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Physics,two wires from copper the lenght of one of them is 10m, and it's mass is 0.1kg. ?

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and the lenght of the other is 40m. and it's mass is 0.2kg.

find the ratio between thier resistance

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  1. Volume of a wire = length * cross sectional area

    mass = volume * density

    call the areas of wires A1 and A2 respectively

    density of copper = d

    For the first wire,

    0.1 = 10 * A1 * d    .....(1)

    for the second wire,

    0.2 = 40 * A2 * d  .....(2)

    divide (2) by (1)

    2 = 4 * A2/A1

    A1 = 2*A2  

    Resistance = resistivity * length / cross sectional area

    For the first wire,

    R1 = rho * 10 / A1 ...(3)

    For the second wire

    R2 = rho * 40 / A2 ...(4)

    Divide (3) by (4)

    R1/R2 = A2 / (4 * A1)

    Substituting A1 = 2*A2

    R1/R2 = A2 / (8 * A2)  = 1/8


  2. What kind of resistance are you talking about?

    Electric, mechanic, strength?

    Electric will be ( length / mass ) ^2

    So, think about the wire not to be round but square section of "n"mm, with "x" ohms (Ω) per meter.  Now using a very good knife, split the wire in the longitudinally in the middle and connect ends, now you have two single length wire with double resistance each one (Ω*2), when mending both, the resistance will be added too (Ω*4), so  your answer is;

    one wire, 10m, 100g = resistance = X ohms.

    other wire, 40m, 200g = resistance = 4X ohms.

    Ratio is  (Length / Mass) ^ 2

    First wire = (10 / 0.1) ^2 = 10,000

    Second wire = (40 / 0.2) ^2 = 40,000

      

      

  3. Volumes..V1/V2=1/2..V=A*L

    >A1*L1/A2*L2=1/2..L2=4*L1..>A1=2*A2

    Now second wire has 4X length ..and 1/2 csa..>8xresistance

    Since R propnl 2 L/A

    oops! jst spotted 2 v good answrs above!

  4. Mass = area of cross-section x length x density.

    Density is the same (both copper).

    Second wire is twice the mass, but 4 times the length.

    Hence it must have half the area.

    Electrical resistance = resistivity x length/area. Resistivity is the same (again, both copper).

    Second one is 4 times the length, half the area, hence 4/½ = 8 times the resistance.

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