$A$ wire $100\,cm$ long and $2.0\,mm$ diameter has a resistance of $0.7\,\Omega$. The electrical resistivity of the material is ...........$ \times 10^{-6}\,\Omega \cdot m$.

$A$ cylindrical resistor is connected to a battery with emf $5 \ V$. The resistance per unit length varies as $\rho(x) = \rho_0 \left(\frac{x}{L}\right)^\alpha$,where $\rho_0$ and $\alpha$ are constants and $x$ is the distance from one end of the resistor. The product $\rho_0 L$ is $10 \ \Omega$,where $L$ is the length of the resistor. If the thermal power generated by the resistor is $20 \ W$,then the value of $\alpha$ is:

$A$ wire of length $1.0\, m$ and diameter $0.6\, cm$ has a resistance of $3.0 \times 10^{-3}\, \Omega$. What will be the resistance between the faces of a disc made of the same material with diameter $2.0\, cm$ and thickness $1.0\, mm$?

An aluminium (resistivity $\rho = 2.2 \times 10^{-8} \Omega \cdot m$) wire of diameter $1.4 \text{ mm}$ is used to make a $4 \Omega$ resistor. The length of the wire is: (in $\text{ m}$)

$n$ conducting wires of same dimensions but having resistivities $1, 2, 3, . . . , n$ are connected in series. The equivalent resistivity of the combination is

The resistivity of alloys is $\rho_{\text{alloy}}$ and the resistivity of constituent metals is $\rho_{\text{metal}}$. Then,usually: