CUET UG Physics Booster Test - 3 Material Properties and Power
📌 Answers are locked once submitted — results and explanations appear at the end.
QUESTION 1 OF 20
Match List I with List II regarding fundamental deviations from basic Ohm's Law.
| List I | List II |
|---|---|
| 1. Linear Ohm's Law behavior | a. A single current value intersects the curve at multiple distinct voltage values |
| 2. Non-linear deviation behavior | b. Changing the voltage polarity radically alters the resulting current magnitude |
| 3. Sign-dependent relation behavior | c. The measured voltage completely ceases to be strictly proportional to current |
| 4. Non-unique V-I relation behavior | d. Voltage scales perfectly and infinitely proportional to current |
QUESTION 2 OF 20
Identify the correct statements analyzing non-ohmic sign-dependent conduction phenomena.
Statements:
1. Forward and reverse biases of the same magnitude can produce vastly different currents.
2. Such asymmetric behavior violates the linear proportionality of Ohm's law.
3. This characteristic prevents the device from dissipating heat.
QUESTION 3 OF 20
If a diode produces a forward current I_0 at +V_0 and a reverse current I_r at -V_0, which condition confirms its non-ohmic nature?
QUESTION 4 OF 20
Identify the correct statements regarding the V-I characteristics of Gallium Arsenide (GaAs).
Statements:
1.An increase in voltage can sometimes cause a decrease in current.
2.A single current value may correspond to multiple voltage values.
3.GaAs exhibits non-linear and non-ohmic V-I characteristics in certain regions.
QUESTION 5 OF 20
A copper wire has a cross-sectional area of 1.0×10^(-7) m^2 and carries a current of 1.5 A. Given:
n=8.5×10^(28) m^(-3)e=1.6×10^(-19) C
Calculate the average drift speed of electrons.
QUESTION 6 OF 20
Comparing the foundational electrical characteristics of diverse bulk materials, insulators like dense ceramic and vulcanized rubber differ fundamentally from standard conductors in that:
QUESTION 7 OF 20
Identify the correct statement regarding the temperature coefficient of resistivity α.
QUESTION 8 OF 20
Identify the mathematically incorrect statement regarding the formula
ρ_T=ρ_0[1+α(T-T_0)]
QUESTION 9 OF 20
Identify the correct statements explaining why metals exhibit a positive temperature coefficient.
Statements:
1. The number density n of free electrons is nearly independent of temperature.
2. Increasing temperature raises the random thermal speed of electrons.
3. Higher thermal speeds lead to more frequent collisions and a smaller relaxation time τ.
QUESTION 10 OF 20
Identify the correct statements regarding metallic resistivity at temperatures far below 0^∘C.
Statements:
1. The simple linear approximation using αfails.
2. The resistivity-temperature graph deviates significantly from a straight line.
3. Resistivity becomes completely independent of temperature.
QUESTION 11 OF 20
For a specialized heating alloy like nichrome, if its measured fractional change in resistivity is extremely small over a substantially large ΔT, its intrinsic temperature coefficient α must mathematically satisfy the condition:
QUESTION 12 OF 20
Why exactly are specialized heavy alloys like manganin and constantan specifically mandated for constructing precision wire bound standard resistors over pure low-resistance metals like pure copper?
QUESTION 13 OF 20
Identify the correct statements regarding microscopic conductivity mechanisms strictly within intrinsic semiconductors.
Statements:
1. The active carrier number density n increases heavily with a physical rise in ambient temperature.
2. This dramatic and rapid increase in mobile n overwhelmingly outpaces the slight subsequent decrease in mean relaxation time τ.
3. The structural lack of bound electrons completely mimics an electrolytic solution.
QUESTION 14 OF 20
If an unknown semiconducting material heavily exhibits a sharp resistance drop such that its overall macroscopic conductivity σ doubles when actively heated from T₁ to T₂, and analytically assuming the relaxation time τ dropped by precisely 20% during this interval, by what exact factor did the intrinsic carrier density n forcefully increase?
QUESTION 15 OF 20
In a closed steady direct current circuit, the measured macroscopic electric potential steadily and uniformly decreases along a uniform wire directly in the path of the current, which fundamentally and physically implies that
QUESTION 16 OF 20
Identify the physically incorrect statement analyzing the potential energy change ΔUₚₒₜ for a drifting charge ΔQ traveling continuously from higher potential location A to lower potential location B.
QUESTION 17 OF 20
Identify the correct statements regarding the microscopic kinetic energy dynamics of drifting lattice electrons.
Statements:
1. A theoretical conservation of total energy utterly without collisions would mistakenly imply ΔK = −ΔUₚₒₜ > 0.
2. Due strictly to continuous fixed-ion collisions, the drifting electrons do not gain macroscopic kinetic energy indefinitely.
3. The energy transferred dynamically during these countless collisions inherently causes the local atoms to vibrate much more vigorously.
QUESTION 18 OF 20
Critically evaluating the foundational heat dissipation relation ΔW = IVΔt, identify the correct statements.
Statements:
1. It conceptually relies on the firm assumption that cascading drifting electrons maintain a steady terminal average velocity due to lattice scattering.
2. It directly represents the exact raw energy aggressively gained by the atomic lattice due to inelastic charge interactions.
3. The analytical quantity ΔW scales completely quadratically with applied voltage if the underlying resistance is assumed perfectly constant.
QUESTION 19 OF 20
The exact physical microscopic mechanism internally connecting the external macroscopic calculated power P = IV to the microscopic vibrating lattice domain is best and fully described as:
QUESTION 20 OF 20
When rigorously determining the optimal transmission voltage for safely moving a massive fixed electrical power P over long cables with significant resistance R₍c₎, why is scaling to a very high voltage highly and practically advantageous?
Test Complete!
Answer Review
1 Match List I with List II regarding fundamental deviations from basic Ohm's Law.
| List I | List II |
|---|---|
| 1. Linear Ohm's Law behavior | a. A single current value intersects the curve at multiple distinct voltage values |
| 2. Non-linear deviation behavior | b. Changing the voltage polarity radically alters the resulting current magnitude |
| 3. Sign-dependent relation behavior | c. The measured voltage completely ceases to be strictly proportional to current |
| 4. Non-unique V-I relation behavior | d. Voltage scales perfectly and infinitely proportional to current |
�� Ohmic conductors show linear V-I behavior. �� Non-linear devices do not maintain proportionality. �� Some devices show polarity-dependent conduction.
Ohm's law states that voltage is directly proportional to current, producing a straight-line V-I graph. Therefore, linear Ohm's law behavior corresponds to voltage remaining proportional to current. Non-linear deviation occurs when this proportionality breaks down. Sign-dependent behavior is observed in devices such as diodes where reversing the voltage polarity changes the current drastically. Non-unique V-I behavior occurs in materials such as GaAs where a single current value may correspond to multiple voltage values. Hence the correct matching is: 1-d, 2-c, 3-b, 4-a
- �� Option A → Non-linear and non-unique behaviors are interchanged.
- �� Option B → Linear and sign-dependent relations are mismatched.
- �� Option D → Multiple physical interpretations are incorrect.
Used – Concept Application
- Application
- Match each type of V-I behavior with its defining characteristic.
- Final Logic
- Linear → proportional, Non-linear → non-proportional, Sign-dependent → polarity sensitive, Non-unique → multiple voltages for same current.
- Non-Unique → GaAs
2 Identify the correct statements analyzing non-ohmic sign-dependent conduction phenomena.
Statements:
1. Forward and reverse biases of the same magnitude can produce vastly different currents.
2. Such asymmetric behavior violates the linear proportionality of Ohm's law.
3. This characteristic prevents the device from dissipating heat.
�� Diodes conduct differently in forward and reverse bias. �� Such behavior is non-ohmic. �� Heat dissipation still occurs.
In sign-dependent devices such as semiconductor diodes, reversing the polarity of the applied voltage changes the current dramatically. A forward bias permits substantial current flow, whereas a reverse bias allows only a very small current. This violates the simple linear relation between voltage and current assumed in Ohm's law. Statement 3 is incorrect because electrical power can still be dissipated as heat whenever current flows through the device.
- �� Option B → Statement 3 is incorrect.
- �� Option C → Statement 1 is correct.
- �� Option D → Statement 3 is incorrect.
Used – NCERT Recall
- Application
- Recall the behavior of semiconductor diodes.
- Final Logic
- Forward and reverse currents differ significantly, indicating non-ohmic behavior.
- Same Voltage, Different Current → Non-Ohmic
3 If a diode produces a forward current I_0 at +V_0 and a reverse current I_r at -V_0, which condition confirms its non-ohmic nature?
�� Ohmic devices show symmetric behavior. �� Diodes show asymmetric conduction. �� Current magnitudes differ for opposite voltages.
For an ohmic conductor, reversing the applied voltage simply reverses the current direction while maintaining the same magnitude. Thus equal positive and negative voltages produce equal current magnitudes. A diode behaves differently. Under forward bias it conducts strongly, while under reverse bias the current is extremely small. Therefore the magnitudes of forward and reverse currents are unequal. Hence the condition ∣I_r∣≠∣I_0∣ indicates non-ohmic behavior.
- �� Option A → Represents ideal symmetric behavior.
- �� Option C → Not generally true for diodes.
- �� Option D → Assumes Ohm's law.
Used – Concept Application
- Application
- Compare diode behavior with ideal ohmic conductors.
- Final Logic
- Unequal current magnitudes for opposite voltages indicate non-ohmic conduction.
- Diode → One-Way Preference
4 Identify the correct statements regarding the V-I characteristics of Gallium Arsenide (GaAs).
Statements:
1.An increase in voltage can sometimes cause a decrease in current.
2.A single current value may correspond to multiple voltage values.
3.GaAs exhibits non-linear and non-ohmic V-I characteristics in certain regions.
- GaAs can exhibit negative differential resistance.
- The V-I relation may be non-unique.
- Its V-I characteristics are non-linear and non-ohmic.
Gallium Arsenide (GaAs) is a semiconductor material that exhibits electrical behavior significantly different from that of ordinary metallic conductors. In certain operating regions, particularly those associated with negative differential resistance, an increase in the applied voltage can lead to a decrease in current. This unusual behavior is observed in devices such as Gunn diodes fabricated using GaAs.
Because the V-I characteristic is non-linear, a single current value may sometimes correspond to more than one voltage value. This means that the voltage-current relationship is not unique throughout the entire operating range. Such behavior differs fundamentally from the straight-line V-I graph observed for ohmic conductors.
Furthermore, GaAs does not obey Ohm's law globally because its resistance is not constant over all applied voltages. Instead, its V-I curve contains non-linear regions where the ratio V/I changes continuously. Therefore, all three statements correctly describe the electrical characteristics of Gallium Arsenide, making Option D the correct answer.
- Option A → Statements 1 and 2 are correct, but Statement 3 is also correct.
- Option B → Statements 2 and 3 are correct, but Statement 1 is also correct.
- Option C → Statements 1 and 3 are correct, but Statement 2 is also correct.
Application
Recall the NCERT discussion on non-ohmic conductors and the special V-I characteristics exhibited by materials such as Gallium Arsenide.
"GaAs: Same Current, Many Voltages" Remember that GaAs can exhibit non-linear, non-unique and non-ohmic V-I characteristics.
5 A copper wire has a cross-sectional area of 1.0×10^(-7) m^2 and carries a current of 1.5 A. Given:
n=8.5×10^(28) m^(-3)e=1.6×10^(-19) C
Calculate the average drift speed of electrons.
�� Use I=neAv_d. �� Solve for drift velocity. �� Drift speed is extremely small.
Using the microscopic current relation, I=neAv_d Therefore, v_d=I/neA Substituting the values, v_d=1.5/(8.5×10^(28))(1.6×10^(-19))(1.0×10^(-7))v_d≈1.1×10^(-3) m/s This extremely small drift speed demonstrates that large electric currents result from the enormous number of free electrons rather than from rapid electron motion.
- �� Option A → Calculation error.
- �� Option C → Overestimated value.
- �� Option D → Approximately the speed of light, physically impossible for drift speed.
Used – Substitution
- Application
- Apply the microscopic current equation.
- Final Logic
- v_d=I/neA=1.1×10^(-3) m/s
- Huge n → Tiny Drift Speed
6 Comparing the foundational electrical characteristics of diverse bulk materials, insulators like dense ceramic and vulcanized rubber differ fundamentally from standard conductors in that:
�� Insulators have extremely high resistivity. �� Electrons are tightly bound to atoms. �� Very few mobile charge carriers exist.
Insulators such as ceramic, rubber, glass, and plastics possess extremely high resistivities compared to metallic conductors. Their resistivities are often 10^(18)times greater than those of metals. This occurs because electrons remain tightly bound to their parent atoms and cannot move freely under ordinary electric fields. Since free charge carriers are practically absent, electrical conduction is extremely difficult. As a result, insulators are widely used for electrical isolation and protection. Therefore, option A correctly describes the fundamental difference between conductors and insulators.
- �� Option B → Insulators do not exhibit ideal Ohmic conduction over all conditions.
- �� Option C → Free electrons are not available in large numbers.
- �� Option D → Temperature dependence varies and is not the defining feature.
Used – NCERT Recall
- Application
- Recall the classification of materials based on resistivity and carrier availability.
- Final Logic
- Strongly bound electrons lead to extremely high resistivity.
- Insulator → Bound Electrons → High Resistivity
7 Identify the correct statement regarding the temperature coefficient of resistivity α.
�� αmeasures sensitivity of resistivity to temperature. �� Valid only over a limited temperature range. �� Most metals have positive α.
The temperature coefficient of resistivity αis used in the approximate relation ρ_T=ρ_0[1+α(T-T_0)] It represents the fractional change in resistivity per unit change in temperature. This approximation works only over a limited range of temperatures where the resistivity-temperature graph is approximately linear. For most metallic conductors, αis positive, meaning resistivity increases as temperature rises.
- �� Option B → No inverse-cube relation exists.
- �� Option C → Metals generally have positive α.
- �� Option D → Resistivity does not become zero at 0^∘C.
Used – Formula Recall
- Application
- Use the standard temperature dependence equation of resistivity.
- Final Logic
- αmeasures fractional resistivity change per degree.
- Alpha → Change per Degree
8 Identify the mathematically incorrect statement regarding the formula
ρ_T=ρ_0[1+α(T-T_0)]
�� The equation is linear. �� Valid only over a limited range. �� It reproduces ρ_0 at T=T_0.
The relation ρ_T=ρ_0[1+α(T-T_0)] is a first-order linear approximation describing how resistivity varies with temperature. It assumes approximately linear behavior around the reference temperature T_0. At T=T_0, ρ_T=ρ_0 which confirms the meaning of the reference resistivity. The equation is not intended to describe resistivity accurately over all temperatures and certainly does not represent a higher-order non-linear polynomial relation.
- �� Option A → Correct definition of ρ_0.
- �� Option C → Correct limitation of the formula.
- �� Option D → Directly follows from the equation.
Used – Formula Recall
- Application
- Examine each statement using the temperature-resistivity equation.
- Final Logic
- The equation is linear, not globally non-linear.
- At T_0→ ρ=ρ_0
9 Identify the correct statements explaining why metals exhibit a positive temperature coefficient.
Statements:
1. The number density n of free electrons is nearly independent of temperature.
2. Increasing temperature raises the random thermal speed of electrons.
3. Higher thermal speeds lead to more frequent collisions and a smaller relaxation time τ.
�� Free-electron density changes little. �� Thermal agitation increases. �� Collision frequency rises.
In metals, the number density of free electrons remains nearly constant with temperature. However, increasing temperature increases lattice vibrations and the random thermal motion of electrons. As the lattice vibrates more intensely, electrons experience more frequent collisions. This reduces the average relaxation time τ. Since conductivity depends on τ, σ=ne^2τ/m a decrease in τreduces conductivity and increases resistivity. Thus all three statements are correct.
- �� Option A → Statement 3 is also correct.
- �� Option B → Statement 1 is also correct.
- �� Option C → Statement 2 is also correct.
Used – Concept Application
- Application
- Connect microscopic collision theory with conductivity.
- Final Logic
- Higher temperature → More collisions → Lower τ→ Higher resistivity.
- Hot Metal → More Collisions → More Resistance
10 Identify the correct statements regarding metallic resistivity at temperatures far below 0^∘C.
Statements:
1. The simple linear approximation using αfails.
2. The resistivity-temperature graph deviates significantly from a straight line.
3. Resistivity becomes completely independent of temperature.
�� Linear approximation breaks down. �� The graph becomes non-linear. �� Resistivity still depends on temperature.
The relation ρ_T=ρ_0[1+α(T-T_0)] is only an approximate linear description valid over a moderate temperature range. At very low temperatures, experimental observations show significant deviations from this linear behavior. The resistivity-temperature curve bends away from a straight line, indicating that a constant value of αcan no longer accurately describe the material. Statement 3 is incorrect because resistivity does not suddenly become completely independent of temperature.
- �� Option B → Statement 3 is incorrect.
- �� Option C → Statement 2 is also correct.
- �� Option D → Statement 3 is incorrect.
Used – NCERT Recall
- Application
- Recall the limitations of the linear temperature-resistivity relation.
- Final Logic
- Low-temperature resistivity behavior becomes non-linear.
- Low Temperature → Linear Law Breaks
11 For a specialized heating alloy like nichrome, if its measured fractional change in resistivity is extremely small over a substantially large ΔT, its intrinsic temperature coefficient α must mathematically satisfy the condition:
�� Nichrome shows very little change in resistivity with temperature. �� Temperature coefficient α measures the rate of change of resistivity with temperature. �� A very small change in resistivity implies α is nearly zero.
The temperature dependence of resistivity for metallic conductors is expressed by the relation: ρ = ρ₀ [1 + α(T − T₀)] where ρ is the resistivity at temperature T, ρ₀ is the resistivity at reference temperature T₀, and α is the temperature coefficient of resistivity. The coefficient α indicates how strongly the resistivity changes with temperature. For ordinary metals such as copper, α has a significant positive value, causing resistivity to increase noticeably with temperature. Nichrome is a special alloy used in heating elements because its resistivity remains nearly constant even when subjected to large temperature variations. If the measured fractional change in resistivity remains extremely small despite a large increase in temperature, the multiplying factor α must itself be very small. Mathematically, only a value of α close to zero can produce negligible changes in resistivity over a substantial temperature range. This property makes nichrome suitable for heating coils, electric irons, and resistance wires because the resistance remains relatively stable during operation.
- �� Option A → A very large α would produce a huge change in resistivity with temperature.
- �� Option B → Nichrome does not possess a massive negative temperature coefficient.
- �� Option D → T₀/T is not the definition of temperature coefficient of resistivity.
Concept Application
- Application
- Use the relation between resistivity and temperature. If resistivity changes very little despite a large ΔT, the coefficient responsible for that change must be extremely small.
- Final Logic
- Very small fractional change in resistivity ⇒ very small α ⇒ α ≈ 0.
Remember: Nichrome → negligible resistivity variation → α nearly zero.
12 Why exactly are specialized heavy alloys like manganin and constantan specifically mandated for constructing precision wire bound standard resistors over pure low-resistance metals like pure copper?
�� Standard resistors require stable resistance values. �� Manganin and constantan have very small temperature coefficients. �� Their resistance remains nearly constant with temperature changes.
Precision resistors are expected to maintain a constant resistance value under varying environmental conditions. According to NCERT, alloys such as manganin and constantan are preferred for constructing standard resistors because their resistivity changes very little with temperature. This property is represented by a very small temperature coefficient of resistivity. When current passes through a resistor, heating occurs due to the Joule heating effect. If the material had a large temperature coefficient, the resistance would change significantly as temperature rises, causing measurement errors. Manganin and constantan overcome this limitation because their resistance remains almost unchanged over a wide temperature range. As a result, they are extensively used in resistance boxes, bridge circuits, laboratory standards, and precision electrical instruments. The stability of resistance is far more important for standard resistors than simply having low resistance. Therefore, these alloys are selected because they provide accurate and reproducible resistance values independent of ordinary temperature fluctuations.
- �� Option A → These alloys have a small, not enormous, temperature coefficient.
- �� Option C → They are conductors, not perfect insulators.
- �� Option D → Every conductor possesses finite resistance depending on dimensions and resistivity.
NCERT Recall
- Application
- Recall the NCERT discussion of materials used in standard resistors and their temperature-independent resistance characteristics.
- Final Logic
- Standard resistors require resistance stability ⇒ manganin and constantan provide nearly constant resistance.
Manganin → Measurement accuracy → Stable resistance.
13 Identify the correct statements regarding microscopic conductivity mechanisms strictly within intrinsic semiconductors.
Statements:
1. The active carrier number density n increases heavily with a physical rise in ambient temperature.
2. This dramatic and rapid increase in mobile n overwhelmingly outpaces the slight subsequent decrease in mean relaxation time τ.
3. The structural lack of bound electrons completely mimics an electrolytic solution.
�� Temperature increases electron-hole pair generation. �� Carrier density rises rapidly with temperature. �� Semiconductors do not behave like electrolytes.
In intrinsic semiconductors, conductivity depends strongly on the concentration of charge carriers. As temperature increases, more electrons gain sufficient energy to cross the forbidden energy gap from the valence band to the conduction band. This process creates additional electron-hole pairs and substantially increases the carrier density n. Although the relaxation time τ may decrease slightly because of increased lattice vibrations, the increase in carrier concentration is much greater. Consequently, the overall conductivity increases with temperature. This behavior is opposite to that observed in metallic conductors, where conductivity generally decreases as temperature rises. The third statement is incorrect because conduction in semiconductors occurs through electrons and holes moving within a crystal lattice. Electrolytic conduction involves ions moving through a solution. Therefore, semiconductor conduction and electrolytic conduction are fundamentally different mechanisms.
- �� Option B → Statement 3 is incorrect.
- �� Option C → Statement 3 is incorrect.
- �� Option D → All three statements are not correct because Statement 3 is false.
Concept Application
- Application
- Analyze each statement using the NCERT explanation of intrinsic semiconductor conductivity and carrier generation.
- Final Logic
- Statements 1 and 2 follow semiconductor theory, whereas Statement 3 incorrectly compares semiconductor conduction to electrolytic conduction.
Increasing temperature → more electron-hole pairs → higher conductivity.
14 If an unknown semiconducting material heavily exhibits a sharp resistance drop such that its overall macroscopic conductivity σ doubles when actively heated from T₁ to T₂, and analytically assuming the relaxation time τ dropped by precisely 20% during this interval, by what exact factor did the intrinsic carrier density n forcefully increase?
�� Conductivity depends on n and τ. �� Conductivity doubles. �� Relaxation time becomes 80% of original value.
For a semiconductor, σ ∝ nτ Let initial conductivity be σ₁ and final conductivity be σ₂. Given: σ₂ = 2σ₁ Also, relaxation time decreases by 20%: τ₂ = 0.8τ₁ Using the conductivity relation: σ₂/σ₁ = (n₂τ₂)/(n₁τ₁) Substituting values: 2 = (n₂ × 0.8τ₁)/(n₁τ₁) 2 = 0.8(n₂/n₁) n₂/n₁ = 2/0.8 n₂/n₁ = 2.5 Therefore, the carrier density must increase by a factor of 2.5. This result illustrates a key NCERT concept: in semiconductors, the rise in carrier concentration with temperature dominates over the reduction in relaxation time, producing an overall increase in conductivity. Unit Verification n₂/n₁ is a ratio of carrier densities and is dimensionless.
- �� Option A → Ignores the decrease in relaxation time.
- �� Option C → Does not satisfy the conductivity relation.
- �� Option D → Produces conductivity larger than the given value.
Substitution
- Application
- Substitute the given conductivity ratio and relaxation-time ratio directly into the conductivity equation.
- Final Logic
- 2 = (n₂/n₁)(0.8) ⇒ n₂/n₁ = 2.5.
σ doubles and τ becomes 0.8 ⇒ carrier factor = 2.5.
15 In a closed steady direct current circuit, the measured macroscopic electric potential steadily and uniformly decreases along a uniform wire directly in the path of the current, which fundamentally and physically implies that
�� Potential decreases along the direction of current. �� Potential gradient creates an electric field. �� Electric field causes electron drift.
According to NCERT, a steady current in a conductor requires the presence of an electric field inside the conductor. The electric field is related to the potential gradient through the relation E = −dV/dx. When the electric potential decreases continuously along a conductor, a non-zero electric field exists within the material. This electric field exerts force on free electrons, causing them to acquire a drift velocity opposite to the field direction. The collective drift of electrons constitutes electric current. The existence of a potential drop also indicates that electrical energy is continuously being converted into thermal energy because of collisions between electrons and lattice ions. Thus, a conductor carrying current is not an equipotential region. Instead, a finite potential gradient must exist to maintain the electric field required for current flow. This principle forms the basis of Ohm's law and explains why resistance causes a voltage drop in practical circuits.
- �� Option A → Current flow requires a non-zero electric field inside the conductor.
- �� Option C → Charges do not gain infinite potential energy.
- �� Option D → Energy is dissipated as heat due to resistance.
Concept Application
- Application
- Use the relationship between electric field and potential gradient to interpret the physical meaning of a voltage drop.
- Final Logic
- Potential decreases along conductor ⇒ electric field exists ⇒ field drives charge drift.
Potential drop always indicates the presence of an electric field.
16 Identify the physically incorrect statement analyzing the potential energy change ΔUₚₒₜ for a drifting charge ΔQ traveling continuously from higher potential location A to lower potential location B.
�� Potential energy decreases when charge moves from higher to lower potential. �� ΔUₚₒₜ is negative for a positive charge. �� Therefore, the change cannot always be positive.
The electric potential energy associated with a charge depends on the electric potential at its location. The change in potential energy is defined as: ΔUₚₒₜ = Ufinal − Uinitial For a charge ΔQ moving from point A to point B, ΔUₚₒₜ = ΔQ[V(B) − V(A)] When the charge moves from a higher potential to a lower potential, V(B) < V(A). Therefore, the quantity [V(B) − V(A)] becomes negative. As a result, the potential energy decreases and ΔUₚₒₜ becomes negative. This decrease in potential energy supplies the energy ultimately converted into heat within the conductor. For a steady current I flowing for time Δt, the charge transferred is ΔQ = IΔt. Using the voltage drop V = V(A) − V(B), ΔUₚₒₜ = −IVΔt Thus, options A, B and D are consistent with NCERT concepts of electric potential energy and energy dissipation. The incorrect statement is the assertion that the change in potential energy is always positive.
- �� Option A → Correct definition of potential energy change.
- �� Option B → Correct mathematical expression for potential energy change.
- �� Option D → Correct relation for steady current and voltage drop.
Concept Application
- Application
- Apply the definition of electric potential energy and determine the sign of ΔUₚₒₜ when potential decreases.
- Final Logic
- Higher potential → lower potential ⇒ potential energy decreases ⇒ ΔUₚₒₜ is negative.
Charge moving to lower potential loses potential energy.
17 Identify the correct statements regarding the microscopic kinetic energy dynamics of drifting lattice electrons.
Statements:
1. A theoretical conservation of total energy utterly without collisions would mistakenly imply ΔK = −ΔUₚₒₜ > 0.
2. Due strictly to continuous fixed-ion collisions, the drifting electrons do not gain macroscopic kinetic energy indefinitely.
3. The energy transferred dynamically during these countless collisions inherently causes the local atoms to vibrate much more vigorously.
�� Electric field accelerates electrons. �� Collisions prevent unlimited kinetic energy gain. �� Energy is transferred to lattice vibrations.
In a conductor, electrons are continuously accelerated by the internal electric field. If collisions were absent, the decrease in potential energy would be completely converted into kinetic energy, giving: ΔK = −ΔUₚₒₜ However, real conductors contain lattice ions that constantly scatter the drifting electrons. These frequent collisions prevent electrons from continuously accelerating and gaining unlimited kinetic energy. Instead, electrons attain a steady average drift velocity. During each collision, part of the energy acquired from the electric field is transferred to the lattice ions. This transferred energy increases the vibrational motion of the atoms in the conductor. The enhanced lattice vibration manifests as heating of the conductor, which is the microscopic origin of Joule heating. Thus, all three statements correctly describe the NCERT explanation of energy transfer and resistance in conducting materials.
- �� Option A → Statement 3 is also correct.
- �� Option B → Statement 1 is also correct.
- �� Option C → Statement 2 is also correct.
NCERT Recall
- Application
- Recall the microscopic explanation of current flow and Joule heating in conductors.
- Final Logic
- Electron acceleration + collisions + lattice heating ⇒ all three statements are correct.
Electric field supplies energy and lattice absorbs it as heat.
18 Critically evaluating the foundational heat dissipation relation ΔW = IVΔt, identify the correct statements.
Statements:
1. It conceptually relies on the firm assumption that cascading drifting electrons maintain a steady terminal average velocity due to lattice scattering.
2. It directly represents the exact raw energy aggressively gained by the atomic lattice due to inelastic charge interactions.
3. The analytical quantity ΔW scales completely quadratically with applied voltage if the underlying resistance is assumed perfectly constant.
�� Steady drift velocity results from repeated collisions. �� Electrical energy is converted into heat. �� For constant resistance, heat dissipation varies as V².
The heat produced in a resistor is given by: ΔW = IVΔt This expression assumes steady current flow, which means electrons possess a constant average drift velocity maintained by repeated collisions with lattice ions. Without these collisions, current would not remain steady. The energy supplied by the electric field is transferred through electron-ion interactions to the atomic lattice. Consequently, ΔW represents the energy converted into thermal energy within the conductor. Using Ohm's law: V = IR For constant resistance, P = IV = V²/R Therefore, ΔW = (V²/R)Δt This shows that heat generated varies as the square of the applied voltage when resistance remains constant. Hence all three statements accurately describe the physical interpretation of Joule heating.
- �� Option A → Statement 3 is also correct.
- �� Option B → Statement 1 is also correct.
- �� Option C → Statement 2 is also correct.
Concept Application
- Application
- Combine Joule's law with Ohm's law and the microscopic model of current flow.
- Final Logic
- Steady drift + lattice heating + V² dependence ⇒ all statements are correct.
Electron collisions convert electrical energy into heat.
19 The exact physical microscopic mechanism internally connecting the external macroscopic calculated power P = IV to the microscopic vibrating lattice domain is best and fully described as:
�� Electrical power is converted into heat. �� Electrons collide with lattice ions. �� Lattice vibrations increase.
Electrical power in a resistor is expressed as: P = IV At the microscopic level, the electric field performs work on free electrons. These electrons gain energy while drifting through the conductor. However, because of frequent collisions with lattice ions, the gained energy is repeatedly transferred to the lattice. This transferred energy increases the vibrational motion of the atoms within the conductor. The enhanced vibrations manifest as thermal energy, causing the conductor to heat up. This process is known as Ohmic loss or Joule heating. Therefore, the macroscopic power consumed by a resistor corresponds directly to the rate at which energy is transferred from drifting electrons to lattice vibrations. This microscopic interpretation provides the physical basis of power dissipation in resistive electrical devices.
- �� Option A → Magnetic flux accumulation is not responsible for Joule heating.
- �� Option B → Electron emission is unrelated to ordinary resistive power dissipation.
- �� Option D → Joule heating is irreversible, not perfectly reversible.
NCERT Recall
- Application
- Recall the microscopic explanation of electrical resistance and power dissipation.
- Final Logic
- Electron-lattice collisions convert electrical energy into thermal energy.
Electrical power ultimately appears as lattice vibrations.
20 When rigorously determining the optimal transmission voltage for safely moving a massive fixed electrical power P over long cables with significant resistance R₍c₎, why is scaling to a very high voltage highly and practically advantageous?
�� Transmission losses occur due to cable resistance. �� Higher voltage means lower current for the same power. �� Lower current greatly reduces heating losses.
For transmission of electrical power, P = VI Therefore, I = P/V Power lost in transmission cables is P₍c₎ = I²R₍c₎ Substituting I = P/V, P₍c₎ = (P/V)²R₍c₎ P₍c₎ = P²R₍c₎/V² This relation shows that transmission losses are inversely proportional to the square of the transmission voltage. If the voltage is doubled, the loss becomes one-fourth. If the voltage is increased ten times, the loss becomes one-hundredth. Therefore, electric power is transmitted at very high voltages using step-up transformers. The reduced current minimizes I²R losses, improves efficiency and decreases unnecessary heating of transmission lines.
- �� Option B → Transmission voltage does not significantly alter power station resistance.
- �� Option C → Transmission losses decrease rather than increase with voltage.
- �� Option D → High voltage does not make copper superconducting.
Substitution
- Application
- Substitute I = P/V into the cable loss equation P₍c₎ = I²R₍c₎.
- Final Logic
- P₍c₎ = P²R₍c₎/V² ⇒ higher voltage ⇒ lower current ⇒ lower losses.
Increase voltage to decrease transmission loss.
