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Physics NEET MCQ
Quiz 1
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Q.1
Choose the only false statement from the following.
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In conductors, the valence and conduction bands overlap.
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Substances with an energy gap of the order of 10 eV are insulators
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The resistivity of a semiconductor increases with increases in temperature.
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The conductivity of a semiconductor increases with increases in temperature
Explanation
Solution: For a semiconductor, raising the temperature thermally excites MORE charge carriers across the energy gap into the conduction band — so conductivity INCREASES and resistivity DECREASES as temperature rises (the opposite of what happens in ordinary metals/conductors). So "resistivity of a semiconductor increases with temperature" is the false statement; the other three (conductors have overlapping bands, a ~10 eV gap means insulator, and semiconductor conductivity rising with temperature) are all correct.
Q.2
In a transistor
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there is 1 p-n junction
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there are 2 p-n junctions
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there are 3 p-n junctions
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there are 4 p-n junctions
Explanation
Solution: A transistor is formed by sandwiching three doped regions (emitter, base, collector) together, creating two p-n junctions: the emitter-base junction and the base-collector junction.
Q.3
In a semiconducting material, the mobilities of electrons and holes are $\mu _e$ and $\mu _h$ respectively. Which of the following is true?
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$\mu _e > $\mu _h$
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$\mu _e < $\mu _h$
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$\mu _e = $\mu _h$
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$\mu _e < 0$; $\mu _h > 0$
Explanation
Solution: In a semiconductor, electrons move through the conduction band relatively freely, while holes move via a slower "vacancy-hopping" mechanism (a valence electron shifting into the hole, effectively moving the hole). Because of this, electron mobility is consistently greater than hole mobility in typical semiconductors: μₑ > μₕ.
Q.4
In a p-n junction photocell, the value of the photo electromotive force produced by monochromatic light is proportional to
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the barrier voltage at the p-n junction.
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the intensity of the light falling on the cell.
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the frequency of the light falling on the cell.
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the voltage applied at the p-n junction.
Explanation
Solution: In a p-n junction photocell, incoming light generates electron-hole pairs, which the junction's built-in field separates to produce a photo-emf. More light intensity means more electron-hole pairs generated per second, so the photo-emf produced is proportional to the intensity of the light falling on the cell (not directly to its frequency or to the junction/applied voltage).
Q.5
The amplifier has voltage gain =1000, the voltage gain (in dB) is
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30 dB
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60 dB
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3 dB
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20 dB
Explanation
Solution: Voltage gain in decibels = 20 × log₁₀(voltage gain). 20 × log₁₀(1000) = 20 × 3 = 60 dB (since log₁₀(1000) = 3).
Q.6
Which of the following is not a process involved in the fabrication of IC?
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Polymerisation
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Diffusion
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Photolithography
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Metallisation
Explanation
Solution: Standard IC fabrication involves steps like photolithography (patterning the wafer), diffusion (introducing dopants), and metallisation (depositing conducting interconnects) — all genuine semiconductor processing steps. Polymerisation (forming long-chain polymer molecules) is a chemistry process unrelated to IC fabrication and is not one of the standard steps.
Q.7
In the depletion layer of an unbiased p-n junction junction diode , there are
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holes
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Mobile ions
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electrons
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immobile ions
Explanation
Solution: When a p-n junction forms, free electrons and holes near the junction diffuse across and recombine, leaving behind the FIXED, ionised dopant atoms they came from — these ions can't move (they're locked in the crystal lattice), unlike the mobile charge carriers that left. This region of immobile ions (with no free carriers) is exactly what makes up the depletion layer.
Q.8
The current in the circuit will be
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5/40 A
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5/50 A
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5/10 A
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5/20 A
Explanation
Solution: The two diodes D1 and D2 point in opposite directions relative to the loop, so for the given 5 V battery polarity only one of them is forward-biased (conducting) while the other is reverse-biased (blocking, i.e. an open circuit that can be dropped from the circuit entirely). Working out the biasing for this battery polarity, D2 (in series with the 30Ω resistor) is the one forward-biased, while D1's branch (with the 20Ω resistor) is reverse-biased and carries no current. Treating the conducting diode as an ideal short (0Ω), the current path is just the battery's own 20Ω in series with D2's 30Ω branch: total resistance = 20 + 30 = 50Ω. Current = V/R = 5/50 A.
Q.9
we have below circuit given For what values of A and B, the output Y will not be 1
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0, 0
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0, 1
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1,0
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1,1
Explanation
Solution: Each gate drawn (curved body with a bubble on the output) is a NOR gate. The first NOR gate computes Q = NOR(A,B). Q is then fed into BOTH inputs of the second NOR gate, giving Y = NOR(Q,Q) = NOT(Q) (since ORing any signal with itself just gives that signal back, and NOR then inverts it). So Y = NOT(NOR(A,B)) = NOT(NOT(A OR B)) = A OR B — this two-NOR-gate combination is simply an OR gate. For an OR gate, Y = 0 (not 1) only when both inputs are 0. So Y is not 1 only when A = 0 and B = 0.
Q.10
You are required to choose the correct one out of the given five responses and mark it as (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: In a transistor, the base is made thin. Reason: A thin base makes the transistor stable
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a
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b
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c
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d
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e
Explanation
Solution: The Assertion is true — a transistor's base region genuinely is made thin (and lightly doped), so that most of the charge carriers injected from the emitter cross it without recombining and reach the collector. But the Reason is not correct: a thin base is used to maximise current transfer efficiency (keeping the current gain close to 1), not to make the transistor "stable" — in fact an overly thin base can make a transistor MORE prone to issues like punch-through at higher voltages. Transistor operating-point stability is normally addressed through the biasing circuit, not base thickness. So: Assertion true, Reason false.
Q.11
You are required to choose the correct one out of the given five responses and mark it as (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: In a common-emitter amplifier, the load resistance of the output circuit is 1000 times the load resistance of the input circuit. If $\alpha = 0.98$, then voltage gain is $49 \times 10^3$ Reason: $\alpha = \frac {\beta}{1 + \beta}$ (symbols have their usual meaning)
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a
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b
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c
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d
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e
Explanation
Solution: First check the Reason: α = β/(1+β) is indeed the standard, correct relation between common-base current gain (α) and common-emitter current gain (β). Rearranged, β = α/(1−α) = 0.98/0.02 = 49 — this correctly follows from the Reason, so the Reason is true. Now the Assertion: voltage gain of a CE amplifier = current gain (β) × (load resistance ratio) = 49 × 1000 = 49,000 = 49 × 10³ — exactly matching the Assertion's claimed value, using β derived directly from the Reason's formula. So both statements are true, and the Reason is exactly what's needed to derive the Assertion's numeric result — it is the correct explanation.
Q.12
You are required to choose the correct one out of the given five responses and mark it a (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: NOT gate is also called inverter circuit. Reason: NOT gate inverts the input
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a
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b
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c
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d
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e
Explanation
Solution: Both statements are true, and directly connected: a NOT gate flips its input (0 becomes 1, and 1 becomes 0) — that inverting behaviour is precisely why it is called an "inverter" circuit. The Reason is the correct explanation of the Assertion.
Q.13
You are required to choose the correct one out of the given five responses and mark it a (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: germanium is preferred over silicon for making semiconductor devices. Reason: The energy gap for germanium is smaller than the energy gap of silicon
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a
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b
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c
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d
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e
Explanation
Solution: The Assertion is false: in modern semiconductor devices, silicon is generally preferred over germanium — silicon works reliably over a much wider temperature range, is far more abundant, and forms a stable oxide layer (SiO₂) that's essential for IC fabrication, among other advantages. The Reason is true on its own: germanium's energy gap (≈0.7 eV) genuinely is smaller than silicon's (≈1.1 eV). So: Reason is true, but the Assertion is false.
Q.14
Paragraph Question In the p-n-p transistor circuit, the collector current is 10 mA. If 90% of the holes reach the collector. Find the emitter current.
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9 mA
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11 mA
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12 mA
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10.9 mA
Explanation
Solution: If 90% of the emitter current's holes reach the collector, the collector current is 90% of the emitter current: Ic = 0.90 × Ie. Ie = Ic / 0.90 = 10 mA / 0.90 ≈ 11.1 mA ≈ 11 mA.
Q.15
What is the base current?
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1 mA
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2 mA
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.1 mA
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None of these
Explanation
Solution (continuing from the same circuit as the previous question, Ic = 10 mA, Ie ≈ 11 mA): Base current is what's left after the collector current is subtracted from the emitter current: Ib = Ie − Ic ≈ 11 − 10 = 1 mA.
Q.16
A semiconductor having electron and hole mobilities $\mu _n$ and $\mu _p$ respectively, if its intrinsic carrier density is $n_i$ , then what will be the value of hole concentration P for which the conductivity will be minimum at a given temperature?
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$n_i \sqrt {\frac {\mu _n}{\mu _p}}$
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$n_i \sqrt {\frac {\mu _p}{\mu _n}}$
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$\frac {1}{n_i} \sqrt {\frac {\mu _p}{\mu _n}}$
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$ \sqrt {\frac {n_i \mu _p}{\mu _n}}$
Explanation
Solution: Conductivity σ = e(nμₙ + pμₚ), with the constraint that the electron-hole product stays fixed at n·p = nᵢ² (the law of mass action) regardless of doping. Substituting n = nᵢ²/p and minimising σ with respect to p (setting dσ/dp = 0) gives: −nᵢ²μₙ/p² + μₚ = 0 → p² = nᵢ²μₙ/μₚ → p = nᵢ√(μₙ/μₚ).
Q.17
The coordination number for a bcc crystal is
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4
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8
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12
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6
Explanation
Solution: In a body-centred cubic (BCC) lattice, each atom is surrounded by 8 nearest neighbours — one at each corner of the cube touches the atom at the body centre. So the coordination number is 8.
Q.18
which of the following is an amorphous solid?
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glass
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diamond
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Salt
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Sugar
Explanation
Solution: Glass is the classic example of an amorphous solid — its constituent particles lack the long-range, repeating order of a crystal lattice. Diamond, salt (NaCl) and sugar (sucrose) are all crystalline solids, with well-defined, ordered lattice structures.
Q.19
Match the column
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p -> iv, q ->i, r -> ii , s->iii
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p -> iii, q ->i, r -> iv , s->ii
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p -> iv, q ->ii, r -> i , s->iii
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p -> iv, q ->i, r -> iii , s->ii
Explanation
Solution: Each column II expression is the direct Boolean definition of one logic gate: (i) Y = complement of (A + B) — this is exactly the definition of NOR, so it matches (q) NOR. (ii) Y = A + B — the plain OR expression, matching (s) OR. (iii) Y = A · B — the plain AND expression, matching (r) AND. (iv) Y = complement of (A · B) — this is exactly the definition of NAND, so it matches (p) NAND. So: p → iv, q → i, r → iii, s → ii.
0 h : 0 m : 1 s
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