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  1. Garba is a traditional dance form that originates from the state of Gujarat in western India. It is an essential part of the state’s cultural identity and is particularly popular during the Navaratri festival, where people perform Garba in colorful attire, forming circles and dancing rhythmically toRead more

    Garba is a traditional dance form that originates from the state of Gujarat in western India. It is an essential part of the state’s cultural identity and is particularly popular during the Navaratri festival, where people perform Garba in colorful attire, forming circles and dancing rhythmically to devotional songs. The dance is a joyful expression of devotion and unity, symbolizing celebration and reverence, especially toward the Goddess Durga.

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  2. Total resistance in the circuit = 100Ω + 50Ω = 150Ω Current through the galvanometer at full scale deflection: I = V/R = 1.5/150 = 0.01A = 10, 000µA Galvanometer has 20 divisions, so: Figure of merit = 10, 000µA/20 = 500 μA/division Answer: (C) 500. For more visit here: https://www.tiwariacademy.comRead more

    Total resistance in the circuit = 100Ω + 50Ω = 150Ω

    Current through the galvanometer at full scale deflection:
    I = V/R = 1.5/150 = 0.01A = 10, 000µA
    Galvanometer has 20 divisions, so:
    Figure of merit = 10, 000µA/20 = 500 μA/division
    Answer: (C) 500.

    For more visit here:
    https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/

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    • 19
  3. Let the shunt resistance be Rs and galvanometer resistance be Rg = 3663Ω ​Given: only 1/34 of total current I passes through the galvanometer. Using current division rule: Ig/Is = Rs/Rg ' Where Ig/I = 1/34' ⇒ Is/I = 33/34 So: 1/33 = Rs/3663 ⇒ Rs = 3663/33 = 111 Ω ≈100Ω Answer: (A) 100 Ω.

    Let the shunt resistance be
    Rs and galvanometer resistance be
    Rg = 3663Ω

    ​Given: only 1/34 of total current I passes through the galvanometer.
    Using current division rule:
    Ig/Is = Rs/Rg ‘ Where Ig/I = 1/34’ ⇒ Is/I = 33/34
    So:
    1/33 = Rs/3663 ⇒ Rs = 3663/33 = 111 Ω ≈100Ω

    Answer: (A) 100 Ω.

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  4. To convert the galvanometer into an ammeter reading up to 10 A, a small shunt resistance must be added in parallel, allowing most current to bypass the galvanometer. The galvanometer allows only 0.1 mA. Using the formula Rs = Ig.Rg/I-Ig, we get Rs ≈ 0.01Ω. Correct answer: (B) 0.01 Ω in parallel. ForRead more

    To convert the galvanometer into an ammeter reading up to 10 A, a small shunt resistance must be added in parallel, allowing most current to bypass the galvanometer. The galvanometer allows only 0.1 mA. Using the formula Rs = Ig.Rg/I-Ig, we get Rs ≈ 0.01Ω.
    Correct answer: (B) 0.01 Ω in parallel.

    For more visit here:
    https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/

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  5. The magnetic force F between two magnetic poles is given by: F = μ0 ⋅m1 ⋅m2/4πr² Here, μ0 = 4π × 10⁻⁷ N/A² ,m = m = 10A\cdotpm, r = 0.1m. Substituting the values: F = 4π × 10⁻⁷ . 10 . 10/4π. (0.1)² = 10⁻⁵/0.01 = 1 × 10⁻³N Answer: (D) 1 × 10⁻³ N So, the magnetic force of attraction is 1 × 10⁻³ N.

    The magnetic force F between two magnetic poles is given by:
    F = μ0 ⋅m1 ⋅m2/4πr²
    Here,
    μ0 = 4π × 10⁻⁷ N/A² ,m = m = 10A\cdotpm,
    r = 0.1m.
    Substituting the values:
    F = 4π × 10⁻⁷ . 10 . 10/4π. (0.1)² = 10⁻⁵/0.01 = 1 × 10⁻³N
    Answer: (D) 1 × 10⁻³ N

    So, the magnetic force of attraction is
    1 × 10⁻³ N.

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