Physics M.C.Q (1 Marks)

Current $I_4$ can be calculated using Kirchhoff’s first law. The sum of current arriving at a junction is equal to the currents leaving the junction.
Here, Currents arriving at a junction: $7A + 5A = 12A$
Currents leaving the junction: $3A + I_4$
Equating both the currents: $12A = 3A +I_4$
$I_4 = 9A$

The resistivities of semiconductors lie between $10^{-6} \Omega m$ and $10^4 \Omega m$.. A semiconductor material has an electrical conductivity value falling between that of a conductor, such as metallic aluminum, and an insulator, such as wood. Silicon and germanium are typical semiconductors.

Potentiometer is an electric instrument used to measure internal resistance, electromotive force $(emf)$, and it is also used for comparing the emfs of different cells. Potentiometer can be used as a variable resistor as well.

Silicon is an example of a non$-$ohmic conductor. It exhibits non$-$linear behaviour. After a certain voltage, the current decreases as the voltage increases. The $V-I$ plot for silicon is non-linear.

Temperature of a cunductor increases that causes resistivity $(r)$ is increases due conductivity $(s)$ is decrease.
$\therefore\sigma=\frac{1}{\rho}$
$\Rightarrow\text{ratio of}\frac{\text{resistivity}}{\text{conductivity}}=\frac{\rho}{\sigma}=\rho^2$ is increase.

According to $Ohm's$ law, the potential difference across a conductor is directly proportional to the current through it.
$V = IR$
Hence the curve is a straight line passing through origin.

Electrolysis of an aqueous solution of table salt $(\text{NaCl}$, or sodium chloride$)$ produces aqueous sodium hydroxide and chlorine, although usually only in minute amounts. $\text{NaCl} ($aq$)$ can be reliably electrolysed to produce hydrogen. Hydrogen gas will be seen to bubble up at the cathode, and chlorine gas will bubble at the anode.
As the electricity from the battery passes through and between the electrodes, the water splits into hydrogen and chlorine gas, which collect as very tiny bubbles around the electrode tips. Hydrogen collects around the cathode and chlorine gas collects around the anode.

Number of cells $(n)=36$; Emf of each cell $(e)=4 V$; Total emf $(E)=88 V$;
Let the number of reversed cells be ' $y$ '
The required equation: $E_{E F F}= n \times e -2 y \times e$
$88=36 \times 4-2 y \times 4$
$88=144-8 y$
$8 y=56$
$y=7$

$\text{Q}=\text{CE}\Big(1-\text{e}^{\frac{-\text{t}}{\text{Rc}}}\Big)$
$\text{C}-500\times10^{-6}\text{F}$
$\text{R}=10^4\text{W}$
$\text{t}=\text{Rc}=10^4\times500\times10^{-6}=5$
$\text{t}=5\text{sec}$
$\text{Q}=\text{C}\in\Big(1-\text{e}^{-\text{t}}\Big)=\text{C}\in\Big(1-\frac{1}{\text{e}}\Big)=0.63\text{c}\in$
Thus, $63\%$ of the maximum charge is deposited in one time constant.

With the help of the figure we can say that the capacitor in the first $5s$ is larger than the charge stored in the next any second.
$\Rightarrow\text{at}\ \text{t}=\infty$
$\text{Q}=\text{Q}_0\Big(1-\text{e}^{-\infty}\Big)=\text{Q}_0=\text{C}\in$
$\therefore\text{t}_\infty-\text{t}_5=\text{C}\in-63=37$
after $t = 5 \sec$., maximum charge is deposited is only $37\%.$

Key concept: Current per unit area $($taken normal to the current$), \frac{\text{I}}{\text{A}}$, is called current density and is denoted by $\vec{\text{J}}$.

The is unti of the currnt density are $\frac{\text{A}}{\text{m}^2}$. The current density is also directed along $E$ and which is also a vector quantity and the relationship is given by
$\vec{\text{J}}=\sigma\vec{\text{E}}=\frac{\vec{\text{E}}}{\rho}$
where $\sigma$ = conductivity and $\rho =$ resistivity of specific resistance of the the surface of wire.

An electrochemical cell is a device which converts chemical energy to electric energy, and maintains the flow of charge in a circuit. There are $2$ types of electrochemical cells $–$ Galvanic cell and Electrolytic cell.

As, $\text{R}=\rho\Big(\frac{\text{I}}{\text{A}}\Big)$
If we calculate $R$ with the help of above relation then,
From option $(a)$ we will get $\text{R}=20\rho\Omega$
From option $(b)$ we will get $\text{R}=\Big(\frac{\rho}{20}\Big)\Omega$
From option $(c)$ we will get $\text{R}=\Big(\frac{\rho}{5}\Big)\Omega$
Clearly, the resistance will be maximum when the battery is connected across $1\text{cm}\times\frac{1}{2}\text{cm}$ faces.

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Blue: $0.25\%$
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