Reasoning: Drift velocity $\mathbf{(}\mathbf{v}_{\mathbf{d}}\mathbf{)}$ is given by the formula: $\mathbf{v}_{\mathbf{d}}\mathbf{= \mu E}$, where $\mathbf{\mu}$ is the electron mobility and $\mathbf{E}$ is the electric field strength. Mobility itself depends on factors like temperature and the material’s properties.
● The length of the wire affects the potential difference across the wire, but not directly the drift velocity of individual electrons within the wire.
● The cross-sectional area affects the current but not the drift velocity of individual electrons.
● The number of free electrons affects the current density, which in turn affects the current, but not directly the drift velocity of an individual electron.
● The electric field strength directly affects the drift velocity.

Reasoning: As explained above, the drift velocity is directly proportional to the electric field strength and the mobility of the electrons in the material.
● The time consumed is related to the total distance traveled by the electrons, but not the velocity itself.
● The number of free electrons affects the current, but not the drift velocity of an individual electron.
● The length of the wire affects the potential difference, but not directly the drift velocity.
● The strength of the electric field directly affects the drift velocity.

Reasoning: Given: $\mathbf{R = 10}\mathbf{\ }\mathbf{\Omega}\mathbf{\ }$, $\mathbf{V = 10}\mathbf{\ }\mathbf{V}$
Using: $\mathbf{P =}$ $\frac{\mathbf{V}^{\mathbf{2}}}{\mathbf{R}}$ $\mathbf{\Longrightarrow}$ $\mathbf{P =}$ $\frac{\mathbf{(10)}^{\mathbf{2}}}{\mathbf{10}}\mathbf{= 10}\mathbf{\ }\mathbf{W}$
Therefore, the power dissipated is $\mathbf{10}\mathbf{\ }\mathbf{W}$.

Reasoning: Power: $\mathbf{1000}\mathbf{\ }\mathbf{W = 1}\mathbf{\ }\mathbf{kW}$
Total operating time: $\mathbf{t = 10}\mathbf{\ }\mathbf{h}\mathbf{\ }\mathbf{/}\mathbf{\ day}\mathbf{\times 10}\mathbf{\ days}\mathbf{= 100}\mathbf{\ }\mathbf{h}$ Electrical energy: $\mathbf{E = Pt}$ $\mathbf{\Longrightarrow}$ $\mathbf{E = 1}\mathbf{\ }\mathbf{kW \times 100}\mathbf{\ }\mathbf{h = 100}\mathbf{\ }\mathbf{kWh}$ Therefore, the refrigerator consumes $\mathbf{100}\mathbf{\ }\mathbf{kWh}$.

Reasoning: Given: $\mathbf{I = 3}\mathbf{\ }\mathbf{A}\mathbf{\ }$, $\mathbf{t = 5}\mathbf{\ }\mathbf{\min}\mathbf{s}\mathbf{= 300}\mathbf{\ }\mathbf{s}\mathbf{\ }$, $\mathbf{W = 900}\mathbf{\ }\mathbf{J}$
Using: $\mathbf{W = VIt}$ Therefore, $\mathbf{V =}$ $\frac{\mathbf{W}}{\mathbf{It}}$ $\mathbf{\Longrightarrow}$ $\mathbf{V =}$ $\frac{\mathbf{900}}{\mathbf{3}\mathbf{\ }\mathbf{\times}\mathbf{\ }\mathbf{300}}\mathbf{= 1}\mathbf{\ }\mathbf{V}$ Hence, the e.m.f. is $\mathbf{1}\mathbf{\ }\mathbf{V}$.

Reasoning: A potentiometer measures e.m.f. by the null method. At the balance point, it draws no current from the source being measured. Therefore, there is no voltage loss due to the internal resistance of the source, making it more accurate than a voltmeter.

Reasoning: The standard metric (SI) unit for any electric charge – whether it is static (at rest) or dynamic (in motion) – is the Coulomb $\mathbf{(}\mathbf{C}\mathbf{)}$. “Dynamic charge” simply refers to electric charge that is moving through a circuit.
● Note: An Ampere $\mathbf{(A)}$ does not measure the amount of charge itself. Instead, it measures the electric current, which is the rate of flow of that charge over a specific period of time.
One Ampere is defined mathematically as one Coulomb of charge passing through a point in one second:
$\mathbf{1\ Ampere =}$ $\frac{\mathbf{1\ Coulomb}}{\mathbf{1\ second}}$

Reasoning: Resistivity is an intrinsic property of a material. It does not depend on the length, diameter, or cross-sectional area of the wire, provided its temperature remains constant. Thus, doubling the length and diameter changes the wire’s resistance but does not change the resistivity. The approximate resistivity of steel is of the order:
$\mathbf{\rho \approx}$ $\mathbf{1 \times}\mathbf{10}^{\mathbf{- 7}}\mathbf{\ }\mathbf{\Omega}\mathbf{\ }\mathbf{m}$

Reasoning: A$\mathbf{\ 100\%}$ increase in current means that the current becomes double:
$\mathbf{I}^{\mathbf{'}}\mathbf{= 2I}$ $\mathbf{\Longrightarrow}$ At constant resistance: $\mathbf{P =}\mathbf{I}^{\mathbf{2}\mathbf{\ }}\mathbf{R}$
Therefore, $\mathbf{P}^{\mathbf{'}}\mathbf{=}{\mathbf{(2I}\mathbf{)}}^{\mathbf{2}\mathbf{\ }}\mathbf{R = 4}\mathbf{I}^{\mathbf{2}\mathbf{\ }}\mathbf{R = 4P}$ The new power is four times the original power. Hence, the increase is $\mathbf{4P - P = 3P}$.
$\mathbf{Percentage\ increase =}\mathbf{300\%}$

Reasoning: Total resistance in series: $\mathbf{R}_{\mathbf{T}}\mathbf{= 2 + 4 = 6}\mathbf{\ }\mathbf{\Omega}$
Current in the circuit: $\mathbf{I =}$ $\frac{\mathbf{V}}{\mathbf{R}_{\mathbf{T}}}\mathbf{=}\frac{\mathbf{6}}{\mathbf{6}}\mathbf{= 1}\mathbf{\ }\mathbf{A}$ Heat dissipated by the $\mathbf{4}\mathbf{\ }\mathbf{\Omega}$ resistor:
$\mathbf{H =}\mathbf{I}^{\mathbf{2}\mathbf{\ }}\mathbf{Rt}$ $\mathbf{\Longrightarrow}$ $\mathbf{H =}\mathbf{(1)}^{\mathbf{2}}\mathbf{\ }\mathbf{(4)}\mathbf{\ }\mathbf{(5) = 20}\mathbf{\ }\mathbf{J}$
Therefore, the heat dissipated is $\mathbf{20}\mathbf{\ }\mathbf{J}$.

Reasoning: While applying Kirchhoff’s voltage law, the sign of the potential change across a resistor depends on the direction in which the loop is traced relative to the current.
Moving in the direction of current gives a potential drop: $\mathbf{-}\mathbf{\ }\mathbf{IR}$ Moving opposite to the direction of current gives a potential rise: $\mathbf{+}\mathbf{\ }\mathbf{IR}$

Reasoning: Given: $\mathbf{V = 10}\mathbf{\ }\mathbf{V}\mathbf{\ }$, $\mathbf{I = 12}\mathbf{\ }\mathbf{mA = 0.012}\mathbf{\ }\mathbf{A}$
Total required resistance: $\mathbf{R}_{\mathbf{T}}\mathbf{=}$ $\frac{\mathbf{V}}{\mathbf{I}}$ $\mathbf{\Longrightarrow}$ $\mathbf{R}_{\mathbf{T}}\mathbf{=}$ $\frac{\mathbf{10}}{\mathbf{0.012}}\mathbf{= 833.3}\mathbf{\ }\mathbf{\Omega}$ The resistance of the LDR is $\mathbf{20}\mathbf{\ }\mathbf{\Omega}$. Therefore, the protective series resistance is: $\mathbf{R}_{\mathbf{s}}\mathbf{=}\mathbf{R}_{\mathbf{T}}\mathbf{-}\mathbf{R}_{\mathbf{LDR}}$ $\mathbf{\Longrightarrow}$ $\mathbf{R}_{\mathbf{s}}\mathbf{= 833.3 - 20 = 813.3}\mathbf{\ }\mathbf{\Omega}$ The nearest value is $\mathbf{813}\mathbf{\ }\mathbf{\Omega}$.

Reasoning: A galvanometer is connected between the two middle points of a Wheatstone bridge. At the balance condition, no current passes through it, and it shows zero deflection. Therefore, it acts as a null detector.

Reasoning: A potentiometer can compare the e.m.f.s or potential differences of two sources by measuring their balancing lengths.
If $\mathbf{E}_{\mathbf{1}}$ and $\mathbf{E}_{\mathbf{2}}$ balance at lengths $\mathbf{l}_{\mathbf{1}}$and $\mathbf{l}_{\mathbf{2}}$, then: $\frac{\mathbf{E}_{\mathbf{1}}}{\mathbf{E}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{l}_{\mathbf{1}}}{\mathbf{l}_{\mathbf{2}}}$ Thus, it is commonly used for comparing two voltages or e.m.f.s.

Reasoning: The potential gradient of a potentiometer wire is: $\mathbf{k =}$ $\frac{\mathbf{V}}{\mathbf{L}}$
A longer wire gives a smaller potential gradient. Therefore, a small voltage produces a measurable balancing length, increasing the sensitivity and accuracy of the potentiometer.

Reasoning: Thermistors are mainly classified into two types:
● NTC thermistor: Its resistance decreases when temperature increases.
● PTC thermistor: Its resistance increases when temperature increases.

Reasoning: The word thermistor is formed from the words thermal and resistor. It is a temperature-sensitive resistor whose resistance changes considerably with a change in temperature.

Reasoning: The current flowing into a junction is equal to the current exiting the junction, according to Kirchhoff’s law. Kirchhoff’s Current Law (KCL) states that the total current entering a junction equals the total current leaving the junction. This follows from the conservation of electric charge.