CUET UG Physics Booster Test – 2 Ohm’s Law and Resistance

According to NCERT, Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference across its ends, provided physical conditions such as temperature remain constant. Mathematically, V = IR, where R is the resistance of the conductor. Since the dimensions and temperature of the conductor remain unchanged, its resistance remains constant. Rearranging the equation gives I = V/R. This expression shows that current is directly proportional to voltage. Therefore, if the potential difference is increased to three times its original value while resistance remains unchanged, the current must also become three times its original value. This direct proportionality is a defining characteristic of an ohmic conductor. NCERT represents this behavior through a straight-line V-I graph passing through the origin. Hence, tripling the applied voltage results in tripling the current, making option D the correct answer.

NCERT explains that for an ohmic conductor maintained at constant temperature, the potential difference across the conductor is directly proportional to the current flowing through it. This relationship is expressed as V = IR. When voltage is plotted against current, the graph obtained is a straight line passing through the origin. Such a graph demonstrates that the ratio V/I remains constant for all measured values. The slope of the V-I graph is equal to the resistance of the conductor. This graphical representation confirms that voltage and current are directly proportional to each other. If the graph were curved or if a single current corresponded to multiple voltages, the conductor would not obey Ohm's law. Therefore, the V-I plot of an ohmic conductor represents a direct proportionality, making option C the correct answer.

For a conductor obeying Ohm's law, the relationship between voltage and current is given by V = IR. Initially, the conductor carries a current of 2.5 A under a potential difference of 15 V. Therefore, its resistance is: R = V/I = 15/2.5 = 6 Ω Since resistance remains unchanged, the same value applies after the current changes. The new current is 5.0 A. Using Ohm's law again: V = IR V = 5.0 × 6 V = 30 V Unit verification: A × Ω = V Therefore, the calculated unit is volt, which is correct. Since the current doubles while resistance remains constant, the voltage must also double. Hence, the new potential difference is 30 V.

The proportionality constant in Ohm's law is known as resistance. NCERT defines resistance as the ratio of potential difference across a conductor to the current flowing through it, expressed mathematically as R = V/I. The SI unit of resistance is the ohm, represented by the symbol Ω. One ohm is the resistance of a conductor when a potential difference of one volt produces a current of one ampere. Therefore, statements 1 and 2 are correct because Ω is the symbol and ohm is the SI unit. Statement 4 is also correct because resistance is defined by the ratio V/I. However, statement 3 is incorrect because resistance depends on the material, dimensions, and temperature of the conductor, not solely on current. Thus, statements 1, 2, and 4 are correct.

According to NCERT, the resistance of a conductor is given by the relation R = ρl/A, where ρ is the resistivity of the material, l is the length of the conductor, and A is its cross-sectional area. This equation shows that resistance is directly proportional to the ratio l/A. If both the length and cross-sectional area are doubled simultaneously, the new resistance becomes: R′ = ρ(2l)/(2A) R′ = ρl/A R′ = R Thus, the resistance remains exactly equal to its original value because the factor of two in the numerator cancels the factor of two in the denominator. Since the ratio l/A remains unchanged, resistance also remains unchanged. Therefore, the correct statement is that the resistance becomes equal to its original value, and resistance is directly proportional to l/A.

According to NCERT, the resistance of a conductor is given by R = ρl/A. For a single rectangular slab, the resistance is R. When two identical slabs are connected end to end, the effective length becomes 2l while the cross-sectional area remains unchanged. Therefore, the new resistance becomes 2R. Thus, the resistance of the combination corresponds to 2R. Using Ohm's law, the current through a single slab under potential difference V is I = V/R. For the combination having resistance 2R, the current becomes I = V/(2R). Hence the correct matching is: single slab resistance → R, double-length combination resistance → 2R, current through a single slab → V/R, and current through the combination → V/(2R). These results directly follow from the dependence of resistance on conductor length discussed in NCERT Current Electricity.

When two identical slabs are connected end to end, the effective length becomes 2l while the cross-sectional area remains unchanged. Therefore, the combination behaves as a single conductor of length 2l. In a series arrangement, the same current passes through each conductor because there is only one path for charge flow. Hence statement 2 is correct. If each slab has a potential difference V across it, the total potential difference across the series combination becomes V + V = 2V. Therefore statement 4 is also correct. However, the total resistance becomes 2R rather than R/2 because resistance is directly proportional to length. Thus statements 1, 2, and 4 are correct.

NCERT states that resistance is directly proportional to the length of a conductor. Therefore, when the length is doubled, the resistance also doubles. According to Ohm's law, V = IR. For the same current, doubling the resistance requires doubling the potential difference. Therefore, the statement that doubling the length halves the required potential difference is incorrect. Statements A and B correctly describe the addition of potential differences in a series arrangement. Statement D is also correct because the total resistance of the combination is 2R and can be written as Rc = (2V)/I. Thus, option C is the incorrect statement.

When a conductor is cut lengthwise into two identical parts, its length remains unchanged while its cross-sectional area becomes A/2. Using the resistance relation R = ρl/A, the resistance of each half becomes: R' = ρl/(A/2) R' = 2R Thus halving the area doubles the resistance. Therefore statements 1 and 3 are correct. A potential difference V may still be applied across each half-slab independently, making statement 4 correct. The important concept from NCERT is that resistance is inversely proportional to cross-sectional area. Hence the resistance increases when area decreases.

The resistance of a conductor is given by R = ρl/A. If the cross-sectional area is reduced to one-third of its original value, the new resistance becomes: R' = ρl/(A/3) R' = 3R Thus, the resistance becomes three times larger. Since the potential difference remains constant, Ohm's law gives: I = V/R The new current is: I' = V/(3R) I' = I/3 Unit verification confirms that current remains measured in amperes. This result demonstrates the inverse relationship between current and resistance. Since reducing the area increases resistance threefold, the current decreases to one-third of its original value.

NCERT explains that the resistance of a conductor depends on its dimensions. Experimental observations show that resistance is directly proportional to the length of the conductor and inversely proportional to its cross-sectional area. These proportionalities are expressed as R ∝ l and R ∝ 1/A. Combining them gives R ∝ l/A. Introducing the proportionality constant ρ, known as the resistivity of the material, leads to the relation R = ρl/A. This equation shows that a longer conductor offers greater opposition to the flow of current because charge carriers must travel a greater distance. Similarly, a larger cross-sectional area provides more paths for charge flow and therefore reduces resistance. The formula is fundamental in NCERT Current Electricity and forms the basis for understanding how conductor dimensions affect electrical resistance.

The resistance of a conductor is related to its resistivity by the formula: R = ρl/A Rearranging, ρ = RA/l Substituting the given values: ρ = (10)(2.0 × 10⁻⁶)/4.0 ρ = 20 × 10⁻⁶/4 ρ = 5.0 × 10⁻⁶ Ω m Unit verification: ρ = (Ω × m²)/m ρ = Ω m Thus, the SI unit is correctly obtained as ohm metre. Resistivity is an intrinsic property of the material and does not depend on the dimensions of the conductor. NCERT defines resistivity as a material constant that characterizes the opposition offered by a material to the flow of electric current. Therefore, the resistivity of the wire is 5.0 × 10⁻⁶ Ω m.

NCERT defines current density as the electric current flowing per unit cross-sectional area taken normal to the direction of current flow. Mathematically, j = I/A Current density may also be interpreted as the amount of charge crossing unit area per second. Therefore, statements 1 and 3 are correct. The symbol used for current density is j, making statement 4 correct. However, current density possesses both magnitude and direction. Its direction is the same as the direction of conventional current. Therefore, it is a vector quantity and not a scalar quantity. Hence statement 2 is incorrect. Current density is an important concept because it relates microscopic charge transport to macroscopic current and is widely used in the study of electrical conduction.

Current density is defined as the amount of electric current flowing through a unit cross-sectional area normal to the direction of current. The mathematical expression is: j = I/A where I is current and A is area. Since the SI unit of current is ampere (A) and the SI unit of area is square metre (m²), the SI unit of current density becomes: A/m² This unit indicates how much current passes through each square metre of cross-sectional area. Current density provides a measure of how concentrated the current flow is within a conductor. A larger current density implies that more charge passes through a given area per second. Therefore, the correct unit is A/m² and it arises from dividing current by area.

NCERT introduces several important electrical quantities and their mathematical expressions. The uniform electric field inside a conductor is given by E = V/l, where V is the potential difference and l is the length. Current density is defined as current per unit cross-sectional area, so j = I/A. Resistance is defined through Ohm's law as R = V/I. Resistivity is related to electric field and current density through the microscopic form of Ohm's law: ρ = E/j These relations connect macroscopic quantities such as voltage and current with microscopic quantities such as current density and electric field. Matching the expressions correctly gives E → V/l, j → I/A, R → V/I, and ρ → E/j. Therefore, option A provides the correct matching.

NCERT derives the relation V = El for a conductor placed in a uniform electric field. Here, E represents the magnitude of the electric field and l is the length of the conductor along the field direction. The relation assumes that the electric field remains uniform throughout the conductor. Therefore, statements 1 and 2 are correct. This equation can be substituted into Ohm's law, V = IR, together with expressions such as R = ρl/A and I = jA, to derive the microscopic form of Ohm's law connecting electric field and current density. Hence statement 4 is also correct. However, statement 3 is incorrect because the equation V = El shows that potential difference is directly proportional to length when E is constant. Therefore, statements 1, 2, and 4 are correct.

NCERT explains that the current density vector j points in the direction of conventional current flow. In an isotropic conductor, the electric field vector E and current density vector j are parallel to each other. Their relationship is expressed by the microscopic form of Ohm's law: E = jρ where ρ is the resistivity of the material. Since both vectors point in the same direction, the statement that j is strictly perpendicular to E is incorrect. Current density is defined as the current flowing through unit area normal to the direction of current flow, making statement C correct. Therefore, the only incorrect statement is option D.

The microscopic form of Ohm's law is derived using several NCERT relations. First, the potential difference across a conductor in a uniform electric field is given by V = El. Second, the resistance of the conductor is R = ρl/A. Third, current density is related to current through I = jA. Substituting these expressions into Ohm's law V = IR gives: El = (jA)(ρl/A) El = jρl E = jρ Thus, the final result is E = jρ, not E = j/ρ. Therefore, statements 1, 2, and 3 are correct, while statement 4 is incorrect. This derivation forms the basis of the microscopic interpretation of electrical conduction in NCERT.

Conductivity σ is a measure of how easily electric charges move through a material. NCERT gives the microscopic form of Ohm's law as: j = σE where j is current density and E is electric field. For a given electric field, current density is directly proportional to conductivity. Therefore, if conductivity increases while the electric field remains unchanged, the current density must increase in the same proportion. A highly conductive material allows more charge carriers to move through a unit area per second, resulting in a larger current density. Metals generally possess high conductivity and therefore exhibit large current densities under the same applied electric field. Hence, option B is correct.

Conductivity and resistivity are related by the expression: σ = 1/ρ This relation shows that conductivity and resistivity are inversely proportional to each other. If the resistivity of a material is low, the denominator in the expression becomes small, resulting in a large conductivity. Such materials allow electric current to flow easily and are good conductors. Metals such as copper and silver have low resistivity and therefore possess high conductivity. Conversely, insulators have extremely high resistivity and consequently very low conductivity. This reciprocal relationship is fundamental to understanding electrical conduction in materials. Therefore, a material with low resistivity will have high conductivity.