NEET 2026 re-exam Physics answer key with solutions: questions 1–45

Friction force at position x is f(x) = µk(x)Mg = (µ0 − αx)Mg, and since µk(L)=0, α = µ0/L. Work by friction (magnitude) = ∫0^L (µ0 − αx)Mg dx = µ0MgL − αMgL²/2 = µ0MgL − (µ0/L)Mg(L²/2) = µ0MgL/2. Matching to nµ0MgL gives n = 1/2.

• (1) n=3 is not reachable by a natural slip: since µk(x) never exceeds µ0 and falls to 0 by x=L, the friction work must be less than µ0MgL (n<1); a value as large as 3 does not come from any correct-or-slipped integration of this decreasing force
• (2) treating the friction coefficient as constant at µ0 across the whole path (ignoring that µk decreases to 0 by x=L) gives W=µ0MgL, i.e. n=1
• (3) assumes µk decreases quadratically with x (average value µ0/3, as for the integral of (1-x/L)²) instead of linearly (average µ0/2), giving n=1/3

Mean free path λ = 1/(√2 π d²n). Given λA = λB/2 and dA = 2dB: λA/λB = (dB²nB)/(dA²nA) = 1/2, so dB²nB/(4dB²nA) = 1/2, giving nA = nB/2.

• (1) forgets to square the diameter ratio in λ∝1/d² (uses dA/dB=2 instead of (dA/dB)²=4), which gives nA=nB
• (2) takes the reciprocal of the correctly derived ratio nA=(1/2)nB, giving nA=2nB
• (3) correctly squares the diameter ratio but wrongly treats the two mean free paths as equal (ignoring that λA=λB/2), giving nA=(1/4)nB

In P, a–b spans a single inductor of value L, so EP = L(dI/dt). In Q, two identical inductors L in parallel between a and b combine to Leq = L/2, so EQ = (L/2)(dI/dt). Hence EP/EQ = L/(L/2) = 2.

• (1) 1/4 does not follow from a single natural mix-up here: EP can only come out as L(dI/dt) or L/2(dI/dt) depending on how a-b is (mis)read, and EQ likewise as L(dI/dt) or L/2(dI/dt), so the smallest ratio a labelling slip can produce is 1/2, not 1/4
• (2) wrongly combines Q's two identical inductors as if in series (Leq=2L) instead of parallel (Leq=L/2), giving EP/EQ=L/(2L)=1/2
• (3) Treats P's single inductor (L) and Q's parallel pair (L/2) as if numerically equal, so EP=EQ, giving EP/EQ=1.

In an open pipe, the number of nodes equals the harmonic number, so for the 5th harmonic n = 5. In a pipe closed at one end, only odd harmonics (1st, 3rd, 5th, 7th, 9th,...) exist and the number of nodes for the kth odd harmonic is (k+1)/2; for the 9th harmonic m = 5. So n/m = 5/5 = 1.

• (1) comes from taking m=9 directly as the node count instead of counting nodes for a closed pipe
• (2) Makes the same node-counting error as option 1 (treating the closed pipe's 9th-harmonic number itself as its node count, m=9) but inverts the fraction to m/n, giving 9/5 instead of 5/9.
• (4) comes from miscounting the open pipe's node count as 3 instead of 5 (missing two internal nodes) while correctly using m=5 for the closed pipe

Flux linkage L = Nφ/i with N = nl turns and φ = BA = µ0niA where A = πr². So L = (nl)(µ0ni·πr²)/i = µ0n²πr²l.

• (2) Drops the π factor from the circular cross-section area A=πr² entirely, giving L=µ0n²r²l instead of µ0πn²r²l.
• (3) Confuses the circular-area factor π with the µ0/(2π) factor from the straight-wire field formula, giving L=(µ0/2π)n²r²l.
• (4) Mistakes the 1/2 from the energy formula U=1/2 LI² as part of L itself, doubling the correct µ0πn²r²l to 2µ0πn²r²l.

s(t) = t² − 6t + 5, v(t) = 2t − 6, which is zero at t = 3 s (particle reverses direction there). Distance covered = |s(3)−s(0)| + |s(6)−s(3)| = |−4−5| + |5−(−4)| = 9+9 = 18 m. Average speed = total distance/time = 18/6 = 3 ms⁻¹.

• (1) 18 m of path length over 6 s gives 3 m/s; 12 m/s would need either a 72 m path or a 1.5 s interval, and no natural mis-step in this motion produces either, so 12 is simply not reachable here
• (2) Confuses average speed with the instantaneous speed at t=6s: v(6)=2(6)−6=6 m/s, matching this option's value.
• (4) wrongly uses net displacement s(6)−s(0)=0 (average velocity, not average speed)

Mass defect Δm = 238.050 − (234.043 + 4.003) = 0.004 u. Q = Δm·c² = 0.004 × 931.5 MeV = 3.726 MeV = 3726 keV.

• (2) Uses a conversion constant of 932.5 MeV/u instead of the given 931.5 MeV/u: 0.004×932.5=3.730 MeV=3730 keV.
• (3) Uses a conversion constant of 934 MeV/u instead of 931.5 MeV/u: 0.004×934=3.736 MeV=3736 keV.
• (4) Uses a conversion constant of 935 MeV/u instead of 931.5 MeV/u: 0.004×935=3.740 MeV=3740 keV.

Power P ∝ (number of photons per second)×(energy per photon). The source and hence photon energy stay fixed, so P ∝ n; as P falls linearly, the photocurrent I (∝ n) falls linearly too. The stopping potential depends only on the photon energy (frequency), which is unchanged, so |V| stays constant.

• (2) Wrongly makes the stopping potential depend on beam intensity (power) rather than photon frequency alone, so |V| is shown falling as intensity falls.
• (3) Wrongly treats photocurrent as intensity-independent, ignoring that current is proportional to photons arriving per second, which falls as power falls.
• (4) ignores that current must fall as fewer photons arrive per second

P and Q (each C) in series give C/2, so each carries voltage V/2 and UP = (1/2)C(V/2)² = CV²/8. S sees the full V, so US = (1/2)CV². Total UT = UP + UQ + US = CV²/8 + CV²/8 + CV²/2 = (3/4)CV². So UP/UT = (CV²/8)/((3/4)CV²) = 1/6.

• (1) Treats P as carrying the full battery voltage V (ignoring the series split with Q): UP=1/2 CV², so UP/UT=(1/2 CV²)/(3/4 CV²)=2/3.
• (2) wrongly treats all three capacitors as identical parallel branches directly across V (ignoring that P and Q are in series), so each stores equal energy and UP/UT=1/3
• (3) Forgets that Q (in series with P) also stores energy and treats the system as just P and S with equal energy each, giving UP/UT=UP/(UP+US)=1/2.

Continuity: 40VP = 20VQ ⇒ VQ = 2VP. Bernoulli (horizontal pipe): PP − PQ = (1/2)ρ(VQ² − VP²) ⇒ 15 = (1/2)(1000)(4VP² − VP²) = 1500VP² ⇒ VP = 0.1 m/s. Rate = APVP = 40×10⁻⁴ m² × 0.1 m/s = 4×10⁻⁴ m³/s = 400 cm³/s.

• (1) 400 cm3/s (=40 cm2 x 10 cm/s) is the consistent rate at both P and Q by continuity; 100 cm3/s is a quarter of this and does not follow from a single natural mis-step in the Bernoulli/continuity calculation
• (2) pairs the area at Q (20 cm2) with the velocity found at P (10 cm/s) instead of matching area and velocity at the same point, giving 20x10=200 cm3/s
• (3) 300 cm³/s is not reachable by one clean Bernoulli/continuity slip; it sits exactly midway between the correct 400 cm³/s and option 2's mismatched 200 cm³/s, suggesting an averaging error rather than a distinct calculation.

Current splits inversely with resistance: since R(ABC) = R(ADC)/2, the arc with smaller resistance (ABC) carries the larger current. With I(ABC) + I(ADC) = I0 and I(ABC).R = I(ADC).2R, we get I(ABC) = 2.I(ADC), so I(ADC) = I0/3 and I(ABC) = 2I0/3. Each semicircular arc contributes µ0.I(arc)/4r at O but in opposite senses, so B0 = µ0(2I0/3)/4r − µ0(I0/3)/4r = µ0I0/12r.

• (2) uses the single-arc formula µ0I/4r with the full current I0 (forgetting to split the current between the two arcs), giving µ0I0/4r
• (3) uses the full-loop centre-field formula µ0I0/(2r) directly with the total current, ignoring that the field here comes from two semicircular arcs carrying split currents
• (4) uses the infinite straight-wire field formula µ0I0/(2πr) instead of the circular-arc formula appropriate to this loop

Terminal velocity vT = 2r²(σ−ρ)g/(9η) = [2r²ρg/(9η)](σ/ρ − 1), which is a straight line in σ/ρ of the form y = mx − m (m>0), crossing zero at σ/ρ = 1 — matching option 1.

• (2) vT=(2r²ρg/9η)(σ/ρ−1) is linear in σ/ρ; a concave-up curve implies vT grows faster than linearly, which this formula rules out.
• (3) The line must cross zero exactly at σ/ρ=1 (vT=0 when σ=ρ, per the (σ/ρ−1) factor); a positive intercept there contradicts the formula.
• (4) shows a discontinuous jump, which contradicts the continuous linear formula

Escape speed ve = √(2GM/R) ∝ 1/√R for fixed mass M. So v2/v1 = √(R1/R2) = √(R1/(R1/2)) = √2.

• (1) Computes v1/v2=√(R2/R1)=1/√2 correctly but reports it as v2/v1, giving 1/√2 instead of √2.
• (2) Wrongly assumes escape speed is independent of planet radius, so v2/v1=1.
• (4) Uses R1/R2=2 directly without taking the square root required by ve∝1/√R, giving v2/v1=2.

By Kepler's third law (law of periods), T² ∝ R³, so T ∝ R^(3/2).

• (1) Would only hold if Kepler's third law read T²∝R instead of the correct T²∝R³; taking a square root of T²∝R gives T∝R^(1/2).
• (3) misremembers Kepler's third law as T∝R² (mixing up which quantity is squared)
• (4) takes T²∝R³ at face value as T∝R³, forgetting to take the square root

Force per length between the wires (attractive, since currents are parallel) is F = µ0(I)(2I)/(2πh) = µ0I²/(πh). Wire A stays on the floor as long as this upward pull does not exceed its weight per length λg: µ0I²/(πh) ≤ λg, so h ≥ µ0I²/(πλg).

• (1) Omits the factor of 2 from wire B's current 2I (treating the force as if both wires carried I), halving the answer to µ0I²/(2πλg).
• (3) Introduces a spurious extra factor of 2 in the force formula (e.g. µ0(I)(2I)/(πh) instead of /(2πh)), doubling the answer to 2µ0I²/(πλg).
• (4) Compounds two factor-of-2 errors in the force formula, quadrupling the correct answer to 4µ0I²/(πλg).

Since |input| = 5 V exceeds the 3 V breakdown, the Zener clamps at 3 V (Vzener=3V across C–B), leaving the remaining 5−3 = 2 V dropped across the resistor R (from B to A). So |VBA| = 2 V.

• (1) equals the Zener voltage itself, not the remaining drop across R
• (3) Halves the correctly found 2 V drop across R (5−3=2 V), giving 1 V instead.
• (4) Wrongly assumes that once the Zener clamps, no current flows through R at all, so it shows 0 V there instead of the actual 2 V drop.

For an adiabatic process, W = nRΔT/(1−γ). With n=1, ΔT=−10K, γ=5/3: W = (1)(8.3)(−10)/(1−5/3) = −83/(−2/3) = 124.5 J.

• (1) treats the monatomic gas as having only 1 degree of freedom (Cv=R/2) instead of 3 (Cv=3R/2), giving W=41.5 J
• (2) forgets the 3/2 factor and uses Cv=R instead of Cv=3R/2, giving W=|nRΔT|=83 J
• (4) uses Cv=2R instead of the correct Cv=3R/2, giving W=166 J

Work function φ = hc/λthreshold. Since λ1 < λ, φ1 > φ = hc/λ (incident photon energy), so no photoelectrons are ejected from cell 1: V1 = 0. Cells 2 and 3 have λ2, λ3 > λ so φ2, φ3 < φ and both eject electrons; since λ3 >> λ2, φ3 << φ2, so by φ = φ0 + eV, cell 3 (smaller work function) gives a larger stopping potential: V3 > V2.

• (2) Wrongly reverses the stopping-potential inequality: cell 3 has the largest excess photon energy (φ3<<φ2) and so the larger V, not cell 2.
• (3) wrongly makes V3 zero, though cell 3 has the largest excess photon energy
• (4) wrongly makes V1 nonzero when photon energy is below the threshold for cell 1

For the electron, pe = √(2meEe). For the photon, pPh = EPh/c. So pe/pPh = √(2meEe)·c/EPh = c√(2me/EPh) (since Ee=EPh=20 eV = 20×1.6×10⁻¹⁹ J). Plugging in: pe/pPh = √(2×9×10⁻³¹/(20×1.6×10⁻¹⁹)) × 3×10⁸ = (3/4×10⁻⁶)×3×10⁸ = 225.

• (1) Inverts the correct ratio pe/pPh=225 to pPh/pe=1/225=2/450.
• (2) is the reciprocal-scaled value, confusing photon and electron momentum roles
• (4) 275 isn't reachable by a single clean slip here: the exact value is 225, and 275 needs an unexplained extra factor of about 1.22 that matches no unit-conversion or √2-type error.

In the optical-bench focal-length experiment, the indices (pointer positions) on the bench scale usually do not coincide exactly with the actual position of the optical element/pin due to the finite size of the uprights, so an 'index correction' (correcting for this offset) must be applied to raw scale readings; this correction is specific to the optical-bench method among the listed experiments.

• (1) meter bridge uses a balance-point/null method, not index correction
• (2) the simple-pendulum experiment concerns timing and length measurement, not index correction
• (4) the resonance-tube experiment needs an end correction for the tube's mouth, not an index correction

Outside the sphere, the field/potential is as if the whole charge Q = ρ(4/3)πR³ were at the centre. Work done on unit charge W = q(VB − VA) = kQ(1/3R − 1/2R) = kQ(−1/6R). |W| = kQ/(6R) = [1/(4πε0)]·[ρ(4/3)πR³]/(6R) = ρR²/(18ε0), so n = 18.

• (1) Uses Q=4πρR³ (dropping the 1/3 from the sphere's volume) and keeps only the kQ/2R potential term (dropping kQ/3R), giving W=ρR²/(2ε0), so n=2.
• (2) Uses Q=4πρR³ instead of the correct Q=(4/3)πρR³ (dropping the 1/3 sphere-volume factor), giving n=6.
• (3) uses only the kQ/(3R) term (from VB) and ignores the kQ/(2R) term (from VA) in the potential difference, giving n=9

Apparent depth viewed from a rarer medium nP looking into denser Q: h1 = (nP/nQ)×2.5 = (1/1.25)×2.5 = 2 cm. Viewed from R into Q: h2 = (nR/nQ)×2.5 = (1.5/1.25)×2.5 = 3 cm. |h1 − h2| = 1 cm.

• (1) Wrongly assumes h1=h2 so the difference is 0; but h1=(1/1.25)×2.5=2 cm and h2=(1.5/1.25)×2.5=3 cm differ because the two refractive-index ratios aren't equal.
• (3) uses the full 5 cm thickness for both h1 and h2 instead of the 2.5 cm half-distance, doubling the difference to 2 cm
• (4) a 3 cm gap does not follow from a single natural slip here: using the full 5 cm thickness for both sides gives 2 cm (option 3), and swapping the two refractive-index ratios still leaves a 1 cm gap, so 3 cm is not reachable this way

By symmetry the x-components of the two particles' momenta cancel at every instant, leaving Pnet = 2mv·cos θ ĵ pointing along y. At θ=±π/2, cos θ=0 so Pnet=0, and it is maximum at θ=0. So the magnitude of P starts at 0, rises to a peak, and returns to 0 — a smooth arch, matching option 3.

• (1) wrongly keeps momentum constant despite the changing angle between the two velocity vectors
• (2) wrongly shows momentum vanishing in the middle instead of at the endpoints
• (4) Wrongly shows P falling linearly and monotonically from θ=π/2 to −π/2, but Pnet=2mv·cosθ is symmetric and even in θ, peaking at θ=0 rather than declining straight through it.

For small changes, ΔR/R = αΔT. Since V = (4/3)πR³, ΔV/V = 3ΔR/R = 3αΔT, so ΔV = 3αΔT×(4/3)πR³ = 4πR³αΔT.

• (1) Uses the hemisphere volume formula (2/3)πR³ instead of the sphere's (4/3)πR³, then correctly applies ΔV=3αΔT×V, giving 2πR³αΔT.
• (2) drops the factor of 4/3 from the volume formula before multiplying by 3
• (4) uses an incorrect sphere volume formula V=2πR³ instead of (4/3)πR³ before applying the correct 3αΔT scaling, giving 6πR³αΔT

For a floating cylinder bobbing vertically, the restoring force per unit displacement gives T = 2π√(ρsl/(ρliquid·g)), i.e. T ∝ 1/√ρliquid for a fixed cork. So T1/T2 = √(ρ2/ρ1); with T2 = 2T1, (1/2) = √(ρ2/ρ1), giving ρ2/ρ1 = 1/4.

• (1) Inverts the correct ratio ρ2/ρ1=1/4 to ρ1/ρ2, reporting 4 instead.
• (2) Wrongly takes T∝ρliquid (direct proportionality) instead of T∝1/√ρliquid; then T2=2T1 directly gives ρ2/ρ1=2.
• (3) is the square root of the correct ratio instead of the ratio itself

Least count = 1 mm/10 = 0.01 cm. The Vernier scale shifting left of zero with the 4th division coinciding means a negative zero error of −(10−4)×0.01 = −0.06 cm, so the zero correction is +0.06 cm. The corrected length = measured reading + zero correction = 1 + 0.06 = 1.06 cm, which is not among the four listed options, so NTA withdrew this question (no option is correct).

• (1) confuses the 0.4 mm zero-error magnitude (4 divisions x 0.1 mm) with 0.4 cm and subtracts it directly, giving 1 - 0.4 = 0.60 cm
• (2) computes the zero-error magnitude as 4x0.01=0.04 cm directly (instead of (10-4)x0.01=0.06 cm) and subtracts it, giving 1 - 0.04 = 0.96 cm
• (3) Ignores the zero-error correction entirely and reports the raw vernier reading of 1.00 cm unchanged.
• (4) adds the wrong zero-error magnitude 4x0.01=0.04 cm (instead of the correct (10-4)x0.01=0.06 cm) to the reading, giving 1.04 cm

IA = (2/5)MR² + (2/5)mr² + m(R+r)² [B parallel-axis shifted by R+r]. IB = (2/5)MR² + M(R+r)² + (2/5)mr² [A parallel-axis shifted by R+r]. Subtracting, the (2/5)MR² and (2/5)mr² terms cancel, leaving IA − IB = m(R+r)² − M(R+r)² = (m−M)(R+r)².

• (1) Keeps the parallel-axis terms right but flips the sign of the mass difference, giving IA−IB=(M−m)(R+r)² instead of the correct (m−M)(R+r)².
• (3) Uses the centre separation (R−r) instead of the correct (R+r), the actual distance between the two sphere centres.
• (4) Wrongly assumes IA and IB are equal, ignoring that shifting the axis by (R+r) changes each sphere's parallel-axis term differently.

For SHM, v = ω√(A²−x²), so v² + ω²x² = ω²A², i.e. (v²/ω²) + x² = A², an ellipse in (x,v). This becomes a circle only when the x and v axes have equal scale factor, i.e. when 1/ω² = 1, so ω²=1. Since ω²=k/m, this gives k = m.

• (1) Solves k/m=1 as k=1/m instead of correctly cross-multiplying to k=m, a division-vs-multiplication slip.
• (3) Squares the correctly found relation k=m into k=m², rather than leaving it linear.
• (4) Takes an unjustified square root of the correct relation k=m, giving k=√m.

For L1 (u=−30cm, f=+10cm): 1/v − 1/(−30) = 1/10 ⇒ 1/v = 1/10 − 1/30 = 2/30 ⇒ v=+15 cm, i.e. 15 cm to the right of L1, which is 15−3=12 cm to the right of L2 — so u' for L2 is +12 cm. For L2 (u'=+12cm, f'=−10cm): 1/v' = 1/f' + 1/u' = −1/10 + 1/12 = −1/60 ⇒ v'=−60 cm. So the final image is 60 cm to the left of the concave lens L2.

• (1) comes from an arithmetic slip in combining the two lens equations
• (3) forgetting the 3 cm gap between the lenses (using u'=15 cm for L2 instead of the correct 12 cm) gives a 30 cm magnitude, but that slip still places the image on the left, not the right, so this option's side label does not match any single natural slip
• (4) Gets the correct 60 cm magnitude from the two-lens calculation but places the final image on the wrong side (right instead of left) of lens L2.

ω = 2×10³ rad/s. XL = ωL = 2×10³×10×10⁻³ = 20 Ω. XC = 1/(ωC) = 1/(2×10³×25×10⁻⁶) = 20 Ω. Since XL = XC, the circuit is at resonance, Z = R = 100 Ω. Current amplitude i0 = V0/Z = 220/100 = 2.2 A.

• (2) comes from using Z=40Ω (adding XL and XC) instead of resonance
• (3) Uses Z=20 Ω (one reactance alone, since XL=XC=20 Ω) instead of the resonant Z=R=100 Ω, giving i0=220/20=11.0 A.
• (4) Misplaces a decimal point in i0=220/100=2.2 A, reporting 22.0 A — tenfold too large.

For a body spinning about a fixed axis through its centre of mass, angular momentum about any point on that same rotation axis is L = r×P + Icmω; since the centre of mass is not translating, P=0 for all such points, so L = Icmω regardless of which point A or B (both lying on the same vertical axis) is used. Hence LA = LB, so LA/LB = 1.

• (1) Wrongly assumes L scales as 1/(distance from O)², computing LA/LB=(OA/OB)²=(1/2)²=1/4.
• (2) Wrongly assumes L scales as 1/(distance from O), computing LA/LB=OA/OB=1/2.
• (4) Wrongly assumes L scales directly with distance from O but takes the ratio as OB/OA=2 instead of OA/OB, giving LA/LB=2.

φ = BA0(1+sin t), so ε = −dφ/dt = −BA0 cos t, and induced current I = −BA0 cos t/R. Power P = I²R = (B²A0²/R)cos²t. Since cos²t has period π and touches zero whenever t = π/2, 3π/2,..., P(t) is a series of arches meeting the axis with cusps — matching option 2's dip-and-rise shape rather than a smooth periodic hump pattern touching zero at multiples of π (that would be sin²t behaviour).

• (1) matches a sin²t-type pattern (zeros at 0, π, 2π) rather than the cos²t pattern here
• (3) Ignores that induced current I=−BA0cos t/R is time-dependent, so P=I²R cannot stay constant.
• (4) ignores that P is bounded and periodic, not monotonically increasing

Point A lies inside the cavity where the point charge Q itself produces a nonzero field, so EA ≠ 0 (the field inside a conductor's bulk is zero, but the cavity is not conductor material). The induced charge distributes uniformly on the outer surface (electrostatic shielding makes the outside field depend only on total enclosed charge and distance from the centre), so since B and C are equidistant from the centre, EB = EC.

• (1) wrongly claims EA=0, but the cavity interior is not conductor bulk so the field there need not vanish
• (3) wrongly claims the exterior fields differ despite equal distances
• (4) is inconsistent both in the EA claim and the EB, EC inequality

Using energy conservation with the exterior potential of the sphere V=kQ/r: (1/2)mv² = q[kQ/R − kQ/(3R)] = kQq(2/3)/R. So v² = (4kQq)/(3Rm) = 4Qq/(3×4πε0×Rm) = Qq/(3πε0mR), giving v = √(Qq/(3πε0mR)).

• (1) Inverts the potential-difference fraction from 2/(3R) to 3/(2R) (flipping numerator and denominator), giving v²=3Qq/(4πε0mR) instead of Qq/(3πε0mR).
• (2) Adds the two potential terms instead of subtracting (1/R+1/3R=4/(3R) instead of 1/R−1/3R=2/(3R)), doubling the coefficient to give √(2Qq/(3πε0mR)).
• (4) Computes 1/R−1/(3R) by subtracting the denominators (3R−R=2R) instead of the fractions, getting 1/(2R) instead of 2/(3R), giving Qq/(4πε0mR).

The Rutherford scattering formula gives N(θ) ∝ 1/sin⁴(θ/2), which is very large for small θ and decreases monotonically and steeply as θ increases toward 180° — matching a steadily falling curve.

• (1) Shows N(θ) rising with θ, the opposite of the actual 1/sin⁴(θ/2) trend which is largest at small θ.
• (2) Wrongly flattens the large-angle scattering probability; N(θ)∝1/sin⁴(θ/2) keeps falling steeply, never levelling off.
• (4) Wrongly shows a peak away from θ=0; N(θ)∝1/sin⁴(θ/2) is largest at θ=0 and falls monotonically from there.

Circuit A (series): Req = R0(1−αΔT) + R0(1+αΔT) = 2R0, independent of ΔT, so IA = V/(2R0) stays constant. Circuit B (parallel): Req = R0(1−α²ΔT²)/2, which decreases as ΔT increases, so IB = V/Req increases with temperature.

• (2) Wrongly assumes the series combination's resistance changes with ΔT, when the +α and −α terms exactly cancel: Req=R0(1−αΔT)+R0(1+αΔT)=2R0.
• (3) Reverses both trends: IA is actually constant (series Req is temperature-independent) and IB actually increases (parallel Req decreases with ΔT).
• (4) Wrongly claims IB is unaffected by ΔT, but the parallel Req=R0(1−α²ΔT²)/2 decreases as ΔT grows, so IB=V/Req increases.

σs = [MT⁻³K⁻⁴], kB = [ML²T⁻²K⁻¹], b = [LK]. So σs·kB⁻¹·b = [MT⁻³K⁻⁴]·[M⁻¹L⁻²T²K]·[LK] = [L⁻¹T⁻¹K⁻²].

• (2) Drops the T⁻¹ factor from σs=[MT⁻³K⁻⁴], giving [L⁻¹K⁻²] instead of [L⁻¹T⁻¹K⁻²].
• (3) Carries one extra power of K⁻¹ beyond the correct exponent, giving [L⁻¹T⁻¹K⁻³] instead of [L⁻¹T⁻¹K⁻²].
• (4) Carries two extra powers of K⁻¹ beyond the correct exponent, giving [L⁻¹T⁻¹K⁻⁴] instead of [L⁻¹T⁻¹K⁻²].

Speed in medium vm = c/√εr = (3×10⁸)/3 = 10⁸ m/s (option 1 correct). Wavelength λm = vm/f0 = 10⁸/(10⁶) = 100 m, not 300 m, so option 2 is incorrect. Option 3 as written uses B0 (a magnetic-field amplitude symbol undefined in the question) divided by v instead of E0/v — the relation should read By = (E0/v)sin(...), so as literally stated option 3 is also incorrect due to this symbol error. Option 4 correctly reads +z propagation from the (kz − ωt) phase.

• (1) is a true statement, so it is not the intended 'incorrect' choice
• (4) is a true statement about propagation direction, so it is not the intended 'incorrect' choice

For a full cycle, ΔU = 0, so net heat supplied equals net work done, which equals the area enclosed by the rectangle: ΔQ = W = (300−100)×(5−2) = 200×3 = 600 J.

• (1) uses ΔV=2 m3 (the volume at state a) as the box width instead of the correct 5-2=3 m3, giving 200x2=400 J
• (2) uses the lower pressure alone (100 N/m2) times the full volume 5 m3, instead of ΔP x ΔV, giving 500 J
• (4) misreads the volume axis as spanning 1 to 5 m3 (width 4) instead of 2 to 5 m3 (width 3), giving 200x4=800 J

Setting Coulomb force equal to centripetal force: ke²/r² = mv²/r, so r = ke²/(mv²) = (9×10⁹×(1.6×10⁻¹⁹)²)/(9×10⁻³¹×25.6×10¹⁰) = (2.304×10⁻²⁸)/(2.304×10⁻¹⁹) = 1×10⁻⁹ m, so x = 1.

• (1) comes from an arithmetic slip in the numerator/denominator division
• (2) 3×10⁻⁹ m isn't reachable by a single clean slip: the calculation reduces exactly to 2.304×10⁻²⁸/2.304×10⁻¹⁹=1×10⁻⁹ m, and no natural rounding of the 9, 1.6, or 25.6 factors produces a mantissa of 3.
• (3) 2×10⁻⁹ m likewise doesn't fall out of any single natural slip in this exact 2.304/2.304=1 calculation.

Minimum wear (no reliance on friction) occurs at the design speed where tan θ = v²/(rg) = (10×10)/(50×10) = 1/5, so θ = tan⁻¹(1/5).

• (2) Uses r=25 m instead of the given 50 m: tanθ=100/(25×10)=2/5, matching this option.
• (3) √3/2≈0.866 isn't reachable from tanθ=v²/(rg)=1/5 by any single natural slip; it looks like an unrelated trig-identity distractor (sin60°/cos30°-shaped).
• (4) 2√3≈3.46 likewise doesn't follow from tanθ=v²/(rg)=1/5 through any identifiable slip with this problem's numbers (v=10, r=50, g=10).

D1 is forward biased, so its depletion region narrows (W1 minimum). D3 is reverse biased, so its depletion region widens (W3 maximum). D2 carries no biasing voltage across it, so its depletion region stays at the unbiased (intermediate) width. Hence W3 > W2 > W1.

• (1) Lists W1>W2>W3, reversing the correct order: forward-biased D1 should have the narrowest depletion region, not the widest.
• (2) Wrongly equates the forward-biased D1's width with the reverse-biased D3's width, though opposite biasing must narrow one and widen the other.
• (4) Wrongly claims the unbiased D2 has the largest depletion region; reverse bias, not zero bias, produces the widest depletion region.

Microwaves are used for warming food (P-II); UV rays for purifying water (Q-I, via germicidal action); gamma rays for treating cancer cells (R-IV, radiotherapy); radio waves for AM/FM communication (S-III). This gives P-II, Q-I, R-IV, S-III.

• (1) assigns microwave to water purification and UV to warming, both swapped
• (2) Cyclically shifts three assignments — microwave to water purification, UV to cancer treatment, gamma to warming food — leaving only radio wave/communication correctly paired.
• (4) reverses the UV and radio-wave assignments relative to the correct pairing

In an elastic collision between two equal masses where one is initially at rest, the velocities are exchanged: A stops and B moves off with the full 10 ms⁻¹. Using energy conservation as B swings up, (1/2)mv² = mgh, so h = v²/(2g) = 100/20 = 5 m.

• (1) 8 m isn't reachable by a single natural slip: with v0=10 m/s and g=10 m/s², clean errors here land on multiples of 2.5 m (2.5, 5, 10); 8 m fits none of these.
• (2) 7 m likewise isn't a multiple of the 2.5 m family that natural slips in this v0=10 m/s, g=10 m/s² calculation produce.
• (4) forgets the factor of 2 in the energy-conservation formula, using h=v²/(4g) instead of v²/(2g), giving 100/40=2.5 m

Each vibrational mode contributes 1 kBT to internal energy (unlike translational/rotational which contribute (1/2)kBT each). Total U per mole = (3/2 + 3/2 + f)RT = (3+f)RT, so CV = (3+f)R and CP = CV+R = (4+f)R. Given CP/CV = (4+f)/(3+f) = 8/7 ⇒ 7(4+f) = 8(3+f) ⇒ 28+7f = 24+8f ⇒ f = 4.

• (2) Check by substituting: f = 3 gives C_V = 6R and C_P = 7R, so γ = 7/6, not 8/7. Counting each vibrational mode as ½k_BT (instead of k_BT) is the usual slip here, and it gives f = 8, not 3.
• (3) Substituting f = 2 gives C_V = 5R and C_P = 6R, so γ = 6/5, which is larger than 8/7: too few modes make γ too big.
• (4) Substituting f = 1 gives C_V = 4R and C_P = 5R, so γ = 5/4. Fewer vibrational modes raise γ; only f = 4 brings it down to 8/7.