The resistance of a rectangular block of copper of dimensions $2 \,mm \times 2 \,mm \times 5 \,m$ between two square faces is $0.02 \,\Omega$. What is the resistivity of copper?
Easy
Download our app for free and get started
(c)
$R=\frac{\rho l}{A}$
$0.02=\frac{\rho(5)}{4 \times 10^{-6}}$
$\rho=\frac{8 \times 10^{-8}}{5}$
$\rho=1.6 \times 10^{-8} \,\Omega- m$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1For the network shown in the figure the value of the current $i$ isView Solution
- 2A uniform metallic wire of length $L$ is mounted in two configurations. In configuration $1$ (triangle), it is an equilateral triangle and a voltage $V$ is applied to corners $A$ and $B$. In configuration $2$ (circle), it is bent in the form of a circle and the potential $V$ is applied at diametrically opposite points $P$ and $Q$. The ratio of the power dissipated in configuration $1$ to configuration $2$ isView Solution
- 3Two identical batteries, each of $e.m.f.$ $2\, volt$ and internal resistance $1.0\, ohm$ are available to produce heat in an external resistance $R = 0.5\,ohm$ by passing a current through it. The maximum Joulean power that can be developed across $R$ using these batteries is ............. $watt$View Solution
- 4View SolutionA constant voltage is applied between the two ends of a metallic wire. If both the length and the radius of the wire are doubled, the rate of heat developed in the wire
- 5View SolutionFind the ratio of currents as measured by ammeter in two cases when the key is open and when the key is closed
- 6A cylindrical resistance is connected across battery $\varepsilon $ . Cylinder has uniform free electron density, mid part of cylinder has larger radius as shown in figure. Then $V_d$ (drift velocity) $V/S$ (distance across the length of the resistance)View Solution
- 7Consider a block of conducting material ofresistivity '$\rho$' shown in the figure. Current '$I$' enters at '$A$' and leaves from '$D$'. We apply superp osition principle to find voltage '$\Delta V$ ' developed between '$B$' and '$C$'. The calculation is done in the following steps:View Solution
$(i)$ Take current '$I$' entering from '$A$' and assume it to spread over a hemispherical surface in the block.
$(ii)$ Calculatefield $E(r)$ at distance '$r$' from $A$ by using Ohm's law $E = \rho j$, where j is the current per unit area at '$r$'.
$(iii)$ From the '$r$' dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
$(iv)$ Repeat $(i), (ii)$ and $(iii)$ for current '$I$' leaving '$D$' and superpose results for '$A$' and '$D$'.For current entering at $A$, the electric field at a distance '$r$'
from $A$ is
- 8In the given circuit, it is observed that the current $I$ is independent of the value of the resistance $R_6$. Then the resistance values must satisfyView Solution
- 9$A$ Wheatstone's bridge is balanced with a resistance of $625\, \Omega$ in the third arm, where $P, Q$ and $S$ are in the $1^{st}, 2^{nd}$ and $4^{th}$ arm respectively. If $P$ and $Q$ are interchanged, the resistance in the third arm has to be increased by $51\,\Omega$ to secure balance. The unknown resistance in the fourth arm is ............. $\Omega$View Solution
- 10A potential difference of $5 \,V$ is applied across a conductor of length $10 \,cm$. If drift velocity of electrons is $2.5 \times 10^{-4} \,m / s$, then electron mobility will be ............ $m ^2 V ^{-1} s ^{-1}$View Solution