Current Electricity, on the exam screen.

The terminal voltage is the electromotive force (emf) minus the drop across the internal resistance: $V = E - Ir = 12 - 0.6 \times 2 = 10.8$ V, option (4). The 1.2 V drop across the internal resistance is what is subtracted, and 12 V would only be the terminal voltage if no current flowed.

The 4 $\Omega$ wire splits equally, so each side of the square is 1 $\Omega$. With the 2 $\Omega$ across diagonal BD and the 2 V battery across A and C, the four 1 $\Omega$ arms form a balanced Wheatstone bridge, so no current flows through the 2 $\Omega$ branch. The effective resistance across AC is 1 $\Omega$, giving $I = E/R = 2/1 = 2$ A.

Why the other options are wrong:

- (2) Assumes far too low a resistance.
- (3) Arbitrary value from mis-combining resistances.
- (4) Would need 0.5 $\Omega$, as if each path from A to C had one 1 $\Omega$ side instead of two.

The heater resistance is fixed, so $P = V^2/R$ gives $P_c = \left(\frac{V}{V_0}\right)^2 P_0 = \left(\frac{200}{220}\right)^2 \times 400 \approx 331$ W. Power scales as the square of voltage, not linearly, so option (2) (unchanged) and option (1) (linear scaling) are wrong, and option (4) over-reduces.

A metre bridge is a Wheatstone network, whose balance (null) condition is symmetric with respect to interchanging the cell and the galvanometer, so the balance point does not move. Away from that point the bridge is unbalanced and the galvanometer deflects; the direction of deflection depends on which side of balance the jockey lies, and it reads zero exactly at the balance point.

Why the other options are wrong:

- (1) Deflection is not restricted to one side; it flips about the null point.
- (2) No deflection occurs only at balance, not everywhere.
- (3) Same error as (1) but for the opposite side.

In circuit (A) the resistors are in SERIES, so $R_A = R_0(1-\alpha\Delta T) + R_0(1+\alpha\Delta T) = 2R_0$ - the two temperature terms cancel exactly and $I_A$ is unchanged. In circuit (B) they are in PARALLEL, giving $R_B = \frac{R_0(1-\alpha^2\Delta T^2)}{2}$, which decreases as $\Delta T$ grows because of the $-\alpha^2\Delta T^2$ term, so $I_B$ increases. NCERT Class 12, Current Electricity (temperature dependence of resistivity).

Why the other options are wrong:

- (2) assumes the series combination must heat up and resist more, missing the exact cancellation between $+\alpha$ and $-\alpha$
- (3) swaps the two circuits, applying the parallel result to (A) and the series result to (B)
- (4) spots the cancellation in series and wrongly assumes the same cancellation holds for reciprocals in parallel

For copper, resistivity rises with temperature and is nearly linear over a limited ordinary-temperature range, but bends appreciably at very low temperatures. The source curve with that behaviour is option (2). Chapter: Current Electricity.

Why the other options are wrong:

- (1) Resistivity falling steeply and then levelling off as temperature rises is how a semiconductor behaves, not a metal like copper.
- (3) This curve shows resistivity decreasing with temperature, but the resistivity of copper increases with temperature.
- (4) A straight line treats copper as linear at every temperature, but the linear relation holds only over a limited range and the real curve bends at low temperatures.

Yellow and violet give digits 4 and 7, brown is the ×10 multiplier, and gold means $\pm$5%. Therefore R = $47 \times 10 = 470$ $\Omega$ with 5% tolerance. Chapter: Current Electricity.

Why the other options are wrong:

- (1) 47 k$\Omega$ would need an orange ($\times 10^{3}$) multiplier band, and 10% tolerance is a silver band, not gold.
- (2) 4.7 k$\Omega$ is $47 \times 10^{2}$ $\Omega$, which needs a red multiplier band; brown only multiplies by 10.
- (4) 470 k$\Omega$ is $47 \times 10^{4}$ $\Omega$, which needs a yellow multiplier band; brown only multiplies by 10.

At balance, R/10 = 3/2, so the 1.5 m resistance wire has R = 15 $\Omega$. Its length per ohm is 1.5/15 = 0.10 m. Chapter: Current Electricity.

Mobility $\mu$ is given by $\mu = \frac{v_d}{E}$, where $v_d$ is the drift velocity and $E$ is the electric field. Substituting the values, $\mu = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = 2.5 \times 10^6\ m^2V^{-1}s^{-1}$. Chapter: Current Electricity.

Metals have a positive temperature coefficient of resistance, while insulators and semiconductors have a negative temperature coefficient of resistance. This is due to the increase in carrier concentration in insulators and semiconductors with temperature. Chapter: Current Electricity.