600 s
Using the formula, m = Zit, we get:

Heat developed in the 20 Ω resistor,

#### Question 23:

Given:
Mass of silver deposited, m = 2.68 g
Time, t = 10 minutes = 600 s
Using the formula, m = Zit, we get:

Heat developed in the 20 Ω resistor,

Let i be the current through the circuit.
Emf of battery, E = 12 V
Voltage drop across the voltameter, V = 10 V
Internal resistance of the battery, r = 2 Ω
Applying Kirchoff's Law in the circuit, we get:

Using the formula m = Zit, we get:

#### Question 24:

Let i be the current through the circuit.
Emf of battery, E = 12 V
Voltage drop across the voltameter, V = 10 V
Internal resistance of the battery, r = 2 Ω
Applying Kirchoff's Law in the circuit, we get:

Using the formula m = Zit, we get:

Surface area of the plate, A = 10 cm2 = 10 × 10−4 m2
Thickness of copper deposited, t = 10 μm = 10−5m
Density of copper = 9000 kg/m3
Volume of copper deposited, V = A(2t)
V = 10 × 10−4 × 2 × 10 × 10−6
= 2 × 102 × 10−10
= 2 × 10−8 m3
Mass of copper deposited, m = Volume × Density = 2 × 10−8 × 9000
m = 18 × 10−5 kg

Using the formula, m = ZQ, we get:
18 × 10−5 = 3 × 10−7 × Q
Q = 6 × 102 C
Energy spent by the cell = Work done by the cell
W = VQ
= 12 × 6 × 102
= 72 × 102 = 7.2 kJ
Let âˆ†θ be the rise in temperature of water. When this energy is used to heat 100 g of water, we have:
7.2 × 103 = 100 × 10−3 × 4200 × âˆ†θ
⇒ âˆ†θ = 17 K

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