Q&A: practice questions with answers

Concept: $\text{Area}=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$.

1. $A(-5,7)$, $B(-4,-5)$, $C(4,5)$.
2. $=\frac{1}{2}|(-5)(-5-5)+(-4)(5-7)+4(7+5)|$
3. $=\frac{1}{2}|50+8+48|=\frac{106}{2}=53$ sq. units.

Trick: Write the coordinates in a column, repeat the first row, and cross-multiply (shoelace method).

Answer: (A) 53 sq. units

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $19.5=3\times 85-2\text{Mean}=255-2\text{Mean}$.
2. $2\text{Mean}=255-19.5=235.5$.
3. $\text{Mean}=117.75\approx 117.8$.

Trick: $\text{Mean}=\frac{3\text{Median}-\text{Mode}}{2}$.

Answer: (C) 117.8

Concept: If adjacent faces have areas $lb$, $bh$, $hl$, then $(lbh{)}^2=lb\times bh\times hl$, so $V=\sqrt{A_1{A}_2{A}_3}$.

1. $V=\sqrt{260\times 104\times 10}=\sqrt{270400}$.
2. $270400=2704\times 100={52}^2\times {10}^2$.
3. $V=52\times 10=520{\text{cm}}^3$.

Trick: Volume = square root of the product of the three face areas.

Answer: (B) 520

Concept: The building (vertical), its shadow (horizontal) and the line from the top of the building to the shadow's end form a right triangle.

1. Required distance = hypotenuse $=\sqrt{8^2+6^2}$.
2. $=\sqrt{64+36}=\sqrt{100}=10\text{m}$.

Trick: 6-8-10 is the 3-4-5 triplet doubled.

Answer: (D) 10 m

Concept: A quadratic with roots $\alpha ,\beta$ is $x^2-(\alpha +\beta )x+\alpha \beta =0$. If the roots are doubled, the sum doubles and the product becomes 4 times.

1. Given $\alpha +\beta =\frac{11}{2}$, $\alpha \beta =\frac{15}{2}$.
2. New roots $2\alpha ,2\beta$: sum $=2\times \frac{11}{2}=11$, product $=4\times \frac{15}{2}=30$.
3. Required equation: $x^2-11x+30=0$.

Trick: Roots × $k$ ⇒ replace $x$ by $\frac{x}{k}$: $\frac{x^2}{4}-\frac{11}{2}·\frac{x}{2}+\frac{15}{2}=0\Rightarrow x^2-11x+30=0$.

Answer: (B) $x^2-11x+30=0$

Concept: Band gap energy ($E_g$) is the energy needed to move an electron from the valence band to the conduction band.

1. Standard values at room temperature: $E_g(\mathrm{G}\mathrm{e})\approx 0.67\text{eV}$, $E_g(\mathrm{S}\mathrm{i})\approx 1.1\text{eV}$, $E_g(\mathrm{G}\mathrm{a}\mathrm{A}\mathrm{s})\approx 1.43\text{eV}$.
2. Hence $E(\mathrm{G}\mathrm{a}\mathrm{A}\mathrm{s})>E(\mathrm{S}\mathrm{i})>E(\mathrm{G}\mathrm{e})$.
3. This is why Ge devices are more temperature-sensitive (smallest gap) and GaAs is used in LEDs/high-frequency devices.

Yaad rakhein: "Ge < Si < GaAs" – 0.7, 1.1, 1.4 eV (roughly increasing by 0.35).

Answer: (A) E(GaAs) > E(Si) > E(Ge)

Concept: NAND output $=\overline{AB}$; NOR output $=\overline{X+Y}$. Use De Morgan's laws to simplify.

1. Both NAND gates have inputs A and B, so each gives $P=\overline{AB}$.
2. The NOR gate receives $P$ and $P$: $Y=\overline{P+P}=\bar{P}$.
3. $Y=\overline{\stackrel{―}{AB}}=AB$ (double negation).

Trick: A NOR gate with both inputs tied together is just a NOT gate. NOT of NAND = AND.

Answer: (B) $AB$

Concept: For an infinitely long straight wire, $B=\frac{{\mu }_{0}I}{2\pi r}$, where $\frac{{\mu }_0}{2\pi }=2\times {10}^{-7}\text{T m/A}$.

1. $r=\frac{{\mu }_{0}I}{2\pi B}=\frac{2\times {10}^{-7}\times 15}{5\times {10}^{-6}}$.
2. $=\frac{3\times {10}^{-6}}{5\times {10}^{-6}}=0.6\text{m}$.

Trick: Remember $B=\frac{2\times {10}^{-7}I}{r}$ – it avoids writing ${\mu }_0$ and $\pi$ every time.

Answer: (A) 0.6 m

Concept: The magnetic field lines around a straight current-carrying wire are concentric circles centred on the wire (right-hand thumb rule).

1. At any point the field is tangent to the circle through that point, i.e. tangential and perpendicular to the radius.
2. Reversing the current reverses the direction of circulation (thumb = current, curled fingers = field).
3. The field is neither radially outward nor inward – that would be the pattern of an electric field around a line charge.

Yaad rakhein: Electric field of a line charge = radial; magnetic field of a line current = circular/tangential.

Answer: (A) tangential to the wire and depends on the direction of the current

Concept: Hysteresis is the lag between cause and effect – the output depends on whether the input is increasing or decreasing.

1. In a magnetic material, flux density $B$ for the same current (same $H$) is different on the rising and falling curves – this forms the B–H loop.
2. The difference in readings for the same input value is called hysteresis error.
3. Frequency, friction and temperature errors do not depend on the direction of change of input.

Yaad rakhein: "Increasing vs decreasing gives different reading" = Hysteresis (the loop).

Answer: (A) Hysteresis

Concept: Repeatability is the closeness of repeated readings of the same input by the same instrument, same observer, same conditions, in a short time.

1. So the instrument should give the same output for repeated applications of the same input value.
2. Any option with "different" breaks the definition – an instrument giving different outputs for the same input is not repeatable.

Yaad rakhein: Repeatability = same–same–same (output, input, conditions); Reproducibility = same output, same input, but different conditions/observer/location.

Answer: (A) same; same

Concept: The 8051 has five interrupt sources plus RESET; the 8052 adds a Timer 2 interrupt.

1. 8051: INT0, Timer 0, INT1, Timer 1, Serial (TI/RI) = 5 sources; with RESET counted, 6 interrupts (vectors 0000H, 0003H, 000BH, 0013H, 001BH, 0023H).
2. 8052: the same plus Timer 2 (vector 002BH) = 6 sources, i.e. 7 with RESET.

Yaad rakhein: Vector addresses jump by 8 bytes: 03, 0B, 13, 1B, 23 (8051), 2B (8052 Timer 2).

Answer: (A) 6; 7

Concept: Assume $T\propto m^a{L}^b{g}^c$ and equate dimensions on both sides.

1. $[T]=[M{]}^a[L{]}^b[L{T}^{-2}{]}^c=M^a{L}^{b+c}{T}^{-2c}$.
2. Compare with $M^0{L}^0{T}^1$: $a=0$, $b+c=0$, $-2c=1$ ⇒ $c=-\frac{1}{2}$, $b=\frac{1}{2}$.
3. So $T\propto L^{1/2}{g}^{-1/2}=\sqrt{\frac{L}{g}}$; the period does not depend on mass.

Trick: Only $\sqrt{L/g}$ has the dimension of time ($\sqrt{\text{m}/({\text{m/s}}^2)}=\text{s}$); $L/g$ gives ${\text{s}}^2$ and $L/\sqrt{g}$ does not give seconds – eliminate by units.

Answer: (C) $\sqrt{\frac{L}{g}}$

Concept: Reproducibility is the degree of closeness with which a given value may be repeatedly measured – the same output for the same input under the same conditions, over a longer period.

1. The definition requires same output for same input under same environmental conditions.
2. Any "different" in the blanks would describe drift or poor reproducibility, not the property itself.

Yaad rakhein: Repeatability = short time; Reproducibility = long time – both: same–same–same.

Answer: (B) same; same; same

Concept: Write the output of each gate and simplify with Boolean laws: $\overline{X+Y}=\overline{X}\overline{Y}$, $\overline{\stackrel{―}{X}}=X$.

1. First stage: NAND gives $P=\overline{AB}$; NOR gives $Q=\overline{A+B}$.
2. Upper NOR: $\overline{P+Q}=\overline{P}\overline{Q}=AB(A+B)=AB$.
3. Lower NOR (inputs $Q$ and $A$): $\overline{Q+A}=\overline{Q}\overline{A}=(A+B)\overline{A}=\overline{A}B$.
4. Final OR: $Y=AB+\overline{A}B=B(A+\overline{A})=B$.

Trick: Quick check with A = 0, B = 1 and A = 1, B = 0: output follows B both times, so $Y=B$.

Answer: (A) B

Concept: $R=\rho \frac{l}{A}$ with $A=\pi r^2$, so $R\propto \frac{1}{r^2}$ for fixed length and material.

1. New radius $=\frac{r}{2}$ ⇒ new area $=\frac{\pi r^2}{4}=\frac{A}{4}$.
2. $R^{\prime }=\rho \frac{l}{A/4}=4R$.

Trick: Radius halved → area quartered → resistance 4 times. (If the wire were stretched to half radius, length would also change and $R$ would become 16R.)

Answer: (B) 4R

Concept: Put the input value at every point and work gate by gate: NOT inverts, AND gives 1 only if all inputs are 1, OR gives 1 if any input is 1.

1. Input Y = 0. NOT gate output $=1$.
2. Upper AND gate: inputs 1 (from NOT) and Y = 0 ⇒ output $=1·0=0$.
3. OR gate: inputs 0 (from AND) and Y = 0 ⇒ $A=0+0=0$.
4. Lower AND gate: both inputs are Y = 0 ⇒ $B=0·0=0$.

Trick: Any AND gate with a 0 input gives 0; an OR gate gives 1 only if some input is 1 – here every line feeding the OR is 0.

Answer: (A) 0, 0

Concept: For a variable force, work $W=\int Fdx$ = area under the F–x graph.

1. $W={\int }_0^{3}6xdx={\left[3x^2\right]}_0^3=3(9)-0=27\text{J}$.
2. Check by area: F–x graph is a straight line from 0 to 18 N; triangle area $=\frac{1}{2}\times 3\times 18=27\text{J}$.

Trick: For $F=kx$, $W=\frac{1}{2}k{x}^2$ (like a spring): $\frac{1}{2}\times 6\times 9=27$.

Answer: (D) 27 J

Concept: A load cell is a force/weight transducer made of an elastic member (column, ring or beam) fitted with strain gauges.

1. The applied load deforms the elastic member; the strain gauges change resistance; a Wheatstone bridge converts this to a voltage proportional to the weight.
2. LVDT measures displacement by magnetic coupling, piezoelectric transducers give charge under dynamic force, capacitive transducers sense change in capacitance – none is the strain-gauge weighing arrangement.

Yaad rakhein: Strain gauge + elastic member for weighing = Load cell (used in weighbridges and digital scales).

Answer: (D) Load cell

Concept: Parallax error occurs when the eye is not placed perpendicular to the scale, so the pointer appears to read a different value.

1. It is an observational error – a sub-type of systematic error (errors due to instrument, environment or observer), because it shifts the reading consistently in one direction for a given viewing angle.
2. Random errors are unpredictable fluctuations; gross errors are human blunders like misreading or wrong recording; instrumental errors arise from the instrument's construction.

Yaad rakhein: Systematic = instrumental + environmental + observational (parallax); it can be removed by a mirror-backed scale.

Answer: (C) Systematic error