Question

If a wire conductor of 0.2ohm resistance is doubled in length, it's resistance becomes

Answer 1

If the length is doubled by stretching the wire then volume of the wire remains same before and after stretching the wire.....so resistance becomes four times i.e. 0.2×4=0.8 ohm

And if the length is doubled by some other means lyk by adding an another length of wire then resistance simply gets doubled i.e. 0.2×2=0.4

Answer 2

Answer:

The new resistance of wire, when its length is doubled is $0.8$ Ω.

Explanation:

Given that,

The initial resistance of wire, $R = 0.2$ Ω.

We know that,

• For any wire of length, $l$ and with cross-sectional area, $A$, the resistance, $R =$ ρ $\frac{l}{A}$.
• Here,ρ being the resistivity and is constant for given material.

Thus,  $R$ ∝ $\frac{l}{A}$.

• When the length of wire ($l_{1} = l$) is doubled by stretching ($l_{2} = 2l$), the area of cross-section is reduced, so as to maintain constant volume. That is,

       $Al =$ constant.

    $A_{1} l_{1} = A_{2} l_{2}$

       $A l = A_{2} (2l)$

       $A_{2} = \frac{A}{2}$

Thus, new resistance, $R_{2}$ $=$ ρ $\frac{l_{2} }{A_{2} }$

                                          $=$ ρ ($\frac{2l}{\frac{A}{2} }$)

                                          $= 4$ ρ $\frac{l}{A}$

                                          $= 4R$

                                          $= 4(0.2)$

                                          $= 0.8$ Ω.

Therefore, when the length of wire is doubled, resistance increases by four times and equal to  $0.8$ Ω.