Q&A: practice questions with answers

Concept: Early effect (base-width modulation): increasing the collector–base reverse bias widens the CB depletion layer into the base, reducing the effective base width.

1. Narrower base ⇒ fewer carriers recombine in the base ⇒ more reach the collector.
2. So $I_C$ rises slightly even with $I_E$ fixed ($\alpha$ increases a little).
3. Option C is the opposite — injection efficiency is not reduced; it is the base width that shrinks.

Trick: Early effect = base narrows as $V_{CB}$ rises ⇒ $I_C$ creeps up.

Answer: (D) The effective base width decreases with collector–base voltage

Concept: With resistance constant, $P=\frac{V^2}{R}$ — power varies as the square of voltage.

1. $V\to 2V$ ⇒ $P^{\prime }=\frac{(2V{)}^2}{R}=4\frac{V^2}{R}=4P$.
2. Doubling $V$ also doubles $I$ ($I=\frac{V}{R}$), and $P=VI$ gives $2\times 2=4$ times.

Trick: Fixed R: double V ⇒ 4× power; halve V ⇒ $\frac{1}{4}$ power.

Answer: (A) four times

Concept: $R=\rho \frac{L}{A}$. Same material ⇒ same $\rho$; equal $R$ ⇒ $\frac{L_1}{A_1}=\frac{L_2}{A_2}$.

1. $\frac{L}{A_1}=\frac{2L}{A_2}\Rightarrow A_2=2A_1$.
2. $A_1:A_2=1:2$.

Trick: For equal resistance, area must grow in the same ratio as length.

Answer: (B) $1:2$

Concept: In a series circuit there is only one path, so the same current flows through every resistor.

1. Voltage divides: $V_i=I{R}_i$ (different for different $R$).
2. Power $P_i=I^2{R}_i$ also differs; resistances are generally different.
3. Only the current $I$ is identical in each element.

Trick: Series ⇒ same current; parallel ⇒ same voltage.

Answer: (C) Current

Concept: Weight $W=mg$ is the gravitational force on a body — a vector directed towards the Earth's centre, measured in newton.

1. Kilogram measures mass, not weight (A wrong).
2. Weight has a definite direction (downward), so C and D are wrong.
3. Hence weight has both magnitude and direction.

Trick: Mass → scalar, kg. Weight → vector (force), newton.

Answer: (B) Weight has both magnitude and direction

Concept: Peak-to-peak voltage $=$ (divisions) $\times$ (volts/div). $V_{rms}=\frac{V_p}{\sqrt{2}}$, $I_{rms}=\frac{V_{rms}}{R}$.

1. $V_{pp}$ across R $=2\times 1=2\text{V}$ ⇒ $V_p=1\text{V}$.
2. $V_{rms}=\frac{1}{\sqrt{2}}\approx 0.707\text{V}$.
3. $I_{rms}=\frac{0.707}{10}\approx 0.071\text{A}$.

Trick: Total height = peak-to-peak; halve it for peak, divide by $\sqrt{2}$ for RMS.

Answer: (B) $0.071\text{A}$

Concept: OR is not available, so use De Morgan: $X+Y=\overline{\stackrel{―}{X}·\stackrel{―}{Y}}$.

1. $ABC$ with 2-input ANDs: $(A·B)·C$ → 2 AND gates. $DE$ → 1 AND gate.
2. OR of the two terms: invert both (2 NOT), AND them (1 AND), invert the result (1 NOT).
3. Total: $2+1+1=4$ AND gates and $2+1=3$ NOT gates.

Trick: An OR built from AND + NOT always costs 1 AND + 3 NOT.

Answer: (D) 4 AND gates and 3 NOT gates

Concept: $W=mg\Rightarrow g=\frac{W}{m}$.

1. $g=\frac{49}{5}=9.8{\text{m/s}}^2$.

Trick: On Earth $g\approx 9.8{\text{m/s}}^2$; a 1 kg mass weighs about 9.8 N.

Answer: (B) $9.8{\text{m/s}}^2$

Concept: $\varphi =\overrightarrow{B}·\overrightarrow{A}=BA\cos \theta$, where $\theta$ is the angle between $\overrightarrow{B}$ and the normal to the surface.

1. Plane perpendicular to the field ⇒ normal is along $\overrightarrow{B}$ ⇒ $\theta =0$, $\cos \theta =1$.
2. Area of loop $A=\pi r^2$, so $\varphi =\pi r^{2}B$ (maximum flux).

Trick: Flux uses the area $\pi r^2$, never the circumference $2\pi r$.

Answer: (D) $\pi r^{2}B$

Concept: Work $W=qV$; power $P=\frac{W}{t}$.

1. $W=3\times 15=45\text{J}$.
2. $P=\frac{45}{3}=15\text{W}$.

Trick: Also $P=VI$ with $I=\frac{q}{t}=1\text{A}$ ⇒ $P=15\times 1=15\text{W}$.

Answer: (C) $15\text{W}$

Concept: In a parallel connection each branch is connected directly across the supply and works independently.

1. If one branch opens, the other branches still have the full supply voltage and keep conducting.
2. Total current falls slightly, but neither the voltage becomes zero nor the battery gets damaged.
3. This is why house wiring is parallel — one bulb fusing does not switch off the rest.

Trick: Parallel: one breaks, rest work. Series: one breaks, all stop.

Answer: (A) Other branches keep working

Concept: In a semiconductor, heating breaks more covalent bonds, creating extra electron–hole pairs. Conductivity $\sigma =ne\mu$ rises because carrier density $n$ increases sharply.

1. More carriers ⇒ higher $\sigma$ ⇒ lower resistivity $\rho =\frac{1}{\sigma }$.
2. Band gap actually decreases slightly with temperature (A wrong); lattice vibrations increase, not stop (B wrong).
3. This is the opposite of metals, where $\rho$ increases with temperature.

Trick: Semiconductor: hotter ⇒ more carriers ⇒ less resistance (negative temperature coefficient).

Answer: (C) More charge carriers are generated

Concept: Port 0 of the 8051 is an open-drain port: it has no internal pull-up resistors, unlike Ports 1, 2 and 3.

1. To use P0 as a general-purpose I/O port, external pull-ups (typically 10 kΩ) must be added.
2. P0 is also multiplexed as AD0–AD7 (low address/data bus) for external memory, but it is not reserved only for that and it can act as output.
3. All four ports are bit-addressable, so D is wrong.

Trick: P0 = open drain = needs external pull-ups.

Answer: (B) It does not contain internal pull-up resistors

Concept: An insulator has almost no free electrons, so it does not let current pass. Plastic (PVC) is the common insulation on wires.

1. Graphite conducts (free electrons in its layered structure) — used in pencils and electrodes.
2. Iron is a metal and mercury is a liquid metal — both are good conductors.
3. Hence plastic is the only insulator in the list.

Trick: Plastic, rubber, glass, mica, dry wood = insulators; all metals + graphite = conductors.

Answer: (A) Plastic

Concept: $F=BIL\sin \theta$; for a conductor perpendicular to the field $\theta ={90}^{\circ }$, $\sin \theta =1$.

1. $F=0.2\times 5\times 1\times 1=1.0\text{N}$.

Trick: Perpendicular ⇒ just multiply $B\times I\times L$.

Answer: (A) $1.0\text{N}$

Concept: Time $=$ (divisions on X-axis) $\times$ (time/div). A positive half cycle is half the period.

1. Half cycle $=2\times 2\text{ms}=4\text{ms}$.
2. Full period $T=2\times 4=8\text{ms}$.

Trick: Half cycle given ⇒ double it for $T$. Then $f=\frac{1}{T}=125\text{Hz}$.

Answer: (B) $8\text{ms}$

Concept: Ohm's law ($V\propto I$) holds only when physical conditions — especially temperature — remain constant.

1. If the conductor heats up, its resistance changes, so the V–I graph bends and $\frac{V}{I}$ is no longer constant.
2. Options A, B and D describe exactly the conditions under which the law does hold (constant temperature, linear V–I, metallic conductor).

Trick: Ohm's law fails for: changing temperature, diodes/semiconductors, electrolytes, gases.

Answer: (C) The temperature of the conductor increases

Concept: From $R=\rho \frac{L}{A}$, $\rho =\frac{RA}{L}$.

1. Units: $\frac{\Omega \times {\text{m}}^2}{\text{m}}=\Omega \text{m}$ (ohm-metre).
2. Ohm is the unit of resistance, not resistivity; 'ohm per metre' would be $\frac{\Omega }{\text{m}}$, which is wrong.

Trick: Resistivity = ohm-metre; conductivity = siemens per metre.

Answer: (C) ohm-metre

Concept: $\varphi =BA\cos \theta$ — flux depends on the field strength $B$, the area $A$ and the orientation (angle $\theta$ between $\overrightarrow{B}$ and the area normal).

1. All three factors appear in the formula, so none alone decides the flux.
2. Options B, C and D each pick only one factor and are therefore incomplete.

Trick: Flux = $B$ × $A$ × $\cos \theta$ — three things, remember all three.

Answer: (A) Area, magnitude of B and its orientation

Concept: Electric field at a point is the force experienced by a unit positive test charge placed there: $\overrightarrow{E}=\frac{\overrightarrow{F}}{q}$, unit N/C (or V/m).

1. 'Energy per unit charge' is electric potential (volt), not field — option B is the tempting wrong one.
2. Force per unit mass is a gravitational field, not electric.

Trick: Field = force ÷ charge; potential = energy ÷ charge.

Answer: (C) Force acting per unit charge