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  1. The energy stored in the battery is given as 7.20 × 10⁴ J, and the supply voltage is 12 V. Using the formula Energy (E) = Charge (Q) × Voltage (V), we find: Q = E/V = (7.20 × 10⁴ J) / (12 V) = 6.0 × 10⁴ C Thus, the correct answer is (a) 6.0 × 10⁴ C. For more visit here: https://www.tiwariacademy.comRead more

    The energy stored in the battery is given as 7.20 × 10⁴ J, and the supply voltage is 12 V. Using the formula Energy (E) = Charge (Q) × Voltage (V), we find:

    Q = E/V = (7.20 × 10⁴ J) / (12 V) = 6.0 × 10⁴ C

    Thus, the correct answer is (a) 6.0 × 10⁴ C.

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  2. Kirchhoff’s junction rule is a reflection of (B) conservation of charge. It states that the total current entering a junction equals the total current leaving, ensuring no charge accumulation. This follows from the principle that charge cannot be created or destroyed at a junction, making conservatiRead more

    Kirchhoff’s junction rule is a reflection of (B) conservation of charge. It states that the total current entering a junction equals the total current leaving, ensuring no charge accumulation. This follows from the principle that charge cannot be created or destroyed at a junction, making conservation of charge the fundamental basis of this rule.

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  3. The temperature dependence of resistivity ρ(T) in semiconductors, insulators and metals is primarily influenced by: (a) Number of charge carriers can change with temperature T (especially in semiconductors and insulators). (b) Time interval between successive collisions depends on T (affecting electRead more

    The temperature dependence of resistivity ρ(T) in semiconductors, insulators and metals is primarily influenced by:
    (a) Number of charge carriers can change with temperature T (especially in semiconductors and insulators).
    (b) Time interval between successive collisions depends on T (affecting electron scattering).
    Thus, the correct answers are (a) and (b).

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  4. When the wire of resistance 2 R is bent into a circle, each semicircular half has resistance R. These two halves are in parallel when measured across a diameter. The effective resistance is given by: 1/Reff = 1/R + 1/R Reff = R /2 Thus, the correct answer is (b) R/2. For more visit here: https://wwwRead more

    When the wire of resistance 2
    R is bent into a circle, each semicircular half has resistance
    R. These two halves are in parallel when measured across a diameter. The

    effective resistance is given by:
    1/Reff = 1/R + 1/R

    Reff = R /2

    Thus, the correct answer is (b) R/2.

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  5. In a series circuit, the current remains the same across all resistors. Given that R₁ carries 2.0 A, R₃ also has 2.0 A. Power dissipated by R₃ is given by P = I² R. Solving 6 = (2)² R₃, we get R₃ = 1.5 ohm. Voltage across R₃ is V = IR = 2 × 1.5 = 3V. Answer: (c) 3 V. For more visit here: https://wwwRead more

    In a series circuit, the current remains the same across all resistors. Given that
    R₁ carries 2.0 A,
    R₃ also has 2.0 A. Power dissipated by
    R₃ is given by
    P = I² R. Solving 6 = (2)² R₃, we get
    R₃ = 1.5 ohm. Voltage across
    R₃ is V = IR = 2 × 1.5 = 3V. Answer: (c) 3 V.

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    https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/

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