Question

A potential difference $V$ is applied across a copper wire of diameter $d$ and length $l$. In column I, the information about the change in one of the physical quantity is given and the effect of this change are mention in column II. Match the entries of column I with the entries of column II.
$$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(A) Only potential difference}V& \text{(P) Current in wire increases.}\\ \text{is doubled across wire.}& \\ \text{(B) Only length}l\text{is doubled keeping potential}& \text{(Q) Electric field inside conductor decreases.}\\ \text{difference and diameter}d\text{constant.}& \\ \text{(C) Only diameter}d\text{is doubled keeping potential}& \text{(R) Drift speed of electron increases.}\\ \text{difference and length}l\text{constant.}& \\ \text{(D) Temperature of the wire is increased}& \text{(S) Resistance of wire will increases.}\\ \text{keeping potential difference constant.}& \\ & \text{(T) Drift speed of electron will decreases.}\end{array}$$
Which of the following option has the correct combination considering column-I and column-II.

• A. A $\to$ P, S
• B. C $\to$ P
• C. B $\to$ R, T
• D. D $\to$ P

Answer 1

The correct option is B C $\to$ P

A $\to$ P,R; B $\to$ Q,S,T; C $\to$ P; D $\to$ Q,S,T
[A] when potential difference $V$ is doubled across - wire :-
$R=\frac{\rho l}{A}$
Resistance Remains Same
From ohm’s Law - $I=\frac{V}{R}$
As $R$ remains same, $V$ is doubled, I increases 2 times.
$\overrightarrow{E}=\rho \overrightarrow{J}$
$J=\frac{I}{A}$
$E=\frac{\rho I}{A}$
$I$ increases, $E$ also increases :
$I=neA{V}_d$
As $I$ increases, Drift velocity $\left[V_d\right]$ also
increases.
A $\to$ P,R
[B] Length is doubled, keeping potential - difference and diameter $d$ constant
Resistance $\left[R\right]=\frac{\rho l}{A}$
As $\rho$ and $A$ are same, $l$ is doubled, $R$ gets double or increases two times.
$I=\frac{V}{R}$
As $V=$ constant
$I\propto \frac{1}{R}$
$R\uparrow I\downarrow$
As current-density $\left[J\right]=\frac{I}{A},\overrightarrow{E}=\rho \overrightarrow{J}$
As $I$ decreases, $J$ decreases and $E$ decreases.
$I=neA{V}_d$
As $I$ decreases, $V_d$ decreases.
B $\to$ Q, S, T
[C] only diameter $d$ is doubled, keeping potential - difference and length constant.
$R=\frac{\rho l}{A}$
$A=\frac{\pi }{4}{d}^2$
diameter gets double, area becomes 4 times. As area increases, $R$ decreases.
$I=\frac{V}{R}$
As $V=$ constant
$I\propto \frac{1}{R}$
$R$ decreases, current increases.
$J=\frac{I}{A}$
$I\uparrow J\uparrow$
$\overrightarrow{E}=\rho \overrightarrow{J}$
Electric - field inside conductor increases.
$J=ne{V}_d$
$\frac{E}{\rho }=ne{V}_d$
$\frac{V}{\rho l}=ne{V}_d$
$V_d$ is independent of area.
C $\to$ P.
[D] Temperature of the wire is increasing keeping potential - difference constant.
$R=R_0[1+\alpha \Delta T]$
Temperature increases resistances increases.
$I=\frac{V}{R}$
$V=$ constant
$I\propto \frac{1}{R}$
$R$ increases, $I$ decreases.
$J=\frac{I}{A}$
$I$ decreases, $J$ decreases.
$\rho ={\rho }_0[1+\alpha \Delta T]$
$T\uparrow \rho \downarrow$
$E=\frac{V}{l}$
As $l\uparrow$ $E\downarrow$
$J=ne{V}_d$
$J$ decreases, $V_d$ decreases
D $\to$ Q, S, T