$A$ steady current $I$ is set up in a wire whose cross-sectional area decreases in the direction of the flow of the current. Then,as we examine the narrowing region,

An electron takes $40 \times 10^3 \ s$ to drift from one end of a metal wire of length $2 \ m$ to its other end. The area of cross-section of the wire is $4 \ mm^2$ and it is carrying a current of $1.6 \ A$. The number density of free electrons in the metal wire is

Assertion : The electric bulb glows immediately when the switch is on.
Reason : The drift velocity of electrons in a metallic wire is very high.

An insulating pipe of cross-section area $A$ contains an electrolyte which has two types of ions with charges $-e$ and $+2e$. $A$ potential difference applied between the ends of the pipe results in the drifting of the two types of ions,having drift speeds $v$ (for $-e$ ion) and $v/4$ (for $+2e$ ion). Both ions have the same number density $n$. The current flowing through the pipe is:

$A$ cylindrical resistor is connected across a battery $\varepsilon$. The cylinder has a uniform free electron density, and the middle part of the cylinder has a larger radius as shown in the figure. Which of the following graphs represents the variation of $V_d$ (drift velocity) with respect to $x$ (distance along the length of the resistor)?

$A$ potential difference $V$ is applied across the ends of a metallic wire of length $l$. The drift velocity will be doubled if: