(a)

Calculate the resistance of a metal wire of length $\displaystyle 2$ m and area of cross-section $\displaystyle 1 \cdot 55 \times 10^{-6} \mathrm{~m}^{2}$. (Resistivity of the metal is $\displaystyle 2 \cdot 8 \times 10^{-8} \Omega \mathrm{~m}$ )

(b)

Why are alloys preferred over pure metals to make the heating elements of electrical heating devices ?

(a) Given : Length of the wire = $\displaystyle 2$ m \[\text { Area of cross section }=1 \cdot 55 \times 10^{-6} \mathrm{~m}^{2} \] Resistivity, \(\displaystyle \rho=2 \cdot 8 \times 10^{-8} \Omega \mathrm{~m}\) \[R=\rho \frac{l}{a} \] \[=\frac{2 \cdot 8 \times 10^{-8} \text { ohm metre } \text { × } 2 \text { metre }}{1 \cdot 55 \times 10^{-6}(\text { metre })^{2}} \Omega \] \[=3.6 \times 10^{-2} \Omega \]

(b) Alloys used for heating element have generally high melting point / high resistivity / Do not get oxidised at high temperature.