Q&A: practice questions with answers

Concept: The Intel 8051 is an 8-bit microcontroller with on-chip ROM, RAM, four 8-bit I/O ports, a serial port and timers.

1. 8051 has two 16-bit timer/counters: Timer 0 (TH0/TL0) and Timer 1 (TH1/TL1).
2. Each is controlled through the TMOD and TCON registers and can work as a timer or an event counter.
3. The 8052 variant has three timers – a common distractor.

Yaad rakhein: 8051 → 2 timers, 8052 → 3 timers.

Answer: (C) two

Concept: In multiplication or division, the result is reported with as many significant figures as the measurement having the fewest significant figures.

1. $3.24$ has 3 significant figures.
2. $0.006$ has only 1 significant figure (leading zeros are not significant).
3. $3.24\times 0.006=0.01944\approx 0.02$ → 1 significant figure.

Trick: Multiply/divide → least sig. figs.; add/subtract → least decimal places.

Answer: (C) 1

Concept: Net work = work by the net force. Friction force $f={\mu }_{k}N={\mu }_{k}mg$ on a horizontal surface, and $W=F\times d$.

1. Normal force $N=mg=5\times 10=50\text{N}$; friction $f=0.2\times 50=10\text{N}$.
2. Net force $=20-10=10\text{N}$ in the direction of motion.
3. Net work $=10\times 10=100\text{J}$ (equals the gain in kinetic energy).

Trick: $W_{\text{net}}=(F-\mu mg)d$; here $200\text{J}$ by the push minus $100\text{J}$ lost to friction.

Answer: (C) 100 J

Concept: PCB stands for Printed Circuit Board – a board of insulating material (fibreglass/epoxy) on which copper tracks are "printed" (etched) to connect electronic components.

1. Components are soldered onto the copper pads; tracks replace loose wires.
2. "Printed" refers to the photo-etching process of the copper pattern, not to pictures.
3. "Control board" is a different, generic term for a controller module, so B is the exact expansion.

Yaad rakhein: PCB = Printed Circuit Board (green board inside every gadget).

Answer: (B) Printed Circuit Board

Concept: Ohm's law: $V=IR$, so at constant resistance the current is directly proportional to the applied voltage, $I\propto V$.

1. $I_1=\frac{V}{R}$; new current $I_2=\frac{2V}{R}=2I_1$.
2. So the current doubles.
3. "Four times" would apply to power ($P=\frac{V^2}{R}$), not to current.

Trick: $R$ fixed → $I$ follows $V$ exactly; $P$ follows $V^2$.

Answer: (A) It doubles

Concept: A p-type semiconductor is made by adding a trivalent (3 valence electrons) impurity such as boron, gallium, indium or aluminium to Si/Ge; each impurity atom creates a hole.

1. Pentavalent atoms (antimony, arsenic, phosphorus) donate electrons → n-type, so B and C are wrong.
2. Option D pairs "three" with phosphorus, which actually has five valence electrons – inconsistent.
3. Only option A gives a trivalent element: gallium (Group 13).

Yaad rakhein: p-type → trivalent (B, Al, Ga, In); n-type → pentavalent (P, As, Sb).

Answer: (A) three; gallium

Concept: Fleming's Right-Hand Rule gives the direction of induced current/EMF in a conductor moving in a magnetic field (generator rule).

1. Thumb = motion of conductor, forefinger = magnetic field, middle finger = induced current.
2. Fleming's Left-Hand Rule is for the force on a current-carrying conductor (motor rule).
3. Lenz's law gives the sense of the induced EMF in terms of opposing the change, but the "rule" asked is the right-hand rule; Ampere's law relates current to its magnetic field.

Yaad rakhein: "GeneRator → Right hand; Motor → Left hand".

Answer: (D) Fleming's Right-Hand Rule

Concept: The quality factor of a resonant circuit is $Q=2\pi \times \frac{\text{maximum energy stored}}{\text{energy dissipated per cycle}}$ – i.e. how small the energy loss is compared with the energy stored.

1. For a series RLC circuit, $Q=\frac{{\omega }_{0}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}$; lower $R$ (less loss) → higher $Q$ and sharper resonance.
2. $Q$ is not the impedance (at resonance $Z=R$), not the stored energy itself, nor the resonance frequency.

Yaad rakhein: High $Q$ = low loss = narrow bandwidth ($\text{BW}=\frac{f_0}{Q}$).

Answer: (D) rate of energy loss relative to energy stored

Concept: Angular momentum $L=mvr$ (or $I\omega$), so its unit is (mass)(velocity)(distance).

1. $\text{kg}\times {\text{m s}}^{-1}\times \text{m}={\text{kg m}}^2{\text{s}}^{-1}$.
2. Check: $I\omega ={\text{kg m}}^2\times {\text{s}}^{-1}$ – same result; it is also $\text{J s}$.
3. ${\text{kg m s}}^{-1}$ is linear momentum; ${\text{kg m}}^2{\text{s}}^{-2}$ is energy/torque.

Trick: Angular momentum = momentum $\times$ one extra metre → ${\text{kg m}}^2{\text{s}}^{-1}$ (same as Planck's constant $h$).

Answer: (A) ${\text{kg m}}^2{\text{s}}^{-1}$

Concept: The 8085 has 8-bit general-purpose registers (A, B, C, D, E, H, L), temporary W and Z, an 8-bit flag register, and two 16-bit registers: the Program Counter (PC) and the Stack Pointer (SP).

1. SP holds the 16-bit memory address of the top of the stack.
2. W and Z are 8-bit temporary registers used internally during instruction execution; the flag register is 8-bit (5 flags used).

Yaad rakhein: 16-bit in 8085 = PC and SP (both hold addresses).

Answer: (D) Stack pointer

Concept: The barrier (built-in) potential of a p-n junction decreases as junction temperature rises – about $2\text{mV}$ per ${}^{\circ }\text{C}$ for both Si and Ge.

1. Higher temperature generates more minority carriers, which partly neutralise the depletion-layer charge, lowering the barrier.
2. So less barrier potential at higher temperature (A) is correct; D states the opposite and B denies any relation.
3. A conducting diode dissipates power, so its junction is hotter, not cooler, than the surroundings (C wrong).

Yaad rakhein: Diode forward voltage falls $\approx 2\text{mV}{/}^{\circ }\text{C}$ – negative temperature coefficient.

Answer: (A) There is less barrier potential at higher junction temperature.

Concept: Accuracy = closeness of a measurement to the true value. Precision = closeness of repeated measurements to each other (repeatability), regardless of the true value.

1. Statement A defines precision exactly as repeatability → correct.
2. B merges the two terms; C describes repeatability but calls it accuracy; D says precision gives the "quoted (true) value", which is accuracy.

Yaad rakhein: Precision = "same again and again"; Accuracy = "hits the bull's-eye".

Answer: (A) The precision is the degree to which the repeated measurement gives the same result.

Concept: The ampere (A) is the SI base unit of electric current. Charge is defined from it: $1\text{C}=1\text{A}\times 1\text{s}$, so the coulomb follows directly from the definition of the ampere.

1. $Q=It$ → coulomb $=\text{A}·\text{s}$ (only ampere and second involved).
2. Volt $=\text{W/A}$, ohm $=\text{V/A}$ and watt $=\text{J/s}$ all need kg and m as well – they are not derived from the ampere alone.

Yaad rakhein: Coulomb = ampere-second.

Answer: (D) Coulomb

Concept: Conductance $G=\frac{1}{R}=\frac{I}{V}$, so $\text{S}=\text{A/V}={\Omega }^{-1}$. Express the volt in base units and invert.

1. $\text{V}=\frac{\text{W}}{\text{A}}=\frac{{\text{kg m}}^2{\text{s}}^{-3}}{\text{A}}={\text{kg m}}^2{\text{s}}^{-3}{\text{A}}^{-1}$.
2. $\text{S}=\frac{\text{A}}{\text{V}}=\frac{\text{A}}{{\text{kg m}}^2{\text{s}}^{-3}{\text{A}}^{-1}}={\text{kg}}^{-1}{\text{m}}^{-2}{\text{s}}^3{\text{A}}^2$.
3. Option C is the ohm (reciprocal of S); B is the volt.

Trick: Ohm $={\text{kg m}}^2{\text{s}}^{-3}{\text{A}}^{-2}$; flip every sign for siemens.

Answer: (A) ${\text{kg}}^{-1}·{\text{m}}^{-2}·{\text{s}}^3·{\text{A}}^2$

Concept: A tungsten filament glows at about $2500\text{–}{3000}^{\circ }\text{C}$. The bulb is filled with chemically inactive gases (argon, nitrogen, sometimes krypton) instead of air.

1. Without oxygen the hot filament cannot oxidise (burn), and the gas pressure slows down evaporation of tungsten atoms – so the filament lasts longer.
2. The gas does not cool the filament usefully, nor raise its temperature; a vacuum alone would let tungsten evaporate faster.

Yaad rakhein: Argon/nitrogen in a bulb = "no burning, less evaporation" = longer filament life.

Answer: (C) To prolong the life of the filament

Concept: In a common-emitter (CE) amplifier the output is taken from the collector: when the input (base) voltage rises, collector current rises and $V_C=V_{CC}-I_C{R}_C$ falls. So the output is inverted.

1. Inversion means a phase shift of ${180}^{\circ }=\pi$ radians between input and output.
2. Common-collector (emitter follower) and common-base configurations give no phase shift ($0^{\circ }$) – so option D belongs to them, not CE.
3. $\frac{\pi }{2}$ or $\frac{3\pi }{2}$ shifts occur in reactive (RC/LC) networks, not in the transistor itself.

Yaad rakhein: CE = inverting amplifier = ${180}^{\circ }$ ($\pi$) phase shift.

Answer: (B) There is a phase shift of $\pi$ between the input and output voltage.

Concept: For multiplication and division, the result keeps the number of significant figures of the least precise quantity.

1. $0.00456$ → 3 significant figures (leading zeros do not count).
2. $0.12$ → 2 significant figures.
3. $0.00456÷0.12=0.038$ → reported to 2 significant figures.

Trick: Leading zeros never count; count from the first non-zero digit.

Answer: (B) 2

Concept: Joule's law: heat produced in a conductor is directly proportional to the square of the current, the resistance and the time.

1. $H=I^{2}Rt$ (in joules when $I$ in A, $R$ in $\Omega$, $t$ in s).
2. Using $V=IR$ it can also be written as $H=VIt=\frac{V^2}{R}t$.
3. Options A, B and C have the wrong power or position of $R$ and $t$.

Trick: "I square R t" — only the current is squared.

Answer: (D) $H=I^{2}Rt$

Concept: Seebeck effect: the emf of a thermocouple is not linear in temperature difference; it is well approximated by a parabola.

1. $E=a\Delta t+b(\Delta t{)}^2$, where $a,b$ depend on the metal pair.
2. The linear term dominates for small $\Delta t$; the square term gives the 'neutral' and 'inversion' temperatures.
3. Option C is only the first-order approximation; A and B are not thermocouple relations.

Trick: Thermo-emf curve is a parabola → linear + square term.

Answer: (D) $E=a(\Delta t)+b(\Delta t{)}^2$

Concept: Fleming's Left-Hand Rule (motor rule): stretch the thumb, forefinger and middle finger mutually perpendicular.

1. Forefinger → magnetic Field (B)
2. Middle finger → Current (I)
3. Thumb → Force/Motion (F)
4. Induced EMF belongs to the Right-Hand (generator) rule, so option B is wrong.

Trick: F-B-I from thumb to middle finger: Force, B-field, I (current). Left = motor, Right = generator.

Answer: (C) Direction of current