Respuesta :
[tex]\huge\underline{\red{A}\green{n}\blue{s}\purple{w}\pink{e}\orange{r} ⟿}[/tex]
(i) 0.4 Ω m^-1
(ii) 0.8 Ω
(iii) 0.5 Ω
Explanation:
(i) Given, V = 2 volt, I = 1 A
Resistance of 5 m length of wire :-
[tex] = > R = \frac{V}{I} = \frac{2}{1} = 2 \: Ω [/tex]
Resistance per unit length of the wire :-
[tex] = > \frac{R}{l} = \frac{2}{5} = 0.4 \: Ω \: {m}^{ - 1} [/tex]
(ii) Resistance of 2m length of the wire~
[tex] = \: 0.4 \: Ω \: {m}^{ - 1} \times 2 \: m = 0.8 \: Ω [/tex]
(iii) When the wire is doubled on itself, its area of cross section becomes twice and the length becomes half. Let a be the initial cross section and ρ be the specific resistance of the material of wire. Then,
[tex]from \ relation \: \: ➪ \: R = ρ \times \frac{1}{a}. \\ \\ we \: have, \\ \\ initial \: resistance \:➪ \: 2 = ρ\times \frac{5}{a}....eqn(i) \\ \\ new \: resistance \: ➪ \: R` = ρ\times \frac{2.5}{2a}....eqn(ii) \\ \\ on \: dividing \: eqn(ii) \: by \: eqn(i)... \\ \\ \frac{R`}{2} = \frac{2.5ρ}{2a} \div \frac{5ρ}{a} = \frac{1}{4} \\ \\ R` = \frac{2}{4} = 0.5 \: Ω[/tex]
Thus, on doubling the wire on itself, its resistance becomes one-fourth.
