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Two wires are made from the same material, wire 2 has twice the length and half of the diameter of wire 1. What is the ratio of resistances of the two wires? a) 1
b) 2
c) 4
d) 8

Respuesta :

anbu40

Answer:

d)8

Explanation:

  • Resistance of a conductor is directly proportional to its length. R ∝ L. If length increases, the resistance also increases.
  • Resistance of a conductor is indirectly proportional to its cross-sectional area. So, the diameter is reduced, the cross-sectional area also reduces and thus the resistance increases.

             [tex]\sf R \alpha \dfrac{1}{A}[/tex]

[tex]\sf Resistance = \dfrac{\rho L }{A}[/tex]

Here, [tex]\sf \rho[/tex] is the resistivity of the material; L is the length of the conductor and A is the cross-sectional area.

Resistance of wire 1:

              [tex]\sf R= \dfrac{\rho * L}{A} = \dfrac{\rho * L}{\pi r^2}[/tex]

Resistance of wire 2:

       [tex]\sf L_2 = 2L\\\\[/tex]

Diameter of wire is half of diameter of wire 1. So, the radius of wire 2 will be half the radius of wire 1.

          [tex]\sf r_1 = \dfrac{r}{2}\\\\\\A_2 = \pi r_1^2 = \pi \left(\dfrac{r}{2}\right)^2\\\\\\A_2=\dfrac{\pi r^2}{4}[/tex]

     [tex]\sf R_2= \dfrac{\rho * L_2}{A_2}\\\\\\R_2=\dfrac{\rho*2L}{\frac{\pi r^2}{4}}=\rho*2L *\dfrac{4}{\pi r^2}\\\\\\R_2= \dfrac{8 *\rho*L}{\pi r^2}\\\\\\\boxed{\bf R_2=8R}[/tex]

          d) 8

       

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