CUET UG Physics Booster Test – 3 Ohm’s Law and Resistance

According to Ohm's law, V = IR. Let the resistance of each conductor be R. A single conductor requires a potential difference V = IR to maintain current I. When three identical conductors are connected in series, their equivalent resistance becomes: Req = R + R + R = 3R To maintain the same current I through the series combination, the required potential difference becomes: V' = I(3R) = 3IR = 3V Thus, the total potential difference across the series combination is three times the original potential difference. NCERT states that in a series arrangement, resistances add directly and the total potential difference equals the sum of the potential differences across individual conductors. Therefore, the required potential difference is 3V.

NCERT explains that Ohm's law is not universally valid for all materials and devices. Certain components, such as semiconductor junctions, electrolytes, and vacuum tubes, exhibit deviations from linear V-I behavior. In such cases, the relationship between voltage and current may depend on the direction of applied voltage and may not remain proportional. The V-I graph is often nonlinear and may not pass through the origin. Such materials are called non-ohmic conductors. Therefore, the relation between V and I can depend on factors such as the sign and magnitude of voltage, making option B correct.

The resistance of a conductor is given by: R = ρl/A When the slab is cut lengthwise into four identical strips, the length remains unchanged while the cross-sectional area of each strip becomes A/4. Therefore: R' = ρl/(A/4) R' = 4R Applying the same potential difference V across one strip: I' = V/R' I' = V/4R Since the original current is I = V/R, I' = I/4 Thus, the current through each strip becomes one-fourth of the original current. This result follows directly from the inverse dependence of resistance on cross-sectional area discussed in NCERT.

Ohm's law states that V = IR, where R is the proportionality constant relating voltage and current. Thus, statement 1 is correct. Resistance depends on both the material and dimensions of the conductor, according to the relation: R = ρl/A Therefore, statement 4 is also correct. Statement 2 is incorrect because geometry significantly affects resistance through length and area. Statement 3 is true individually, but the answer pattern provided in the question identifies only statements 1 and 4 as the intended correct set. Therefore, option D is the accepted answer.

According to NCERT, the resistance of a conductor is given by: R = ρl/A If the length is doubled while the area remains unchanged, the resistance becomes: R' = ρ(2l)/A = 2R Thus, doubling the length doubles the resistance. If the cross-sectional area is halved while the length remains constant: R'' = ρl/(A/2) = 2R Thus, halving the area also doubles the resistance. These relationships demonstrate the dependence of resistance on the geometry of the conductor. Therefore, both blanks are filled by the word "Doubles."

According to NCERT, resistance is given by: R = ρl/A For the original slab, resistance remains R. When two identical slabs are connected end to end, the effective length becomes 2l while area remains unchanged, giving resistance 2R. If the length is reduced to l/2 while area remains A, resistance becomes R/2 because resistance is directly proportional to length. Finally, if length becomes 2l and area becomes A/2 simultaneously, the resistance becomes: R' = ρ(2l)/(A/2) R' = 4ρl/A R' = 4R Thus, the correct matching is Original slab → R, Double length → 2R, Half length → R/2, and Double length with Half area → 4R.

When identical slabs are connected end to end, the effective length of the conductor increases while the cross-sectional area remains unchanged. Therefore, statement 1 is correct. In a series arrangement, the total potential difference equals the sum of the potential differences across each individual slab, making statement 2 correct. Since resistance is proportional to length, doubling the length doubles the resistance, so statement 4 is also correct. However, statement 3 is incorrect because the current remains the same throughout a series circuit. There is only one path available for charge flow, so the current through every section is identical. Thus, statements 1, 2, and 4 are correct.

For N identical slabs connected in series, the effective length becomes Nl while the cross-sectional area remains unchanged. Since resistance is directly proportional to length, the equivalent resistance becomes NR. The total potential difference across the combination is the sum of the potential differences across all slabs, resulting in NV if each slab has voltage V across it. However, the current in a series circuit remains the same through every component because charge has only one available path. Therefore, the statement that current becomes N times the current through a single slab is incorrect. Hence, option C is the incorrect statement.

A lengthwise division keeps the length of the conductor unchanged while changing only the cross-sectional area. This allows the area dependence of resistance to be studied independently, making statement 1 correct. According to the relation R = ρl/A, resistance is inversely proportional to area. Therefore, halving the cross-sectional area doubles the resistance rather than halving it. Thus, statement 2 is incorrect. Statement 3 correctly describes the inverse proportionality between resistance and area. To compare the electrical behavior of the original slab and the divided portions, the same potential difference is considered across the conductors, making statement 4 correct. Therefore, statements 1, 3, and 4 are correct.

The resistance of a conductor is given by: R = ρl/A If the conductor is divided lengthwise into n identical strips, the length remains unchanged while the area of each strip becomes A/n. Therefore, the resistance of each strip becomes: R' = ρl/(A/n) R' = nR Applying the same potential difference V across each strip: I' = V/R' I' = V/(nR) Since the original current is I = V/R, I' = I/n Therefore, each strip carries one-nth of the original current. This result follows directly from the inverse relationship between current and resistance for a fixed applied voltage.

NCERT derives the resistance formula by combining two experimentally observed proportionalities: resistance is directly proportional to the length of the conductor and inversely proportional to its cross-sectional area. Thus, R ∝ l/A Introducing a proportionality constant gives: R = ρ(l/A) where ρ is called the resistivity of the material. Resistivity is an intrinsic property of the material and depends on factors such as temperature and the nature of the material, but not on the dimensions of the conductor. Two conductors made of the same material may have different resistances due to differences in length and area, yet they possess the same resistivity. Therefore, the generalized formula correctly expresses the dependence of resistance on geometry while identifying resistivity as a dimension-independent material constant.

The resistance of a conductor is given by: R = ρl/A To determine the length, rearrange the equation: l = RA/ρ Substituting the given values: l = (0.05)(3.0 × 10⁻⁶)/(1.5 × 10⁻⁸) l = (1.5 × 10⁻⁷)/(1.5 × 10⁻⁸) l = 10 m Unit verification: (Ω × m²)/(Ω m) = m Thus, the obtained unit is metre, confirming the correctness of the calculation. The length of the conductor is therefore 10 m. This problem demonstrates the practical application of the resistance-resistivity relation discussed in NCERT.

NCERT defines current density as the current flowing per unit cross-sectional area normal to the direction of current flow. Mathematically, j = I/A Therefore, statement 1 is correct. The microscopic form of Ohm's law gives: j = σE which shows that current density is directly proportional to the electric field for a given conductivity. Hence statement 2 is correct. Since current density possesses both magnitude and direction, it is a vector quantity directed along the electric field in an isotropic conductor, making statement 3 correct. However, statement 4 is incorrect because the SI unit of current density is A/m². The unit V/m corresponds to electric field. Therefore, statements 1, 2, and 3 are correct.

Current density is defined as the electric current flowing through unit area normal to the direction of current flow. The mathematical expression is: j = I/A Since current is measured in amperes (A) and area is measured in square metres (m²), the SI unit of current density becomes: A/m² This unit indicates how much current is concentrated within a given cross-sectional area. A larger value of current density means that more current passes through each square metre of area. Therefore, current density is a measure of the spatial concentration of electric current within a conductor. This interpretation is central to the microscopic description of current flow presented in NCERT.

Several important electrical quantities can be expressed using fundamental relations discussed in NCERT. The electric field inside a uniform conductor is given by: E = V/l Thus, V/l corresponds to electric field. Resistance is defined by Ohm's law as: R = V/I Current density is defined as: j = I/A Finally, the microscopic form of Ohm's law gives: ρ = E/j Therefore, the correct matching is: V/l → E V/I → R I/A → j E/j → ρ These expressions connect macroscopic quantities such as voltage and current with microscopic quantities such as electric field and current density.

NCERT derives the microscopic form of Ohm's law by substituting V = El into the macroscopic equation V = IR. Here, V is expressed in terms of the electric field E and conductor length l. This replacement converts a macroscopic quantity into a microscopic field quantity, making statement 1 correct. The relation V = El is valid only when the electric field is uniform throughout the conductor, so statement 2 is also correct. During the derivation, substituting R = ρl/A and I = jA leads to: El = (jA)(ρl/A) The factors l and A cancel, resulting in: E = jρ Thus, the dimensional dependence on length disappears from the final equation, making statement 3 correct. However, statement 4 is incorrect because E is directly proportional to current density j, not inversely proportional. Therefore, statements 1, 2, and 3 are correct.

In an isotropic conductor, the electric field E and current density j are directed along the same line. The vector form of Ohm's law is: j = σE where σ is the conductivity of the material. Since both E and j possess magnitude and direction, they are treated as vectors. Their directions coincide because the electric field drives the motion of charge carriers. The corresponding scalar form may be written as E = jρ, where ρ is resistivity. Therefore, the statement that j is strictly orthogonal or perpendicular to E is incorrect. Such a condition would contradict the physical basis of electrical conduction described in NCERT.

To derive the microscopic form of Ohm's law, NCERT begins with: V = IR The substitutions used are: V = El I = jA R = ρl/A Substituting these relations gives: El = (jA)(ρl/A) After simplification: E = jρ Thus, statements 1, 2, and 3 are correct because they are the actual substitutions used during the derivation. Statement 4 is not considered one of the substitutions; rather, it is a consequence of the algebraic simplification. Therefore, the correct set of statements consists of 1, 2, and 3 only.

The microscopic form of Ohm's law is: j = σE where: σ = 5.0 × 10⁷ S/m E = 2.0 V/m Substituting: j = (5.0 × 10⁷)(2.0) j = 1.0 × 10⁸ A/m² Unit verification: (S/m)(V/m) = (A/V·m)(V/m) = A/m² Thus, the calculated unit is ampere per square metre, which is the SI unit of current density. The result shows that a highly conductive material develops a large current density when subjected to a given electric field. Therefore, the current density is 1.0 × 10⁸ A/m².

Conductivity and resistivity are related through: σ = 1/ρ A perfect insulator does not permit the flow of electric current. Therefore, its resistivity must be extremely large. In the ideal theoretical limit, the resistivity approaches infinity. Since conductivity is the reciprocal of resistivity, an infinitely large resistivity corresponds to zero conductivity. This relationship explains why insulating materials such as rubber, glass, and mica possess very low conductivity and very high resistivity compared to metals. Thus, the limiting case of a perfect insulator is represented by conductivity approaching zero and resistivity approaching infinity.