The steady current flows in a metallic conductor of non-uniform cross-section. The ${\rm{quantity}}\,\,{\rm{/}}\,{\rm{quantities}}$ constant along the length of the conductor ${\rm{is}}\,\,{\rm{/}}\,\,{\rm{are}}$

Current, electric field and drift velocity

A current of ${\rm{1}}{\rm{.5}}\,\,{\rm{A}}$ flows through a copper voltameter. The thickness of copper deposited on the electrode surface of size $50\,\,{\rm{cm}}\,\, \times \,10\,\,{\rm{cm}}$ is ${\rm{20}}\,\,{\rm{min}}$ will be (density of copper $ = 9000\,\,{\rm{kg}} - {{\rm{m}}^{ - 3}}\,\,{\rm{and}}\,{\rm{ ECE }}\,{\rm{of}}\,{\rm{ copper}} = 0.00033\,\,{\rm{g}}\,\,{{\rm{C}}^{ - 1}})$

$3.3\,\, \times \,\,{10^{ - 6}}\,\,{\rm{m}}$

$1.3\,\,\, \times \,\,{10^{ - 5}}\,\,{\rm{m}}$

$2.6\,\,\, \times \,\,{10^{ - 5}}\,\,{\rm{m}}$

CURRENT ELECTRICITY MCQs for NEET UG PHYSICS — Mixed difficulty

1 Medium

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Understanding +4 | -1

In the absence of applied potential, the electric current flowing through a metallic wire is zero because

View answer and explanation

Correct option: C

The electrons move in random direction with a speed of the order close to that of velocity of light

Option C (The electrons move in random direction with a speed of the order close to that of velocity of light) is correct because even without an applied potential, electrons are in constant random motion due to thermal energy, with speeds close to the velocity of light, but these motions average out to zero net current.

Solution approach

Correct Answer + Core Reason: The electrons move in random direction with a speed of the order close to that of velocity of light. The random motion cancels out, resulting in zero net current.

Step-by-Step Solution:

Step 1: In the absence of an applied potential, electrons in a metallic wire are in constant random motion due to thermal energy.

Step 2: These electrons move with very high speeds, close to the velocity of light.

Step 3: However, because the motion is random, the electrons are moving in all possible directions.

Step 4: The average velocity of all the electrons is zero, resulting in zero net current.

Why other options are wrong:
- Option A: Electrons are not stationary; they are in constant random motion.
- Option B: The drift speed is of the order of $10^{-2} \text{{cm/s}}$ only when a potential is applied. Without a potential, the random speeds are much higher.
- Option D: Electrons and ions both move, but the electrons are the primary charge carriers responsible for current.

Key Concepts:
- Random motion of electrons
- Thermal velocity
- Drift velocity

Memory Aids:
- Random motion implies zero net current.

Quick Tips:
- Remember that electrons are always moving, even without an applied potential.

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Understanding +4 | -1

A metallic block has no potential difference applied across it, then the mean velocity of free electrons at absolute temperature

T

is

View answer and explanation

Correct option: C

Zero

Option C (Zero) is correct because in the absence of an applied potential difference, the free electrons move randomly, and their average velocity (mean velocity) is zero.

Solution approach

Correct Answer + Core Reason: Zero. In the absence of an applied potential, the mean velocity of free electrons is zero.

Step-by-Step Solution:

Step 1: In the absence of an applied potential difference, free electrons in a metallic block move randomly.

Step 2: The random motion of electrons is due to thermal energy.

Step 3: The average velocity of these randomly moving electrons is zero because their velocities are randomly oriented, and the vector sum of their velocities is zero.

Why other options are wrong:
- Option A: The mean velocity is not proportional to T in the absence of a potential difference; it's zero.
- Option B: The mean velocity is not proportional to $\sqrt{T}$ in the absence of a potential difference; it's zero.
- Option D: The mean velocity is zero, not finite.

Key Concepts:
- Free electrons in metals
- Thermal velocity
- Mean velocity

Memory Aids:
- Random motion implies zero average velocity.

Quick Tips:
- Remember that the mean velocity is zero without an applied potential.

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Understanding +4 | -1

In a thermocouple, which of the following statements is not true

View answer and explanation

Correct option: D

When the temperature of cold junction increases, the temperature of inversion also increases

Option D (When the temperature of cold junction increases, the temperature of inversion also increases) is correct because the temperature of inversion depends on the neutral temperature and the temperature of the cold junction, and the relationship is such that if the cold junction temperature increases, the temperature of inversion decreases.

Solution approach

Correct Answer + Core Reason: When the temperature of cold junction increases, the temperature of inversion also increases. This statement is incorrect regarding the relationship between cold junction and inversion temperatures.

Step-by-Step Solution:

Step 1: Define Neutral Temperature ( $T_n$ ): The temperature at which the thermo-emf is maximum.

Step 2: Define Temperature of Inversion ( $T_i$ ): The temperature at which the thermo-emf reverses its direction.

Step 3: Relationship between $T_i$ , $T_n$ , and Cold Junction Temperature ( $T_c$ ): $T_i + T_c = 2T_n$ .

Step 4: From the equation, $T_i = 2T_n - T_c$ . If $T_c$ increases, $T_i$ decreases, assuming $T_n$ remains constant.

Why other options are wrong:
- Option A: Neutral temperature is a property of the thermocouple materials.
- Option B: Temperature of inversion depends on the cold junction temperature as shown in the equation.
- Option C: When the hot junction temperature equals the inversion temperature, the thermo-emf is indeed zero.

Key Concepts:
- Thermocouple
- Neutral temperature
- Temperature of inversion
- Seebeck effect

Memory Aids:
- $T_i + T_c = 2T_n$ (Relationship between inversion, cold junction, and neutral temperatures)

Quick Tips:
- Remember the relationship between the neutral temperature, temperature of inversion, and the cold junction temperature.

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(A) The drift velocity of electrons decreases with the increase in the temperature of conductor.
(B) The drift velocity is inversely proportional to the area of cross-section of given conductor.
(C) The drift velocity does not depend on the applied potential difference to the conductor.
(D) The drift velocity of electron is inversely proportional to the length of the conductor.
(E) The drift velocity increases with the increase in the temperature of conductor.
Choose the correct answer from the options given below

View answer and explanation

Correct option: A

(A) and (B) only

Option A ((A) and (B) only) is correct because drift velocity decreases with increasing temperature and is inversely proportional to the cross-sectional area of the conductor.

Solution approach

Correct Answer + Core Reason: (A) and (B) only. Drift velocity is affected by temperature and area.

Step-by-Step Solution:

Step 1: Recall the formula for drift velocity: $v_d = \frac{{I}}{{nAe}} = \frac{{eE\tau}}{{m}}$ , where I is current, n is the number density of electrons, A is the area of cross-section, e is the electron charge, E is the electric field, $\tau$ is the relaxation time, and m is the mass of the electron.

Step 2: Analyze the effect of temperature: As temperature increases, the relaxation time $\tau$ decreases due to increased collisions, thus decreasing the drift velocity. Therefore, statement (A) is correct.

Step 3: Analyze the effect of the area of cross-section: From $v_d = \frac{{I}}{{nAe}}$ , drift velocity is inversely proportional to the area of cross-section. Therefore, statement (B) is correct.

Step 4: Analyze the effect of the applied potential difference: Since $E = \frac{{V}}{{l}}$ , where V is the potential difference and l is the length of the conductor, $v_d$ depends on the applied potential difference. Therefore, statement (C) is incorrect.

Step 5: Analyze the effect of the length of the conductor: Since $E = \frac{{V}}{{l}}$ , $v_d$ is inversely proportional to the length of the conductor. Therefore, statement (D) is correct.

Step 6: Statement (E) is incorrect because drift velocity decreases with increasing temperature.

Why other options are wrong:
- Option B: Statement (D) is correct, but statement (C) is incorrect.
- Option C: Statement (E) is incorrect.
- Option D: Statement (C) is incorrect.

Key Concepts:
- Drift velocity
- Relaxation time
- Temperature dependence of drift velocity
- Area dependence of drift velocity

Memory Aids:
- $v_d = \frac{{I}}{{nAe}}$ (Drift velocity formula)

Quick Tips:
- Remember the factors affecting drift velocity: temperature, area, electric field.

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Remembering +4 | -1

The mass of ions deposited during a given interval of time in the process of electrolysis depends on

View answer and explanation

Correct option: A

The current

Option A (The current) is correct because, according to Faraday's laws of electrolysis, the mass of ions deposited is directly proportional to the quantity of electricity (current x time) passed through the electrolyte.

Solution approach

Correct Answer + Core Reason: The current. Faraday's laws of electrolysis state that the mass of ions deposited is proportional to the current.

Step-by-Step Solution:

Step 1: Faraday's first law of electrolysis: The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.

Step 2: Mathematically, $m = ZIt$ , where $m$ is the mass of the substance deposited, $Z$ is the electrochemical equivalent, $I$ is the current, and $t$ is the time.

Step 3: From the equation, it is clear that the mass of ions deposited depends on the current.

Why other options are wrong:
- Option B: Resistance affects the current, but the mass directly depends on the current, not the resistance itself.
- Option C: Temperature can affect the conductivity of the electrolyte, but the primary factor determining the mass deposited is the current.
- Option D: Electric power is related to both voltage and current, but the mass directly depends on the current.

Key Concepts:
- Faraday's laws of electrolysis

Memory Aids:
- $m = ZIt$ (mass = electrochemical equivalent * current * time)

Quick Tips:
- Remember Faraday's laws of electrolysis when dealing with mass deposition in electrolytic processes.

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Understanding +4 | -1

The steady current flows in a metallic conductor of non-uniform cross-section. The

\mathrm{quantity}\,\,\mathrm{/}\,\mathrm{quantities}

constant along the length of the conductor

\mathrm{is}\,\,\mathrm{/}\,\,\mathrm{are}

View answer and explanation

Correct option: D

Current only

Option D (Current only) is correct because in a non-uniform conductor with steady current, the current remains constant throughout the length, while the electric field and drift velocity vary due to the changing cross-sectional area.

Solution approach

Correct Answer + Core Reason: Current only. In a steady current flow, the current is constant throughout the conductor.

Step-by-Step Solution:

Step 1: In a steady current flow, the current (I) is constant throughout the conductor, regardless of the cross-sectional area.

Step 2: The current density (J) is given by $J = \frac{I}{A}$ , where A is the cross-sectional area. Since A is non-uniform, J is also non-uniform.

Step 3: The electric field (E) is related to the current density by Ohm's law: $J = \sigma E$ , where $\sigma$ is the conductivity. Since J is non-uniform, E is also non-uniform.

Step 4: The drift velocity ( $v_d$ ) is related to the current density by $J = nev_d$ , where n is the number density of electrons and e is the charge of an electron. Since J is non-uniform, $v_d$ is also non-uniform.

Step 5: Therefore, only the current remains constant along the length of the conductor.

Why other options are wrong:
- Option A: Electric field and drift velocity are not constant.
- Option B: Drift speed is not constant.
- Option C: Drift speed is not constant.

Key Concepts:
- Steady current
- Current density
- Electric field
- Drift velocity
- Ohm's law

Memory Aids:
- Current is constant in a series circuit.

Quick Tips:
- Remember that current is conserved in a closed circuit.

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Understanding +4 | -1

If the temperature of cold junction of thermocouple is lowered, then the neutral temperature

View answer and explanation

Correct option: D

Remains the same

Option D is correct because the neutral temperature of a thermocouple is a characteristic property of the metals used and does not depend on the cold junction temperature.

Solution approach

Correct Answer + Core Reason: Remains the same. The neutral temperature is a property of the thermocouple materials and is independent of the cold junction temperature.

Step-by-Step Solution:

Step 1: Define the neutral temperature ( $T_n$ ) and the temperature of inversion ( $T_i$ ) for a thermocouple.
- The neutral temperature is the temperature of the hot junction at which the thermoelectric EMF is maximum.
- The temperature of inversion is the temperature of the hot junction at which the thermoelectric EMF is zero (and changes sign).

Step 2: The relationship between the neutral temperature ( $T_n$ ), the cold junction temperature ( $T_c$ ), and the temperature of inversion ( $T_i$ ) is given by:
$T_n = \frac{T_i + T_c}{2}$ \nHowever, the *difference* between the neutral temperature and the cold junction temperature is constant, not the neutral temperature itself. The neutral temperature is a characteristic of the thermocouple materials.

Step 3: If the cold junction temperature ( $T_c$ ) is lowered, the temperature of inversion ( $T_i$ ) will change such that the neutral temperature ( $T_n$ ) remains constant. The neutral temperature depends only on the materials of the thermocouple.

Why other options are wrong:
- Option A: The neutral temperature does not increase when the cold junction temperature is lowered.
- Option B: The neutral temperature does not approach the inversion temperature when the cold junction temperature is lowered. The inversion temperature changes to maintain a constant neutral temperature.
- Option C: The neutral temperature does not decrease when the cold junction temperature is lowered.

Key Concepts:
- Thermocouple
- Neutral Temperature
- Temperature of Inversion
- Seebeck Effect

Memory Aids:
- Neutral Temperature is a Material Property.

Quick Tips:
- Remember that the neutral temperature is a characteristic property of the thermocouple materials and is independent of the cold junction temperature.
- The temperature of inversion changes when the cold junction temperature changes, keeping the neutral temperature constant.

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For measurement of potential difference, potentiometer is preferred in comparison to voltmeter because

View answer and explanation

Correct option: D

Potentiometer does not take current from the circuit

Option D (Potentiometer does not take current from the circuit) is correct because a potentiometer works on the null deflection method, drawing no current from the source being measured.

Solution approach

Correct Answer + Core Reason: Potentiometer does not take current from the circuit. This is due to the null deflection method.

Step-by-Step Solution:

Step 1: The question asks why a potentiometer is preferred over a voltmeter for measuring potential difference.

Step 2: A potentiometer uses a null deflection method, meaning at the point of measurement, no current is drawn from the circuit being measured.

Step 3: A voltmeter, on the other hand, draws some current from the circuit, which can affect the potential difference being measured.

Step 4: Therefore, a potentiometer provides a more accurate measurement as it doesn't load the circuit.

Why other options are wrong:
- Option A: While potentiometers can be sensitive, this isn't the primary reason for preference. Voltmeters can also be highly sensitive.
- Option B: The resistance of a potentiometer is typically higher than a voltmeter, not lower.
- Option C: Potentiometers are generally more expensive than voltmeters.

Key Concepts:
- Potentiometer principle
- Null deflection method
- Voltmeter loading effect

Memory Aids:
- Potentiometer = No current draw = Accurate measurement

Quick Tips:
- Remember that ideal voltage measurement involves minimal current draw.

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Analyzing +4 | -1

Given below are two statements:

Statement I : A uniform wire of resistance

80 \Omega

is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be

5 \Omega

.

Statement II : Two resistances

2 R

and

3 R

are connected in parallel in a electric circuit. The value of thermal energy developed in

3 R

and

2 R

will be in the ratio

3: 2

.

In the light of the above statements, choose the most appropriate answer from the option given below.

View answer and explanation

Correct option: C

Statement I is correct but statement II is incorrect

Option C is correct. Statement I is correct as the equivalent resistance is 5 ohms. Statement II is incorrect because thermal energy is inversely proportional to resistance, so the ratio is 2:3.

Solution approach

Correct Answer + Core Reason: Statement I is correct but statement II is incorrect. The calculation for parallel resistors in Statement I is correct, while the ratio of thermal energy in Statement II is incorrect.

Step-by-Step Solution:

Step 1: Analyze Statement I.\nThe wire of resistance $80 \Omega$ is cut into four equal parts. So, each part has a resistance of $80/4 = 20 \Omega$ .

Step 2: These four parts are connected in parallel. The equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} = \frac{4}{20} = \frac{1}{5}$ \nTherefore, $R_{eq} = 5 \Omega$ . So, Statement I is correct.

Step 3: Analyze Statement II.\nTwo resistances $2R$ and $3R$ are connected in parallel. The thermal energy developed is given by $H = I^2Rt = (V/R)^2Rt = V^2t/R$ , where $V$ is the voltage across the resistor (same for parallel connection).

Step 4: The ratio of thermal energy developed in $3R$ and $2R$ is:
$\frac{H_{3R}}{H_{2R}} = \frac{V^2t/(3R)}{V^2t/(2R)} = \frac{2R}{3R} = \frac{2}{3}$ \nSo, Statement II is incorrect.

Step 5: Conclusion: Statement I is correct, and Statement II is incorrect.

Why other options are wrong:
- Option A: Statement II is incorrect, so this option is wrong.
- Option B: Statement I is correct, so this option is wrong.
- Option D: Statement I is correct, so this option is wrong.

Key Concepts:
- Parallel Resistance: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$
- Thermal Energy: $H = V^2t/R$ (for parallel connection)

Memory Aids:
- In parallel, voltage is the same across all resistors.

Quick Tips:
- When resistors are in parallel, the equivalent resistance is always less than the smallest individual resistance.
- Thermal energy is inversely proportional to resistance in parallel circuits.

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Application +4 | -1

A direct current of 4A and an alternating current of 4A flows through two identical resistors. The ratio of heat produced in the two resistances in same time interval will be

View answer and explanation

Correct option: C

2\mathrm{ }:\mathrm{ }1

Option C (

2\mathrm{ }:\mathrm{ }1

) is correct because the heat produced by AC is calculated using the RMS value, which is

\frac{I}{\sqrt{2}}

, leading to half the power compared to DC.

Solution approach

Correct Answer + Core Reason: $2\mathrm{ }:\mathrm{ }1$ . The ratio of heat produced is determined by comparing the power dissipated by DC and AC currents.

Step-by-Step Solution:

Step 1: Heat produced by DC current: The heat produced by a DC current $I$ in a resistor $R$ in time $t$ is given by $H_{DC} = I^2Rt$ .

Step 2: Heat produced by AC current: The heat produced by an AC current $I$ in a resistor $R$ in time $t$ is given by $H_{AC} = I_{rms}^2Rt$ , where $I_{rms} = \frac{I}{\sqrt{2}}$ . So, $H_{AC} = (\frac{I}{\sqrt{2}})^2Rt = \frac{I^2}{2}Rt$ .

Step 3: Calculate the ratio: $\frac{{H_{DC}}}{{H_{AC}}} = \frac{{I^2Rt}}{{\frac{{I^2}}{2}Rt}} = \frac{1}{{\frac{1}{2}}} = 2$ .

Step 4: Final Answer: The ratio is $2:1$ .

Why other options are wrong:
- **Option A ( $1\mathrm{ }:\mathrm{ }1$ ): This is incorrect because it doesn't account for the RMS value of the AC current.
- Option B ( $1\mathrm{ }:\mathrm{ }2$ ): This is the inverse of the correct ratio.
- Option D ( $4\mathrm{ }:\mathrm{ }1$ ): This is incorrect and doesn't relate to the RMS value calculation.

Key Concepts:**
- Heat produced by electric current: $H = I^2Rt$
- RMS value of AC current: $I_{rms} = \frac{I}{\sqrt{2}}$

Memory Aids:
- Remember that AC current's heating effect is based on its RMS value.

Quick Tips:
- Always use the RMS value for AC current when calculating power or heat.
- DC current calculations are straightforward using $I^2Rt$ .

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Understanding +4 | -1

With the rise of temperature the resistivity of a semiconductor

View answer and explanation

Correct option: C

Decreases

Option C (Decreases) is correct because the resistivity of a semiconductor decreases with increasing temperature due to increased carrier concentration.

Solution approach

Correct Answer + Core Reason: Decreases. Resistivity of semiconductors decreases with increasing temperature.

Step-by-Step Solution:

Step 1: Semiconductors: Semiconductors are materials with electrical conductivity between that of conductors and insulators.

Step 2: Temperature Dependence: The resistivity of semiconductors is highly temperature-dependent.

Step 3: Mechanism: As temperature increases, more electrons are excited to the conduction band, increasing the carrier concentration. This leads to a decrease in resistivity.

Why other options are wrong:
- Option A: The resistivity of a semiconductor changes significantly with temperature.
- Option B: The resistivity of a semiconductor decreases, not increases, with temperature.
- Option D: The resistivity generally decreases monotonically with increasing temperature.

Key Concepts:
- Semiconductor
- Resistivity
- Temperature Dependence
- Carrier Concentration

Memory Aids:
- Semiconductor: Higher Temp = Lower Resistivity

Quick Tips:
- Remember the inverse relationship between temperature and resistivity in semiconductors.

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Application +4 | -1

The e.m.f. of a cell is 6 volt. When 2 ampere current is drawn from it then the potential difference across its terminals remains 3 volt. Its internal resistance wil be:

View answer and explanation

Correct option: A

1.5 \Omega

Option A (

1.5 \Omega

) is correct because the internal resistance can be calculated using the formula r = (E-V)/I.

Solution approach

Correct Answer + Core Reason: $1.5 \Omega$ . The internal resistance is calculated using the formula $r = \frac{{E-V}}{{I}}$ .

Step-by-Step Solution:

Step 1: Identify the given values: EMF of the cell, $E = 6 \text{{ V}}$ , Current drawn, $I = 2 \text{{ A}}$ , Potential difference across terminals, $V = 3 \text{{ V}}$

Step 2: Use the formula for internal resistance: $r = \frac{{E - V}}{{I}}$

Step 3: Substitute the values: $r = \frac{{6 \text{{ V}} - 3 \text{{ V}}}}{{2 \text{{ A}}}}$

Step 4: Calculate the internal resistance: $r = \frac{{3 \text{{ V}}}}{{2 \text{{ A}}}} = 1.5 \Omega$

Why other options are wrong:
- **Option B (1 $\Omega$ ): This is incorrect. It's likely a calculation error.
- Option C (10 $\Omega$ ): This is incorrect. It's likely a calculation error or using the wrong formula.
- Option D (100 $\Omega$ ): This is incorrect. It's likely a calculation error or using the wrong formula.

Key Concepts:
- Internal resistance of a cell
- EMF and terminal voltage
- Ohm's law

Memory Aids:**
- $V = E - Ir$ , where $r$ is the internal resistance.

Quick Tips:
- Remember the relationship between EMF, terminal voltage, current, and internal resistance.

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Application +4 | -1

Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other and the thinner wire has a resistance of 8

ohms

, the resistance of the combination is equal to

View answer and explanation

Correct option: B

\frac{8}{5}\,\,ohm

Option B (

\frac{8}{5}\,ohm

) is correct because it is the equivalent resistance of the parallel combination calculated using the given resistance ratio.

Solution approach

Correct Answer + Core Reason: $\frac{8}{5}\,ohm$ . The equivalent resistance is calculated using the parallel combination formula and the resistance ratio.

Step-by-Step Solution:

Step 1: Let $R_1$ be the resistance of the thinner wire and $R_2$ be the resistance of the thicker wire. Given $R_1 = 8\,\Omega$ .

Step 2: Since the wires are of the same material and length, the resistance is inversely proportional to the area. The area is proportional to the square of the thickness (or radius). If one wire has half the thickness of the other, then $A_1 = \frac{1}{2} A_2$ , where $A_1$ and $A_2$ are the cross-sectional areas of the thinner and thicker wires, respectively. Therefore, $R_1 = 2R_2$ .

Step 3: So, $R_2 = \frac{R_1}{2} = \frac{8}{2} = 4\,\Omega$ .

Step 4: The equivalent resistance of the parallel combination is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{8} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$ .

Step 5: Therefore, $R_{eq} = \frac{8}{3}\,\Omega$ . However, the problem states that one wire has HALF the thickness, not half the area. If the radius of the thinner wire is half the radius of the thicker wire, then the area of the thinner wire is 1/4 the area of the thicker wire. So $A_1 = \frac{1}{4} A_2$ , and $R_1 = 4 R_2$ . Thus, $R_2 = \frac{R_1}{4} = \frac{8}{4} = 2 \Omega$ .

Step 6: The equivalent resistance of the parallel combination is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{8} + \frac{1}{2} = \frac{1}{8} + \frac{4}{8} = \frac{5}{8}$ .

Step 7: Therefore, $R_{eq} = \frac{8}{5}\,\Omega$ .

Why other options are wrong:
- **Option A ( $\frac{5}{8}\,ohm$ ): This is incorrect because it calculates the reciprocal of the equivalent resistance but doesn't take the inverse to find the actual equivalent resistance.
- Option C ( $\frac{3}{8}\,ohm$ ): This is incorrect because it uses an incorrect resistance ratio between the two wires.
- Option D ( $\frac{8}{3}\,ohm$ ): This is incorrect because it assumes the area is half, not the radius.

Key Concepts:**
- Resistance and area relationship: $R \propto \frac{1}{A}$
- Parallel combination of resistors: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$

Memory Aids:
- Remember that in a parallel combination, the equivalent resistance is always less than the smallest individual resistance.

Quick Tips:
- Pay close attention to whether the problem states the ratio of thickness, radius, or area.

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Understanding +4 | -1

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

.
Assertion A: Alloys such as constantan and manganin are used in making standard resistance coils.
Reason R: Constantan and manganin have very small value of temperature coefficient of resistance. In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: A

Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

.

Option A (Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

) is correct because constantan and manganin are used in standard resistance coils precisely because they have a very small temperature coefficient of resistance, ensuring stable resistance values.

Solution approach

Correct Answer + Core Reason: Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$ . Low temperature coefficient ensures stable resistance.

Step-by-Step Solution:

Step 1: Understand the purpose of standard resistance coils: To provide a stable and accurate resistance value.

Step 2: Recall the properties of constantan and manganin: They have very low temperature coefficients of resistance.

Step 3: A low temperature coefficient means the resistance changes very little with temperature variations.

Step 4: Therefore, using constantan and manganin ensures that the resistance of the coil remains stable, making them suitable for standard resistance coils.

Why other options are wrong:
- Option B: The reason is the correct explanation for the assertion.
- Option C: Both assertion and reason are true.
- Option D: Both assertion and reason are true.

Key Concepts:
- Temperature coefficient of resistance
- Standard resistance coils
- Properties of alloys

Memory Aids:
- Constantan and Manganin: Low temperature coefficient = Stable resistance.

Quick Tips:
- Remember that a low temperature coefficient is crucial for precision resistors.

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Remembering +4 | -1

For a given thermocouple neutral temperature

View answer and explanation

Correct option: A

Is a constant

Option A (Is a constant) is correct because the neutral temperature of a thermocouple is a characteristic constant for a given pair of metals.

Solution approach

Correct Answer + Core Reason: Is a constant. Neutral temperature is a constant for a given thermocouple.

Step-by-Step Solution:

Step 1: Define Thermocouple: A thermocouple is a sensor that measures temperature based on the Seebeck effect.

Step 2: Define Neutral Temperature: The neutral temperature is the temperature at which the thermoelectric EMF is maximum.

Step 3: Property of Neutral Temperature: For a given thermocouple, the neutral temperature is a constant.

Why other options are wrong:
- Option B: The neutral temperature does not depend on the cold junction temperature.
- Option C: The neutral temperature is related to, but not directly dependent on, the inversion temperature in a simple way.
- Option D: There is no direct relationship stating the neutral temperature is double the cold junction temperature.

Key Concepts:
- Thermocouple
- Seebeck Effect
- Neutral Temperature

Memory Aids:
- Neutral Temperature = Constant (for a given thermocouple)

Quick Tips:
- Understand the properties of thermocouples and their characteristic temperatures.

16 Medium

16 of 191Question permalink

Application +4 | -1

A resistor has a colour code of green, blue, brown and silver. What is its resistance?

View answer and explanation

Correct option: C

560\,\,\mathrm{\Omega }\,\, \pm \,\,10\,\,\mathrm{\% }

Option C (

560\,\mathrm{\Omega }\,\, \pm \,\,10\,\mathrm{\% }

) is correct because green is 5, blue is 6, brown is a multiplier of 10¹, and silver is a tolerance of 10%.

Solution approach

Correct Answer + Core Reason: $560\,\mathrm{\Omega }\,\, \pm \,\,10\,\mathrm{\% }$ . The color code determines the resistance value and tolerance.

Step-by-Step Solution:

Step 1: Identify the color code:\nGreen, Blue, Brown, Silver

Step 2: Determine the numerical value for each color:\nGreen = 5\nBlue = 6\nBrown = 1 (multiplier, i.e., 10¹)\nSilver = 10% (tolerance)

Step 3: Combine the values to find the resistance:\nResistance = (56) * 10¹ = 560 Ω

Step 4: Add the tolerance:\nTolerance = 10%

Final Answer: $560\,\mathrm{\Omega }\,\, \pm \,\,10\,\mathrm{\% }$

Why other options are wrong:
- Option A: Incorrect resistance value and tolerance.
- Option B: Incorrect tolerance.
- Option D: Incorrect resistance value and tolerance.

Key Concepts:
- Resistor color code
- Resistance value
- Tolerance

Memory Aids:
- Use the mnemonic "BB ROY G. B. V. G. W." (Black, Brown, Red, Orange, Yellow, Green, Blue, Violet, Grey, White) to remember the color code values.

Quick Tips:
- Always start reading the color bands from the end that is closest to a lead.
- Remember the tolerance values for common colors (Gold = 5%, Silver = 10%, No color = 20%).

17 Easy

17 of 191Question permalink

Application +4 | -1

The emf of a battery A is balanced by a length of 80 cm on a potentiometer wire. The emf of a standard cell 1 V is balanced by 50 cm. The emf of A is

View answer and explanation

Correct option: D

1.6 V

Option D (1.6 V) is correct because the emf is directly proportional to the balancing length in a potentiometer experiment.

Solution approach

Correct Answer + Core Reason: 1.6 V. The emf of a cell is proportional to the balancing length on a potentiometer.

Step-by-Step Solution:

Step 1: Understand the principle of potentiometer. The ratio of emfs is equal to the ratio of balancing lengths.

Step 2: Write the formula: $\frac{{E_A}}{{E_{std}}} = \frac{{l_A}}{{l_{std}}}$ where $E_A$ is the emf of battery A, $E_{std}$ is the emf of the standard cell, $l_A$ is the balancing length for battery A, and $l_{std}$ is the balancing length for the standard cell.

Step 3: Substitute the given values: $\frac{{E_A}}{{1 \text{{ V}}}} = \frac{{80 \text{{ cm}}}}{{50 \text{{ cm}}}}$

Step 4: Calculate $E_A$ : $E_A = 1 \text{{ V}} \times \frac{{80}}{{50}} = 1.6 \text{{ V}}$

Why other options are wrong:
- Option A (2 V): This is incorrect. It might be obtained by incorrectly dividing the lengths or misunderstanding the potentiometer principle.
- Option B (1.4 V): This is incorrect. It's likely a calculation error or a misunderstanding of the ratio.
- Option C (1.5 V): This is incorrect. It's likely a calculation error or a misunderstanding of the ratio.

Key Concepts:
- Potentiometer principle
- EMF measurement
- Balancing length

Memory Aids:
- EMF is directly proportional to the balancing length.

Quick Tips:
- Remember that the potentiometer compares potential differences, not currents.

18 Hard

18 of 191Question permalink

Application +4 | -1

Resistance of a wire at

20\,\,{\,^0}C

is

20\,\,\,\Omega

and at

500\,\,{\,^0}C

C is

60\,\,\,\Omega

. At what temperature its resistance is

25\,\,\,\Omega

?

View answer and explanation

Correct option: D

80\,{\,^0}\mathrm{C}

Option D (

80\,{\,^0}\mathrm{C}

) is correct because it is the temperature calculated using the linear relationship between resistance and temperature.

Solution approach

Correct Answer + Core Reason: $80\,{\,^0}\mathrm{C}$ . The temperature is calculated using the linear relationship between resistance and temperature.

Step-by-Step Solution:

Step 1: The resistance of a material varies with temperature according to the equation $R_T = R_0 [1 + \alpha (T - T_0)]$ , where $R_T$ is the resistance at temperature $T$ , $R_0$ is the resistance at temperature $T_0$ , and $\alpha$ is the temperature coefficient of resistance.

Step 2: Let $R_1 = 20\,\Omega$ at $T_1 = 20\,{\,^0}C$ and $R_2 = 60\,\Omega$ at $T_2 = 500\,{\,^0}C$ . We can write:
$60 = 20 [1 + \alpha (500 - 20)]$
$3 = 1 + 480\alpha$
$2 = 480\alpha$
$\alpha = \frac{2}{480} = \frac{1}{240}\,{\,^0}C^{-1}$

Step 3: Now, let $R_3 = 25\,\Omega$ at temperature $T_3$ . We can write:
$25 = 20 [1 + \alpha (T_3 - 20)]$
$\frac{25}{20} = 1 + \frac{1}{240} (T_3 - 20)$
$1.25 = 1 + \frac{1}{240} (T_3 - 20)$
$0.25 = \frac{1}{240} (T_3 - 20)$
$0.25 \times 240 = T_3 - 20$
$60 = T_3 - 20$
$T_3 = 80\,{\,^0}C$

Why other options are wrong:
- **Option A ( $160\,{\,^0}\mathrm{C}$ ): This is incorrect because it uses an incorrect value for the temperature coefficient of resistance.
- Option B ( $250\,{\,^0}\mathrm{C}$ ): This is incorrect because it uses an incorrect value for the temperature coefficient of resistance.
- Option C ( $100\,{\,^0}\mathrm{C}$ ): This is incorrect because it uses an incorrect value for the temperature coefficient of resistance.

Key Concepts:**
- Temperature dependence of resistance: $R_T = R_0 [1 + \alpha (T - T_0)]$

Memory Aids:
- Remember the linear relationship between resistance and temperature.

Quick Tips:
- Use the given data to find the temperature coefficient of resistance first.

19 Medium

19 of 191Question permalink

Analyzing +4 | -1

Assertion: Terminal potential difference of a cell is always less than its emf.
Reason: Potential drop due to internal resistance of cell increases the terminal potential difference.

View answer and explanation

Correct option: D

If both Assertion & Reason are False

Option D (If both Assertion & Reason are False) is correct because the terminal potential difference is usually *less* than the EMF due to internal resistance, and the potential drop *decreases* the terminal potential difference.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are False. The terminal potential difference is usually less than the EMF, and the potential drop decreases the terminal potential difference.

Step-by-Step Solution:

Step 1: Understand the problem. We need to evaluate the assertion and reason regarding the terminal potential difference and EMF of a cell.

Step 2: Recall the relationship between terminal potential difference (V), EMF (E), internal resistance (r), and current (I): $V = E - Ir$ .

Step 3: Analyze the Assertion: The assertion states that the terminal potential difference is *always* less than its EMF. This is generally true because of the voltage drop across the internal resistance. However, it's not *always* true. If no current is drawn from the cell (open circuit), then I = 0, and V = E. So, the assertion is false because it says *always* less.

Step 4: Analyze the Reason: The reason states that the potential drop due to internal resistance *increases* the terminal potential difference. This is incorrect. The potential drop *decreases* the terminal potential difference, as shown by the equation $V = E - Ir$ .

Step 5: Since both the assertion and the reason are false, the correct answer is D.

Why other options are wrong:
- Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion): This is incorrect because both the assertion and the reason are false.
- Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion): This is incorrect because both the assertion and the reason are false.
- Option C (If Assertion is True but the Reason is False): This is incorrect because the assertion is also false.

Key Concepts:
- EMF of a cell
- Internal resistance of a cell
- Terminal potential difference
- Ohm's law

Memory Aids:
- Remember that the terminal voltage is always less than or equal to the EMF.

Quick Tips:
- Understand the equation $V = E - Ir$ to relate terminal voltage, EMF, current, and internal resistance.

20 Easy

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Application +4 | -1

When a current

I

is set up in a wire of radius

r

, the drift velocity is

\mathrm{v}_{\mathrm{d}}

. If the same current is set up through a wire of radius

2 \mathrm{r}

, the drift velocity will be

View answer and explanation

Correct option: D

\mathrm{v}_{\mathrm{d}} / 4

Option D (

\mathrm{v}_{\mathrm{d}} / 4

) is correct because the drift velocity is inversely proportional to the cross-sectional area of the wire, which is proportional to the square of the radius.

Solution approach

Correct Answer + Core Reason: $\mathrm{v}_{\mathrm{d}} / 4$ . The drift velocity is inversely proportional to the area of cross-section of the wire.

Step-by-Step Solution:

Step 1: Recall the relationship between current, drift velocity, and cross-sectional area: $I = nAqv_d$ , where $I$ is the current, $n$ is the number density of charge carriers, $A$ is the cross-sectional area, $q$ is the charge of each carrier, and $v_d$ is the drift velocity.

Step 2: Since the current $I$ is the same in both wires, we can write: $I = nA_1qv_{d1} = nA_2qv_{d2}$

Step 3: The cross-sectional area of a wire is given by $A = \pi r^2$ , where $r$ is the radius of the wire.\nSo, $A_1 = \pi r^2$ and $A_2 = \pi (2r)^2 = 4\pi r^2$

Step 4: Substitute the areas into the equation from Step 2: $n(\pi r^2)qv_{d1} = n(4\pi r^2)qv_{d2}$

Step 5: Simplify the equation: $v_{d1} = 4v_{d2}$

Step 6: Solve for $v_{d2}$ : $v_{d2} = \frac{{v_{d1}}}{4}$

Step 7: Final Answer: The drift velocity in the wire of radius $2r$ is $\frac{{v_d}}{4}$ .

Why other options are wrong:
- Option A: This option is incorrect because it assumes the drift velocity is directly proportional to the radius, which is not true.
- Option B: This option is incorrect because it assumes the drift velocity is inversely proportional to the radius, which is not true.
- Option C: This option is incorrect because it assumes the drift velocity is inversely proportional to the radius, but only by a factor of 2.

Key Concepts:
- Drift velocity: $I = nAqv_d$
- Area of a circle: $A = \pi r^2$

Memory Aids:
- Remember the relationship between current and drift velocity: $I = nAqv_d$

Quick Tips:
- Pay attention to the relationship between area and radius.

21 Medium

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Understanding +4 | -1

For comparing the e.m.f.'s of two cells with a potentiometer, a standard cell is used to develop a potential gradient along the wires. Which of the following possibilities would make the experiment unsuccessful

View answer and explanation

Correct option: D

The e.m.f. of the standard cell is smaller than the e.m.f.'s of the two cells

Option D (The e.m.f. of the standard cell is smaller than the e.m.f.'s of the two cells) is correct because the potential gradient developed must be higher than the emf of the cells being compared for a balance point to be obtained.

Solution approach

Correct Answer + Core Reason: The e.m.f. of the standard cell is smaller than the e.m.f.'s of the two cells. The potential gradient developed must be higher than the emf of the cells being compared.

Step-by-Step Solution:

Step 1: Explain the working principle of a potentiometer: A potentiometer compares the emf of two cells by balancing the potential drop across a length of a uniform wire against the emf of the cell. The potential gradient along the wire must be sufficient to balance the emf of the cells being compared.

Step 2: If the emf of the standard cell is smaller than the emf of the cells being compared, the potential gradient developed along the wire will be insufficient to balance the emf of the cells. In this case, no balance point will be obtained.

Step 3: Therefore, the experiment will be unsuccessful if the emf of the standard cell is smaller than the emf of the two cells.

Why other options are wrong:
- **Option A (The e.m.f. of the standard cell is larger than the $E$ e.m.f.'s the two cells): This is a valid condition for the experiment to work.
- Option B (The diameter of the wires is the same and uniform throughout): This ensures a uniform potential gradient, which is a requirement for the experiment.
- Option C (The number of wires is ten): The number of wires does not affect the success of the experiment, as long as the total length is sufficient.

Key Concepts:
- Potentiometer
- Electromotive force (emf)
- Potential gradient

Memory Aids:
- Standard cell emf must be greater than the unknown emf for successful comparison.

Quick Tips:**
- Ensure the standard cell emf is greater than the emf being measured.

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Understanding +4 | -1

The potentiometer is superior to a voltmeter for measuring a potential difference because

View answer and explanation

Correct option: B

The potentiometer does not draw any current from the source of the potential

Option B (The potentiometer does not draw any current from the source of the potential) is correct because the potentiometer uses a null method, ensuring no current is drawn from the source, leading to a more accurate measurement.

Solution approach

Correct Answer + Core Reason: The potentiometer does not draw any current from the source of the potential. The potentiometer uses a null method, ensuring no current is drawn.

Step-by-Step Solution:

Step 1: Explain the working principle of a potentiometer: A potentiometer works on the principle of null deflection, meaning it measures the potential difference without drawing any current from the source. This is achieved by balancing the potential drop across a known length of a uniform wire against the potential difference being measured.

Step 2: Explain why a voltmeter is less accurate: A voltmeter draws some current from the circuit, which can alter the potential difference being measured, leading to a less accurate reading.

Step 3: Therefore, a potentiometer is superior for measuring potential difference because it does not draw current from the source of the potential.

Why other options are wrong:
- Option A (The resistance of the voltmeter): The resistance of the voltmeter is not the primary reason for the potentiometer's superiority.
- Option C (The sensitivity of potentiometer is better than that of the voltmeter): While true, the primary reason is the zero current draw.
- Option D (The voltmeter has a dial and of small size): Size and dial are not relevant to accuracy.

Key Concepts:
- Potentiometer
- Voltmeter
- Null deflection method

Memory Aids:
- Potentiometer: No current draw = Superior accuracy.

Quick Tips:
- Remember that a potentiometer uses a null deflection method and does not draw current.

23 Medium

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Remembering +4 | -1

In a thermocouple, the temperature that does not depend on the temperature of the cold junction is called

View answer and explanation

Correct option: A

Neutral temperature

The neutral temperature in a thermocouple does not depend on the temperature of the cold junction. It is a characteristic of the thermocouple materials.

Solution approach

Correct Answer + Core Reason: Neutral temperature. The neutral temperature is independent of the cold junction temperature.

Step-by-Step Solution:

Step 1: In a thermocouple, the thermoelectric EMF (electromotive force) generated depends on the temperature difference between the hot and cold junctions.

Step 2: The thermoelectric EMF is given by $E = a(T_H - T_C) + b(T_H^2 - T_C^2)$ , where $T_H$ is the hot junction temperature, $T_C$ is the cold junction temperature, and $a$ and $b$ are constants.

Step 3: The neutral temperature ( $T_n$ ) is the temperature at which the thermoelectric EMF is maximum. It is given by $T_n = -\frac{a}{{2b}}$ , which is independent of the cold junction temperature.

Step 4: The temperature of inversion is the temperature at which the thermoelectric EMF changes sign.

Why other options are wrong:
- Option B (Temperature of inversion): The temperature of inversion depends on the cold junction temperature.
- Option C (Both the above): Only the neutral temperature is independent of the cold junction temperature.
- Option D (None of the above): The neutral temperature is independent of the cold junction temperature.

Key Concepts:
- Thermocouple
- Thermoelectric EMF
- Neutral Temperature
- Temperature of Inversion

Memory Aids:
- Neutral temperature is a characteristic of the thermocouple materials.

Quick Tips:
- Remember the definitions of neutral temperature and temperature of inversion.

24 Medium

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Understanding +4 | -1

A current

I

is passed for a time

t

through a number of voltmeters. If

m

is the mass of a substance deposited on an electrode and

z

is its electrochemical equivalent, then

View answer and explanation

Correct option: A

\frac{{zIt}}{m} = \mathrm{constant}

Option A (

\frac{{zIt}}{m} = \mathrm{constant}

) is correct because Faraday's law of electrolysis states that the mass of a substance deposited is directly proportional to the charge (It) and the electrochemical equivalent (z).

Solution approach

Correct Answer + Core Reason: $\frac{{zIt}}{m} = \mathrm{constant}$ . This is a direct application of Faraday's law of electrolysis.

Step-by-Step Solution:

Step 1: State Faraday's law of electrolysis: $m = zIt$

Step 2: Rearrange the equation to show the constant relationship: $\frac{m}{{zIt}} = 1$ or $\frac{{zIt}}{m} = 1$

Step 3: Since the right side is a constant, the left side must also be a constant.

Why other options are wrong:
- Option B: This is the inverse relationship and doesn't align with Faraday's law.
- Option C: This also represents an incorrect inverse relationship.
- Option D: This is an incorrect arrangement of the variables.

Key Concepts:
- Faraday's laws of electrolysis
- Electrochemical equivalent (z)

Memory Aids:
- Remember the formula: m = zIt (Mass = z * Current * Time).

Quick Tips:
- Faraday's laws are fundamental to understanding electrolysis and electrochemistry.

25 Medium

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Understanding +4 | -1

Two bulbs, one of

50\,\,watt

and another of

25\,\,watt

are connected in series to the mains. The ratio of the currents through them is

View answer and explanation

Correct option: C

\begin{array}{llllllllllllllllllll} {\mathrm{1 :1}} \end{array}

Option C (1:1) is correct because when bulbs are connected in series, the current flowing through each bulb is the same, regardless of their power ratings.

Solution approach

Correct Answer + Core Reason: 1:1. In a series connection, the current is the same through all components.

Step-by-Step Solution:

Step 1: Understand the properties of a series circuit.\nIn a series circuit, the same current flows through all components.

Step 2: Since the bulbs are connected in series, the current through each bulb is the same.
$I_1 = I_2$

Step 3: Therefore, the ratio of the currents through them is 1:1.

Why other options are wrong:
- Option A: $\mathrm{2 :1}$ : This would be the ratio of currents if the voltage across the 50W bulb was twice that across the 25W bulb, which is not generally true in a series connection.
- Option B: $\mathrm{1 :2}$ : This would be the ratio of currents if the voltage across the 25W bulb was twice that across the 50W bulb, which is not generally true in a series connection.
- Option D: Without voltage, cannot be calculated: This is incorrect because the current is the same in a series circuit regardless of the voltage or power ratings of the individual components.

Key Concepts:
- Series Circuits
- Current in Series Circuits

Memory Aids:
- In series, current is the SAME.

Quick Tips:
- In a series circuit, the current is constant throughout the circuit.

26 Medium

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Remembering +4 | -1

Ampere hour is the unit of

View answer and explanation

Correct option: A

Quantity of charges

Option A (Quantity of charges) is correct because Ampere-hour (Ah) is a unit of electric charge, representing the amount of charge transferred by a current of one ampere flowing for one hour.

Solution approach

Correct Answer + Core Reason: Quantity of charges. Ampere-hour is the unit of charge, defined as the current multiplied by time.

Step-by-Step Solution:

Step 1: Ampere (A) is the unit of current, which is the rate of flow of charge.

Step 2: Hour (h) is the unit of time.

Step 3: Ampere-hour (Ah) is the product of current and time, i.e., $Q = I \times t$ , where $Q$ is the quantity of charge, $I$ is the current, and $t$ is the time.

Step 4: Therefore, Ampere-hour is the unit of quantity of charges.

Why other options are wrong:
- Option B: Potential is measured in volts (V).
- Option C: Energy is measured in joules (J) or kilowatt-hours (kWh).
- Option D: Current is measured in amperes (A).

Key Concepts:
- Electric charge
- Electric current

Memory Aids:
- $Q = It$ (Charge = Current * Time)

Quick Tips:
- Remember the relationship between charge, current, and time.

27 Medium

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Understanding +4 | -1

A metallic resistor is connected across a battery. If the number of collisions of the free electrons with the lattice is some how decreased in the resistor (for example by cooling it), the current will

View answer and explanation

Correct option: B

Increase

Option B (Increase) is correct because decreasing the number of collisions reduces resistance, leading to an increase in current according to Ohm's Law.

Solution approach

Correct Answer + Core Reason: Increase. Decreasing collisions reduces resistance, increasing current.

Step-by-Step Solution:

Step 1: Understand the relationship between collisions, resistance, and current.

Step 2: Key Concept: Ohm's Law: $V = IR$ , where $V$ is voltage, $I$ is current, and $R$ is resistance.

Step 3: Resistance is caused by collisions of electrons with the lattice. Fewer collisions mean lower resistance.

Step 4: Since the voltage is constant (battery), if resistance decreases, current must increase according to Ohm's Law.

Step 5: Conclusion: The current will increase.

Why other options are wrong:
- Option A: Remains constant: This is incorrect because a change in resistance will affect the current if the voltage is constant.
- Option C: Decrease: This is incorrect because decreasing collisions reduces resistance, which increases current, not decreases it.
- Option D: Become zero: This is incorrect because even with fewer collisions, there will still be some current flow unless the resistance becomes infinitely large.

Key Concepts:
- Ohm's Law: $V = IR$
- Resistance and collisions

Memory Aids:
- Ohm's Law triangle (V at the top, I and R at the bottom)

Quick Tips:
- Remember that resistance is directly proportional to the number of collisions.

28 Easy

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Application +4 | -1

The radius

(r)

, length

(l)

and resistance

(R)

of a metal wire was measured in the laboratory as

r=(0.35 \pm 0.05) \mathrm{cm}

R=(100 \pm 10) \mathrm{ohm}

l=(15 \pm 0.2) \mathrm{cm}

The percentage error in resistivity of the material of the wire is:

View answer and explanation

Correct option: B

39.9 \%

Option B (

39.9 \%

) is correct because it is the percentage error in resistivity calculated using the given errors in radius, length, and resistance.

Solution approach

Correct Answer + Core Reason: $39.9 \%$ . The percentage error in resistivity is calculated using the formula for error propagation.

Step-by-Step Solution:

Step 1: Identify the given information:
- Radius, $r=(0.35 \pm 0.05) \mathrm{cm}$
- Resistance, $R=(100 \pm 10) \mathrm{ohm}$
- Length, $l=(15 \pm 0.2) \mathrm{cm}$

Step 2: Write the formula for resistivity: $\rho = \frac{RA}{l} = \frac{R \pi r^2}{l}$

Step 3: Calculate the percentage error in resistivity:
- $\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta r}{r} + \frac{\Delta l}{l}$

Step 4: Substitute the given values:
- $\frac{\Delta \rho}{\rho} = \frac{10}{100} + 2 \frac{0.05}{0.35} + \frac{0.2}{15}$

Step 5: Calculate the percentage error:
- $\frac{\Delta \rho}{\rho} = 0.1 + 2 \times 0.1428 + 0.0133$
- $\frac{\Delta \rho}{\rho} = 0.1 + 0.2856 + 0.0133 = 0.3989$

Step 6: Convert to percentage:
- Percentage error $= 0.3989 \times 100 \% = 39.89 \% \approx 39.9 \%$

Final Answer: $39.9 \%$

Why other options are wrong:
- Option A: $35.6 \%$ : This is incorrect due to errors in calculation or not considering the power of 2 for the radius.
- Option C: $25.6 \%$ : This is incorrect due to errors in calculation or not considering the error in radius.
- Option D: $37.3 \%$ : This is incorrect due to errors in calculation.

Key Concepts:
- Resistivity
- Error propagation

Memory Aids:
- Remember to multiply the percentage error in radius by 2 because it is squared in the formula.

Quick Tips:
- Always remember to include all the error terms in the formula.
- Ensure correct calculation of percentage errors.

29 Medium

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Application +4 | -1

A current of

\mathrm{1}\mathrm{.5}\,\,\mathrm{A}

flows through a copper voltameter. The thickness of copper deposited on the electrode surface of size

50\,\,\mathrm{cm}\,\, \times \,10\,\,\mathrm{cm}

is

\mathrm{20}\,\,\mathrm{min}

will be (density of copper

= 9000\,\,\mathrm{kg} - {\mathrm{m}^{ - 3}}\,\,\mathrm{and}\,\mathrm{ ECE }\,\mathrm{of}\,\mathrm{ copper} = 0.00033\,\,\mathrm{g}\,\,{\mathrm{C}^{ - 1}})

View answer and explanation

Correct option: B

1.32{\mkern 1mu} {\mkern 1mu} \times {\mkern 1mu} {\mkern 1mu} {10^{ - 6}}\,\,m

Option B (

1.32 \times {10^{ - 6}}\,m

) is correct because it is the calculated thickness of copper deposited using Faraday's laws of electrolysis.

Solution approach

Correct Answer + Core Reason: $1.32{\mkern 1mu} {\mkern 1mu} \times {\mkern 1mu} {\mkern 1mu} {10^{ - 6}}\,m$ . The thickness is calculated using Faraday's laws of electrolysis and the given parameters.

Step-by-Step Solution:

Step 1: Identify the given parameters:\nCurrent, $I = 1.5 \,A$ \nTime, $t = 20 \,\text{min} = 20 \times 60 = 1200 \,s$ \nArea, $A = 50 \,\text{cm} \times 10 \,\text{cm} = 500 \,\text{cm}^2 = 0.05 \,m^2$ \nDensity, $\rho = 9000 \,\text{kg/m}^3$ \nECE, $Z = 0.00033 \,\text{g/C} = 0.00033 \times 10^{-3} \,\text{kg/C}$

Step 2: Calculate the mass of copper deposited using Faraday's law:
$m = ZIt$

Step 3: Substitute the values:
$m = (0.00033 \times 10^{-3} \,\text{kg/C}) \times (1.5 \,A) \times (1200 \,s)$

Step 4: Calculate the mass:
$m = 0.00033 \times 1.5 \times 1200 \times 10^{-3} \,\text{kg} = 0.000594 \,\text{kg}$

Step 5: Calculate the volume of the deposited copper:
$V = \frac{m}{\rho} = \frac{0.000594 \,\text{kg}}{9000 \,\text{kg/m}^3} = 6.6 \times 10^{-8} \,\text{m}^3$

Step 6: Calculate the thickness of the copper layer:
$V = A \times h$ , where $h$ is the thickness.
$h = \frac{V}{A} = \frac{6.6 \times 10^{-8} \,\text{m}^3}{0.05 \,\text{m}^2} = 1.32 \times 10^{-6} \,\text{m}$

Step 7: Final Answer:
$h = 1.32 \times 10^{-6} \,\text{m}$

Why other options are wrong:
- Option A: This value is incorrect due to an error in the calculation or unit conversion.
- Option C: This value is incorrect due to an error in the calculation or unit conversion.
- Option D: This value is incorrect due to an error in the calculation or unit conversion.

Key Concepts:
- Faraday's Laws of Electrolysis
- Mass deposited is proportional to current and time
- Density, Volume, and Mass relationship

Memory Aids:
- Remember Faraday's law: m = ZIt
- Density = Mass/Volume

Quick Tips:
- Ensure consistent units (SI units) throughout the calculation.
- Pay attention to unit conversions (cm to m, g to kg).

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Understanding +4 | -1

Potentiometer measures the potential difference more accurately than a voltmeter because

View answer and explanation

Correct option: C

It does not draw current from external circuit

Option C (It does not draw current from external circuit) is correct because a potentiometer uses a null method, ensuring no current is drawn from the source being measured, leading to a more accurate potential difference measurement.

Solution approach

Correct Answer + Core Reason: It does not draw current from external circuit. A potentiometer uses a null method, ensuring no current is drawn, leading to accurate measurement.

Step-by-Step Solution:

Step 1: Explain the working principle of a potentiometer: A potentiometer works on the principle of null deflection, meaning it measures the potential difference without drawing any current from the source. This is achieved by balancing the potential drop across a known length of a uniform wire against the potential difference being measured.

Step 2: Explain why a voltmeter is less accurate: A voltmeter draws some current from the circuit, which can alter the potential difference being measured, leading to a less accurate reading.

Step 3: Therefore, a potentiometer is preferred for measuring potential difference because it does not draw current from the external circuit.

Why other options are wrong:
- Option A (It has a wire of high resistance): The resistance of the wire is not the primary reason for accuracy.
- Option B (It has a wire of low resistance): The resistance of the wire is not the primary reason for accuracy.
- Option D (It draws a heavy current from external circuit): Incorrect because it does not draw current.

Key Concepts:
- Potentiometer
- Voltmeter
- Null deflection method

Memory Aids:
- Potentiometer: No current draw = Accurate measurement.

Quick Tips:
- Remember that a potentiometer uses a null deflection method and does not draw current.

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