Q&A: practice questions with answers

Concept: In series, the same current flows through both resistors and the voltage divides in proportion to resistance: $V_2=V\frac{R_2}{R_1+R_2}$.

1. $I=\frac{12}{10+20}=0.4\text{A}$.
2. $V_2=I{R}_2=0.4\times 20=8\text{V}$ (and $V_1=4\text{V}$; $4+8=12$ ✓).

Trick: Voltage splits in the ratio of resistances 10 : 20 = 1 : 2, so 12 V → 4 V and 8 V.

Answer: (C) 8 V

Concept: Power is the rate of doing work: $P=\frac{W}{t}$, and work done by a force is $W=F\times d$.

1. Substitute $W=F\times d$ into $P=\frac{W}{t}$.
2. $P=\frac{F\times d}{t}$, i.e. $P=F\times \frac{d}{t}=F\times v$.
3. Options A, B and C place $t$ or $d$ wrongly and do not give the unit watt ($\text{J/s}$).

Trick: Power = Force × Velocity, and velocity = distance/time.

Answer: (D) $P=F\times \frac{d}{t}$

Concept: Classically the ampere was defined through the magnetic force between two parallel current-carrying conductors.

1. One ampere is that constant current which, flowing in two infinitely long parallel conductors 1 m apart in vacuum, produces a force of $2\times {10}^{-7}\text{N}$ per metre of length between them.
2. Since 2019 the SI fixes the ampere via the elementary charge $e$, but the force-between-conductors definition is the one asked in exams.
3. Flux (weber), potential difference (volt) and force between charges (coulomb's law) define other quantities.

Yaad rakhein: Ampere ↔ force between two parallel wires = $2\times {10}^{-7}$ N/m.

Answer: (A) the force between two current-carrying conductors

Concept: Motional emf $e=Blv\sin \theta$, where $\theta$ is the angle between the conductor's motion and the field.

1. The emf is maximum when $\sin \theta =1$, i.e. the conductor cuts the field lines at ${90}^{\circ }$ (perpendicular).
2. Moving parallel to the field lines cuts no flux, so $e=0$ (option C is the minimum case).
3. A zero field gives zero emf; non-uniformity is not required.

Trick: Maximum flux cutting = perpendicular motion; Fleming's right-hand rule needs field, motion and emf mutually perpendicular.

Answer: (D) the conductor is perpendicular to the magnetic field lines

Concept: An electric motor converts electrical energy into mechanical energy using the force on a current-carrying conductor placed in a magnetic field ($F=BIl$).

1. Current in the coil + external magnetic field → force (Fleming's left-hand rule) → torque → rotation.
2. No heat, gravity or light interaction is involved in the basic principle; those options are distractors.

Yaad rakhein: Motor = left-hand rule (force); Generator = right-hand rule (emf).

Answer: (A) magnetic field and current-carrying conductors

Concept: An LVDT (Linear Variable Differential Transformer) is an inductive transducer whose output voltage varies linearly with core displacement within its rated range.

1. Output = difference of the two secondary voltages; for small core movements it is directly proportional to displacement.
2. Standard text statement: the output is practically linear for displacements up to about 5 mm; beyond that the curve bends.
3. "Exponential" is wrong – an LVDT is prized precisely for its linearity.

Yaad rakhein: LVDT = Linear, range about ±5 mm, infinite resolution, no friction.

Answer: (D) linear; 5 mm

Concept: The SI unit of pressure is the pascal ($1\text{Pa}=1{\text{N/m}}^2$); the bar is a convenient metric unit close to atmospheric pressure.

1. $1\text{bar}={10}^5\text{Pa}=100\text{kPa}$.
2. Compare: $1\text{atm}=1.013\times {10}^5\text{Pa}=1.013\text{bar}$; $1\text{bar}\approx 14.5\text{psi}$.
3. Option B (${10}^6{\text{N/m}}^2$) is 10 bar; option C is far too small.

Yaad rakhein: bar → ${10}^5$ Pa, atm → $1.013\times {10}^5$ Pa.

Answer: (A) ${10}^5$ Pa

Concept: Only quantities with the same dimensions can be added, so each term of $v=at+b{t}^2$ must have the dimensions of velocity $[\mathrm{L}{\mathrm{T}}^{-1}]$.

1. $[b{t}^2]=[v]=[\mathrm{L}{\mathrm{T}}^{-1}]$.
2. $[b]=\frac{[\mathrm{L}{\mathrm{T}}^{-1}]}{[{\mathrm{T}}^2]}=[\mathrm{L}{\mathrm{T}}^{-3}]$.
3. Similarly $[a]=[\mathrm{L}{\mathrm{T}}^{-2}]$ (option D is the dimension of $a$, not $b$).

Trick: Principle of homogeneity – divide the dimension of the left side by the dimension of the multiplying power of $t$.

Answer: (C) $[\mathrm{L}{\mathrm{T}}^{-3}]$

Concept: Fleming's right-hand rule gives the direction of current induced in a conductor that moves through a magnetic field (thumb = motion, forefinger = field, middle finger = current).

1. The rule has three inputs: field, motion and current. The loop here is stationary, so there is no "motion" direction to point the thumb along.
2. Hence Fleming's right-hand rule cannot assign any clockwise or anticlockwise direction, and within the frame of this rule no motional current arises.
3. Note: a current due to the changing field is governed by Faraday's law and Lenz's law, not by Fleming's rule – that is why options C and D cannot be decided here.

Yaad rakhein: Fleming's right hand = motional emf (needs movement); changing field without motion = Faraday/Lenz.

Answer: (B) There is no induced current since the loop is stationary

Concept: $P=\frac{W}{t}$, with work in joules and time in seconds giving watts.

1. $P=\frac{3000\text{J}}{15\text{s}}=200\text{W}$.
2. Option B ($5$ W) is the inverted ratio $\frac{15}{3}$; option A would need $6750$ J.

Trick: $3000÷15$: $15\times 200=3000$.

Answer: (C) 200 W

Concept: Electrical power $P=VI$, so $I=\frac{P}{V}$.

1. $I=\frac{60\text{W}}{120\text{V}}=0.5\text{A}$.
2. Bulb resistance (for interest): $R=\frac{V}{I}=\frac{120}{0.5}=240\Omega$.

Trick: Watts ÷ Volts = Amperes.

Answer: (A) 0.5 A

Concept: Average power $P=V_{rms}{I}_{rms}\cos \varphi$. In a pure capacitor the current leads the voltage by ${90}^{\circ }$, so $\cos \varphi =\cos {90}^{\circ }=0$.

1. Energy stored in the capacitor during one quarter cycle is returned to the source in the next quarter.
2. Net energy consumed over a full cycle $=0$; only reactive power ($VI\sin \varphi$) flows.
3. The same holds for a pure inductor.

Yaad rakhein: Pure L or pure C → power factor 0 → "wattless" current, zero real power.

Answer: (C) zero

Concept: For a long straight current-carrying wire, $B=\frac{{\mu }_{0}I}{2\pi r}$.

1. $B\propto I$ – directly proportional to the current (not its square).
2. $B\propto \frac{1}{r}$ – inversely proportional to the distance (option A says "inverse of distance", which is the same idea but the question's wording wants the direct proportionality; option B is wrong).
3. Doubling the current doubles the field; doubling the distance halves it.

Yaad rakhein: Straight wire: $B\propto \frac{I}{r}$; circular coil centre: $B\propto \frac{nI}{r}$.

Answer: (D) the current through the conductor

Concept: The figure shows two NOT gates feeding an AND gate: $Y=\overline{A}·\overline{B}$. By De Morgan's theorem $\overline{A}·\overline{B}=\overline{A+B}$.

1. Truth table: output is 1 only when $A=0$ and $B=0$ (both inverted inputs become 1).
2. "Output 1 only when all inputs are 0" is exactly the NOR gate.
3. A NAND gate would be an OR with inverted inputs ($\overline{A}+\overline{B}=\overline{AB}$).

Yaad rakhein: Bubbled-input AND = NOR; bubbled-input OR = NAND.

Answer: (C) NOR GATE

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

1. New voltage $=3V$ → $P^{\prime }=\frac{(3V{)}^2}{R}=\frac{9V^2}{R}=9P$.
2. Option C (3 times) would be true for $P=VI$ with fixed current, but here $R$ is fixed so $I$ also triples.

Trick: Fixed $R$: $V$ × $n$ ⇒ $P$ × $n^2$.

Answer: (A) It increases by nine times.

Concept: A common thermistor is an NTC (negative temperature coefficient) device made of semiconductor oxides: its resistance falls as temperature rises.

1. Higher temperature → more charge carriers → resistance decreases.
2. With a fixed supply voltage, $I=\frac{V}{R}$ → current increases.
3. Options A/D describe a metal (PTC) resistor, not the usual thermistor.

Yaad rakhein: Thermistor (NTC): hot → low R → high I.

Answer: (B) decrease; increase

Concept: Potential energy converts fully into kinetic energy: $mgh=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{2gh}$ (independent of mass).

1. $v=\sqrt{2\times 9.8\times 10}=\sqrt{196}=14\text{m/s}$.
2. The 2 kg mass does not affect the speed.
3. With $g=10$: $\sqrt{200}\approx 14.1$ m/s – still option C.

Trick: $v=\sqrt{2gh}$; for $h=10$ m, $v\approx 14$ m/s.

Answer: (C) 14 m/s

Concept: An $n$-bit ADC divides the input into $2^n$ levels; the quantisation error is at most $\pm \frac{1}{2}$ LSB.

1. Taking the reference amplitude as 1 V: step (1 LSB) $=\frac{1}{2^8}=\frac{1}{256}\approx 3.9\text{mV}$.
2. Maximum quantisation error $=\frac{1}{2}\text{LSB}\approx 1.95\text{mV}\approx 2\text{mV}$.
3. Among the options only 2 mV is of this order; 0.1–0.4 mV would need 10–12 bits.

Yaad rakhein: Resolution $=\frac{V_{ref}}{2^n}$, error $=\pm \frac{1}{2}$ LSB.

Answer: (C) 2

Concept: In addition/subtraction the result is reported to the least precise decimal place among the numbers (not the least number of significant figures).

1. Write both in plain form: $1.23456\times {10}^3=1234.56$ (hundredths place); $1.234\times {10}^2=123.4$ (tenths place).
2. Sum $=1357.96$; least precise place is the tenths → $1358.0$.
3. $1358.0$ has 5 significant figures.

Trick: Add → match decimal places; multiply → match significant figures.

Answer: (D) 5

Concept: $W=Fd\cos \theta$, where $\theta$ is the angle between force and displacement.

1. $W=20\times 13\times \cos {30}^{\circ }=260\times \frac{\sqrt{3}}{2}$.
2. $=260\times 0.866\approx 225\text{J}$.
3. Option C (260 J) ignores the angle ($\cos \theta =1$).

Trick: $\cos {30}^{\circ }\approx 0.87$ – take 87% of $Fd$.

Answer: (B) 225 Joules