Q&A: practice questions with answers

Concept: $P(M\cup Ph)=P(M)+P(Ph)-P(M\cap Ph)$.

1. Given $P(M)=\frac{3}{4}$, $P(M\cap Ph)=\frac{13}{36}$, $P(M\cup Ph)=\frac{5}{6}$.
2. $P(Ph)=\frac{5}{6}-\frac{3}{4}+\frac{13}{36}=\frac{30-27+13}{36}=\frac{16}{36}$.
3. $=\frac{4}{9}$.

Trick: Take LCM 36 for all three fractions at once: $30,27,13$.

Answer: (B) $\frac{4}{9}$

Concept: Powers first, then multiplication, then addition/subtraction left to right.

1. $3^3=27$, $5^2=25$, ${\left(\frac{26}{13}\right)}^2=2^2=4$, $0\times 9=0$.
2. Expression $=27-25+4-8+0$.
3. $27-25=2$; $2+4=6$; $6-8=-2$.

Trick: Simplify $\frac{26}{13}=2$ inside the bracket before squaring.

Answer: (D) $-2$

Concept: Force per unit length between two parallel conductors: $\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$, i.e. $F\propto \frac{I_1{I}_2}{d}$ for fixed length.

1. Initially $I_1=I_2=I$, distance $d$: $F_1=3\text{N}$.
2. New currents $2I$ and $2I$ → product becomes $4$ times; distance $\frac{d}{2}$ → force becomes $2$ times.
3. $F_2=3\times 4\times 2=24\text{N}$.

Trick: Doubling both currents ×4, halving the gap ×2 → total ×8. Like currents attract, opposite currents repel.

Answer: (B) 24 N

Concept: In a BJT the three regions are doped differently to suit their jobs.

1. Emitter is most heavily doped – it must inject a large number of charge carriers into the base.
2. Base is very lightly doped (and thin) so that most carriers cross it without recombining.
3. Collector is moderately doped and largest in size to collect carriers and dissipate heat.
4. Order: Base < Collector < Emitter.

Yaad rakhein: Doping: E > C > B; Size: C > E > B; Thickness: base is the thinnest.

Answer: (D) Base < collector < emitter

Concept: Physical construction of a bipolar junction transistor.

1. The base is made very thin (a few microns) and lightly doped so that carriers injected by the emitter pass through to the collector with little recombination – this gives a high current gain.
2. The emitter is moderately sized and heavily doped; the collector is the largest region, to collect carriers and dissipate the heat produced at the collector junction.
3. Hence options A, B and C are wrong.

Yaad rakhein: Thin + lightly doped base = heart of transistor action; collector is biggest (heat), emitter is most doped (carriers).

Answer: (D) Base region is much thinner as compared to emitter and collector.

Concept: LVDT (Linear Variable Differential Transformer) is a displacement transducer whose core moves without touching the windings.

1. Because there is no friction or mechanical contact between core and coils, the LVDT shows low hysteresis.
2. Low hysteresis means the output for the same displacement is practically identical each time → excellent repeatability under all conditions.
3. Other advantages: high sensitivity, infinite resolution, ruggedness, low power consumption.

Yaad rakhein: LVDT – no contact → no friction → low hysteresis → excellent repeatability; disadvantage – sensitive to stray magnetic fields.

Answer: (B) low; excellent

Concept: Work-energy theorem: net work done on a body equals the change in its kinetic energy, $W_{\text{net}}=\Delta K=K_f-K_i$.

1. If the net work is positive, kinetic energy increases (e.g. a falling ball); if negative, kinetic energy decreases (e.g. friction stopping a sliding block).
2. So work done can either increase or decrease kinetic energy → option A.
3. Options B and C ignore negative work; option D is false – kinetic energy alone is not conserved (only total mechanical energy is, and that too only for conservative forces).

Yaad rakhein: $W=\Delta K$: sign of work = sign of change in K.E.

Answer: (A) Work done on an object can either increase or decrease its kinetic energy

Concept: Magnetic field lines form closed loops – they have no beginning or end.

1. Outside the magnet the lines go from the North pole to the South pole.
2. To close the loop, inside the magnet the lines run from the South pole to the North pole.
3. Inside, the lines are nearly parallel to the magnetic axis and crowded (strong field), not random or perpendicular.

Yaad rakhein: Outside N → S, inside S → N; this is what makes magnetic lines closed curves (unlike electric field lines).

Answer: (D) from South Pole to North Pole

Concept: A force is conservative if the work it does around any closed path is zero (gravity, spring force, electrostatic force). Non-conservative forces (friction, air drag, viscous force) do not satisfy this.

1. Work done by friction is always against motion, so around a closed loop it keeps adding up (negative) instead of cancelling – it depends on the length of the path travelled.
2. Hence the closed-loop work of a non-conservative force is path-dependent and generally non-zero; "always zero" and "independent of path" describe conservative forces.
3. Option D ("negative") is true for friction but not a general definition – e.g. an applied push can do positive work.

Yaad rakhein: Conservative → closed-loop work $=0$, path-independent; Non-conservative → path-dependent.

Answer: (C) dependent on the path

Concept: In a dimensionally consistent equation both sides have the same dimensions, and the argument of a trigonometric function must be dimensionless.

1. $x=A\cos (\omega t+\varphi )$: $\cos (\dots )$ is a pure number, so $A$ must carry the dimension of $x$ → $A$ has dimensions of length (it is the amplitude).
2. The argument $\omega t+\varphi$ must be dimensionless (an angle): $\omega$ in $\text{rad/s}$ × $t$ in s → dimensionless; hence $\varphi$ is dimensionless too.
3. Options B and D give wrong dimensions to $A$ or $\varphi$; option A wrongly gives $\omega t$ a dimension.

Yaad rakhein: Arguments of sin, cos, log, exp are always dimensionless; the coefficient in front carries the physical dimension.

Answer: (C) A has dimensions of length and ωt is dimensionless

Concept: Faraday's law – EMF is induced only when the magnetic flux $\varphi =BA\cos \theta$ through the coil changes.

1. Moving the coil parallel to the field lines does not change the number of lines passing through it → flux constant → no EMF.
2. Changing flux (A), rotating the coil (C – $\theta$ changes) and moving the coil perpendicular to the lines in a non-uniform field (D – the coil cuts field lines) all produce an induced EMF.

Yaad rakhein: No flux change = no EMF. Motion along the field lines never cuts them.

Answer: (B) The coil is moved parallel to the magnetic field lines.

Concept: Faraday's law gives the magnitude of induced EMF; Lenz's law gives its direction.

1. Lenz's law: the induced current flows in such a direction that it opposes the change (in flux/motion) that produced it – this fixes the direction of induced current.
2. Faraday's law: $\epsilon =-\frac{d\varphi }{dt}$ (magnitude; the minus sign is Lenz's law). Coulomb's law – force between charges; Ampere's law – magnetic field due to current.
3. For a straight conductor the direction is found practically by Fleming's right-hand rule, which is consistent with Lenz's law.

Yaad rakhein: Faraday = how much, Lenz = which way; Lenz's law is a consequence of energy conservation.

Answer: (A) Lenz's Law

Concept: The number of address lines decides the addressable memory: $n$ lines → $2^n$ locations.

1. 8085 is an 8-bit microprocessor with a 16-bit address bus (A₀–A₁₅; the lower 8 lines AD₀–AD₇ are multiplexed with data).
2. Addressable memory $=2^{16}=65536$ bytes $=64$ kilobytes.
3. 8 lines would give only $2^8=256$ bytes – far too small.

Yaad rakhein: 8085: 8-bit data, 16-bit address, 64 KB memory, 40-pin IC, 3 MHz clock.

Answer: (D) 16; 64 kilobytes

Concept: Dynamic characteristics of instruments describe how they respond to changing inputs.

1. Dead time is the time required by an instrument to begin to respond to a change in the input (measured) quantity – the output stays unchanged during this interval.
2. It is different from dead zone (the largest change of input to which the instrument does not respond) and from lag (delay in reaching the final value).
3. Responses always refer to the input (measurand), so options with "output" are wrong.

Yaad rakhein: Dead time = delay before response starts; Dead zone = input range with no response; Speed of response = how fast it follows.

Answer: (C) begin; input

Concept: Response time is the time a display element takes to switch between states; it is a key difference between LCDs and LEDs.

1. LCDs work by physically re-orienting liquid-crystal molecules, which is slow: typical response times are 100 ms to 300 ms.
2. LEDs are semiconductor junctions that switch electronically, with response times below 100 ns – about a million times faster.
3. This is why LCDs are used for low-power static displays (watches, meters) while LEDs suit fast indicators and high-speed signalling.

Yaad rakhein: LCD – milliseconds (100–300 ms), low power; LED – nanoseconds (<100 ns), more power.

Answer: (D) 100 ms; 300 ms; 100 ns

Concept: Eddy currents are circulating currents induced in a bulk conductor by a changing magnetic field; they cause heating and damping.

1. Induction furnace: a high-frequency alternating field induces strong eddy currents in the metal charge; the I²R heat melts it – eddy currents are used beneficially.
2. In transformer cores and motors, eddy currents are an unwanted loss, so cores are laminated to reduce them.
3. Electric heaters work by direct resistive heating of a coil, not by induced currents.
4. Other useful applications: magnetic braking in trains, moving-coil galvanometer damping, speedometers, induction cooktops, metal detectors.

Yaad rakhein: Eddy currents – useful in furnace, brakes, damping, speedometer; harmful in transformer/motor cores (hence lamination).

Answer: (D) Induction furnaces

Concept: Band gap $E_g$ = energy difference between the valence band and the conduction band.

1. In metals the valence and conduction bands overlap (or the conduction band is partly filled), so $E_g=0\text{eV}$ – electrons move freely, giving high conductivity.
2. Semiconductors: $E_g\approx 0.7\text{eV}$ (germanium), $1.1\text{eV}$ (silicon); insulators: $E_g>3\text{eV}$ (e.g. diamond $\approx 6\text{eV}$).

Yaad rakhein: Metal 0 eV, Ge 0.7 eV, Si 1.1 eV, insulator >3 eV (diamond ~6 eV).

Answer: (A) 0 eV

Concept: In a CRT with electrostatic deflection, the spot on the screen moves in proportion to the voltage applied to the deflecting plates.

1. Deflection sensitivity $s=\frac{D}{E}$, i.e. deflection on the screen per volt of deflecting voltage, measured in mm/V (typically 0.1–1 mm/V).
2. Its reciprocal $\frac{E}{D}$ (V/mm) is called the deflection factor.
3. $s=\frac{L{l}_d}{2d{E}_a}$ shows that $s$ increases with longer plates and distance to screen, and decreases with higher accelerating voltage $E_a$.

Yaad rakhein: Sensitivity = output (deflection) ÷ input (voltage) = $\frac{D}{E}$; deflection factor is its inverse.

Answer: (D) $s=\frac{D}{E}$

Concept: Field inside a long solenoid: $B={\mu }_{0}nI$ (or $\mu nI$ with a core), where $n$ = turns per unit length.

1. $B$ increases with turns per unit length $n$ and current $I$.
2. A soft-iron core raises $\mu$ and hence $B$ many times.
3. The total length of the solenoid does not appear in the formula; only the turn density $n$ matters, so length by itself has no effect.

Yaad rakhein: $B={\mu }_{0}nI$ – only $\mu$, $n$ and $I$; 'length' and 'radius' are absent.

Answer: (B) Length of the solenoid

Concept: $R=\rho \frac{l}{A}$ and for metals $\rho$ rises with temperature.

1. For most metals, higher temperature → more lattice vibration → more collisions → resistance increases. Statement A is true.
2. B is false: $R\propto l$. C is false: $R\propto \rho$ (directly). D is false: $R\propto \frac{1}{A}$.

Yaad rakhein: Long and thin wire = high resistance; hot metal = higher resistance (opposite for semiconductors).

Answer: (A) Resistance increases with an increase in the temperature for most metals.