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Two wires of the same length and of the same materials have different diameters, d, and d2 (d,<d2) determine which wire has greater (i) conductivity? (ii) Conductance? 8. a. (i) State the fundamental laws of conservation of electric charges. (ii) What will happened to a

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Question 5:
The resistance of a wire is R=ρLA=4ρLπd2R = \rho \frac{L}{A} = \frac{4\rho L}{\pi d^2}.

i) If its length is doubled (L′=2LL' = 2L):

  1. Calculate the new resistance R′R'.

    R′=4ρ(2L)πd2=2(4ρLπd2)=2RR' = \frac{4\rho (2L)}{\pi d^2} = 2 \left( \frac{4\rho L}{\pi d^2} \right) = 2R

    The resistance will be 2R\boxed{2R}.

    ii) If its diameter is halved (d′=d/2d' = d/2):

  2. Calculate the new resistance R′′R''.

    R′′=4ρLπ(d/2)2=4ρLπ(d2/4)=16(4ρLπd2)=16RR'' = \frac{4\rho L}{\pi (d/2)^2} = \frac{4\rho L}{\pi (d^2/4)} = 16 \left( \frac{4\rho L}{\pi d^2} \right) = 16R

    The resistance will be 16R\boxed{16R}.

    Question 6:

  3. Evaluate the expression.

    The expression Ω=1ϵ0μ0\Omega = \frac{1}{\sqrt{\epsilon_0 \mu_0}} represents the speed of light in vacuum (cc).

    c=1ϵ0μ0≈3×108 m/sc = \frac{1}{\sqrt{\epsilon_0 \mu_0}} \approx 3 \times 10^8 \text{ m/s}
  4. State its unit.

    The unit of speed is meters per second.

    The expression evaluates to 3×108 m/s\boxed{3 \times 10^8 \text{ m/s}}.
    Its unit is meters per second (m/s)\boxed{\text{meters per second (m/s)}}.

    Question 7:
    Given: Two wires of the same length (LL) and same material (resistivity ρ\rho, conductivity σ\sigma). They have different diameters d1d_1 and d2d_2, with d1<d2d_1 < d_2.

    i) Which wire has greater conductivity?

  5. Compare conductivities.

    Conductivity (σ\sigma) is an intrinsic property of the material. Since both wires are made of the same material, their conductivities are equal.
    The wire with greater conductivity is neither, they have the same conductivity\boxed{\text{neither, they have the same conductivity}}.

    ii) Which wire has greater conductance?

  6. Compare conductances.

    Conductance (GG) is the reciprocal of resistance (RR), G=1R=AρL=πd24ρLG = \frac{1}{R} = \frac{A}{\rho L} = \frac{\pi d^2}{4\rho L}.
    Since G∝d2G \propto d^2 and d1<d2d_1 < d_2, the wire with the larger diameter (d2d_2) will have greater conductance.
    The wire with greater conductance is the wire with diameter d2\boxed{\text{the wire with diameter } d_2}.

    Question 8:

    a) (i) State the fundamental laws of conservation of electric charges.
    The law of conservation of electric charge states that the net electric charge of an isolated system remains constant. Charge can neither be created nor destroyed, but it can be transferred from one object to another.

    a) (ii) What will happened to a
    This question is incomplete and cannot be answered.

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