Current electricity practice · Mixed difficulty

Option B ($R \, \, \ll \, \, r$) is correct because maximum current is obtained when the external resistance is much smaller than the internal resistance in a parallel combination of cells.

Option D ($\mathrm{50 }\,\mathrm{W, 200 }\,\mathrm{W}$) is correct because when bulbs are in series, the power decreases, and when in parallel, the power increases.

Option B ($\alpha-\frac{\beta^2}{3 \gamma}$) is correct because it is the minimum value of the current obtained by differentiating the current equation with respect to time and setting it to zero.

Option A is correct because statements I and II are correct and statement II provides the correct explanation for statement I. Addition and subtraction require same units, and multiplication/division can be done with different units.

Option B ($\frac{{n{e^2}\tau }}{{{m_e}}}$) is correct because the electrical conductivity is directly proportional to the free electron density, the square of the electron charge, and the relaxation time, and inversely proportional to the electron mass.

Option A ($1.7 \times 10^{-4} \mathrm{~V/m}$) is correct because it is calculated using the charging equation for a capacitor and the relationship between electric field and voltage.

Option C (Positive and negative ions) is correct because in electrolysis, the current is carried by the movement of positive and negative ions within the electrolyte.

Option C (Decreases) is correct because semiconductors have a negative temperature coefficient of resistance.

Option C ($6 \,\Omega$) is correct because when the bridge is balanced, the ratio of resistances in opposite arms must be equal. The equivalent resistance of S in parallel with 6 ohms is used in the Wheatstone bridge equation.

Option B (only $25W$ bulb will fuse) is correct because the $25W$ bulb has higher resistance and will dissipate more power, exceeding its rating.

Option A is correct because both the assertion and reason are true, and the reason correctly explains why the electric field outside a current-carrying wire is zero.

The correct answer is A because the electric field outside a steady current-carrying wire is zero due to the net charge being zero, which cancels out the electric field.

Option D (Equal to sum of component resistors) is correct because the equivalent resistance in a series circuit is the sum of all individual resistances.

Option A ($R_B > R_A$) is correct because resistance is inversely proportional to the slope of the V-I graph, and conductor B has a smaller slope.

Option B (The resistance of an incandescent lamp is greater when the lamp is switched off) is the wrong statement because the resistance of the lamp increases with temperature, hence it's lower when switched off.

Option B (If both Assertion and Reason are true but Reason is not correct explanation of Assertion) is correct because the assertion is true based on the power formula, and the reason is true but doesn't directly explain the 4% power increase.

Option C (Statement-1 is true, Statement-2 is false) is correct because when the diameter is halved, the resistance becomes 16 times, and resistance increases with length.

Option D (${10^{8}}\,\mathrm{V}{\mathrm{m}^{ - 1}}$) is correct because it is the potential gradient calculated using the formula $k = \frac{{I\rho }}{A}$.

Option D (57600 km) is correct because it accurately calculates the height above the Earth's surface where gravity is 1% of its surface value using the inverse square law.

Option B (The potential drop across the $\mathrm{60}\,\mathrm{W}$ bulb is greater than the potential drop across the $\mathrm{200}\,\mathrm{W}$ bulb) is correct because in a series circuit, the voltage drop is proportional to the resistance, and the 60W bulb has higher resistance.

Option B (Is the same as the outside value) is correct because the current is the same throughout a series circuit, including the copper voltameter.

Option A ($1.4 \mathrm{~V}$) is correct because the voltage drop across R is the emf minus the voltage drop across the internal resistance.

Option A (High specific resistance and low melting point) is correct because a fuse needs to generate heat quickly (high resistance) and melt easily (low melting point) to break the circuit.

Option C (The drift speed decreases on moving from A to B) is correct because drift velocity is inversely proportional to the cross-sectional area.

Option A (Length) is correct because the length of the fuse wire is designed to be as short as possible to minimize resistance; the material, radius, and current rating are crucial factors.

Option C (Temperature coefficient of resistance is negligible) is correct because constantan wire is used for standard resistances due to its negligible temperature coefficient.

The correct answer is 500 because the average translational kinetic energy of N2 gas molecules at 500°C equals the KE of an electron accelerated through 0.1V.

Option C (Manganin) is correct because manganin has a low temperature coefficient of resistance, making it suitable for potentiometer wires.

Option D (If both Assertion & Reason are False) is correct. There is an electric field inside a current-carrying wire, and the net charge is zero.

Option D is correct. When cells are connected in a loop, the potential drop across any number of cells is zero due to the closed-loop configuration.