Question

$\left( 1 \right)$ Charge $Q$ passing through a cross-section of conductor at an instant is given by \[Q = \left( {0.5{t^2} + t} \right)\;{\text{C}}\] where $t$ is in second. Current through the conductor at $t = 1\;{\text{s}}$ is
\[\left( a \right)\,\,{\text{Zero}}\]
$\left( b \right)\,\,1A$
$\left( c \right)\,\,2A$
$\left( d \right)\,\, - 1A$

$\left( 2 \right)$ If the radius of the cross-section of the conductor is increased by $0.1\% $ keeping volume constant, then the percentage change in the resistance of the conductor is

Answer 1

Hint: Here in first question we have to find the current in specified time and for this we will use the formula of current, which will be given by $I = \dfrac{{dQ}}{{dt}}$ and for the second question we have to find the percentage change in the resistance, and for this firstly the resistance will be calculated by using the formula of resistance which is given by $R = \rho \dfrac{l}{A}$ and then its percentage will be calculated.

Complete step by step answer:
$\left( 1 \right)$ We know from the question that the equation of charge passes through the conductor is \[Q = \left( {0.5{t^2} + t} \right)\;{\text{C}}\].
We know that the current in the conductor is equal to the rate of flow of charge in the conductor. We have,
$ \Rightarrow I = \dfrac{{dQ}}{{dt}}$
Now we substitute the equation of $Q$ in the above expression.
\[ \Rightarrow I = \dfrac{d}{{dt}}\left( {0.5{t^2} + t} \right)\]
Now on solving it, we get
\[ \Rightarrow 0.5\left( {2t} \right) + 1\]
Solving it more, we get
\[ \Rightarrow t + 1\]
Now we substitute the value of $t$ as $1s$ . We get the value of current flowing through the conductor at $1s$ ,
\[ \Rightarrow I = 1 + 1\]
And on solving it, we get
$ \Rightarrow 2{\text{A}}$
Hence, the current flowing through the conductor at $t = 1\;{\text{s}}$ is $2\;{\text{A}}$
Therefore, option $\left( c \right)$ is the correct option.
$\left( 2 \right)$
From the question, we know that the percentage increment of the radius of the cross-section is $\Delta r = 0.1\% $ when the volume is constant.
The final radius of the cross-section of the conductor is calculated as:
$ \Rightarrow \dfrac{{r' - r}}{r} \times 100 = 0.1$
And on solving it, we get
$ \Rightarrow r' = r + \dfrac{{0.1}}{{100}}r$
Further solving more, we get
$ \Rightarrow r' = 1.001r$
Let us assume the initial resistance of the conductor is $R$ and we can express it as,
$ \Rightarrow R = \rho \dfrac{l}{A}$
Here, $\rho $ is the resistivity, $l$ is the initial length, and $A$ is the initial cross-sectional area.
Now we rewrite the above expression.
\[ \Rightarrow R = \rho \dfrac{l}{A} \cdot \dfrac{A}{A}\]
Now on solving it, we get
\[ \Rightarrow R = \rho \dfrac{V}{{{A^2}}}\]
And area can be written as,
\[ \Rightarrow R = \rho \dfrac{V}{{{{\left( {\pi {r^2}} \right)}^2}}}\]
Since the resistivity is constant for the same material and we know that volume is constant, we can write the above equation as,
$ \Rightarrow R \propto \dfrac{1}{{{r^4}}}$
The ratio of the initial and final resistance of the conductor is expressed as,
$ \Rightarrow \dfrac{R}{{R'}} = \dfrac{{{{r'}^4}}}{{{r^4}}}$
Substitute the values in the above equation.
\[ \Rightarrow \dfrac{{R'}}{R} = \dfrac{{{r^4}}}{{{{\left( {1.001r} \right)}^4}}}\]
Now on solving it, we get

\[ \Rightarrow \dfrac{{R'}}{R} - 1 = 0.996 - 1\]
Percentage change in the resistance will be calculated as,
\[ \Rightarrow \left( {\dfrac{{R' - R}}{R}} \right) \times 100 = - 0.4\% \]
Hence, the percentage change in the resistance of the conductor is $ - 0.4\% $

Note: The rate of flow of the charge through a conductor is termed as current. The resistance of a conductor is the tendency to oppose the current flowing through it. Here the negative sign in the resistance indicates the decrement in the percentage.