NEET 2026, on the exam screen.

In the new system the speed of light itself is 1 unit, so distance = speed $\times$ time = $1 \times t$. Converting the time, 6 min 40 s = 360 + 40 = 400 s, giving a distance of 400 new units, so (4) is correct. Options $3\times10^{8}$ and $3\times10^{10}$ wrongly re-insert the SI value of c even though c is now unity, and 500 comes from mis-converting the time.

Young's modulus is stress/strain $= \frac{FL}{A\,\Delta L}$ (II); compressibility is the reciprocal of bulk modulus $= -\frac{1}{\Delta P}\frac{\Delta V}{V}$ (III); bulk modulus $= -P\frac{V}{\Delta V}$ (IV); and Poisson's ratio is lateral strain over longitudinal strain $= \frac{\Delta d}{\Delta L}\frac{L}{d}$ (I). Hence P-II, Q-III, R-IV, S-I, which is option (4).

An ideal diode has zero forward resistance and infinite reverse resistance, so only the branches whose diodes are forward biased conduct; here that leaves the 4 $\Omega$ and 2 $\Omega$ resistors in parallel across the 10 V battery. The current is $I = \frac{10}{2} + \frac{10}{4} = \frac{15}{2}$ A, giving (4). The other options wrongly include the reverse-biased 3 $\Omega$ and 5 $\Omega$ branches that actually carry no current.

With uniform angular acceleration the number of revolutions equals the average angular speed times the time: $\frac{600+1200}{2} = 900$ revolutions per minute (rpm) over $\frac{10}{60}$ min = $\frac{1}{6}$ min, giving $900 \times \frac{1}{6} = 150$ revolutions, option (3). Equivalently, using $\omega_2^2 = \omega_1^2 + 2\alpha\theta$ gives $\theta = 300\pi$ rad, so $\theta/2\pi = 150$.

The velocity in simple harmonic motion (SHM) is $v = A\omega\cos(\omega t + \phi)$, so the kinetic energy $K = \tfrac{1}{2}mA^2\omega^2\cos^2(\omega t + \phi)$ is proportional to $\cos^2(\omega t + \phi)$. This is always non-negative and repeats with period $T/2$, producing two identical humps between 0 and $T$ that fall to zero at the extreme positions. Graph (3) shows exactly this, so the answer is (3).

The terminal voltage is the electromotive force (emf) minus the drop across the internal resistance: $V = E - Ir = 12 - 0.6 \times 2 = 10.8$ V, option (4). The 1.2 V drop across the internal resistance is what is subtracted, and 12 V would only be the terminal voltage if no current flowed.

Since $v_{rms} = \sqrt{\frac{3RT}{M}}$, at the same temperature $v_{rms} \propto \frac{1}{\sqrt{M}}$. Therefore $\frac{v_{rms}^{Ar}}{v_{rms}^{Cl}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}} = \sqrt{\frac{70}{40}} = \frac{\sqrt{7}}{2}$, option (1). The 2:1 mass ratio of the gases does not affect the rms speed, which depends only on temperature and molar mass.

When the ray inside an equilateral prism ($A = 60^{\circ}$) travels parallel to the base, the prism is in the minimum-deviation condition where the angle of incidence equals the angle of emergence ($i = e$). Using $i + e = A + \delta$ gives $\delta = 2i - A = 2(50^{\circ}) - 60^{\circ} = 40^{\circ}$, option (3).

$E = h\nu$ gives the energy of a photon (IV); diffraction and interference demonstrate the wave nature of light (III); $\lambda = h/p$ is the de Broglie wavelength of a particle (I); and the Compton effect confirms the particle nature of light (II). Hence P-IV, Q-III, R-I, S-II, which is option (4).

For the hydrogen atom the magnitude of the total energy equals half the magnitude of the potential energy, so $\frac{KQ^2}{2r} = 3.4$ eV. Solving, $r = \frac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times3.4\times1.6\times10^{-19}} \approx 2.1\times10^{-10}$ m, option (3). This is the $n=2$ Bohr orbit radius, i.e. $4a_0 \approx 2.1$ angstrom.

The box stays with the trolley only while static friction can supply the force needed to accelerate it. The maximum friction is $\mu mg$, so the greatest common acceleration is $a = \mu g = 0.12 \times 10 = 1.2$ m/s$^2$, option (4). The mass cancels out, and beyond this acceleration the box slips relative to the trolley.

The four 10 $\mu F$ capacitors are in series, giving $\frac{10}{4} = 2.5\ \mu F$, and this branch is in parallel with $C_5 = 2.5\ \mu F$, so $C_{eq} = 2.5 + 2.5 = 5\ \mu F$. Each branch has 50 V across it, so the series branch carries $q = 2.5 \times 50 = 125\ \mu C$ (the same charge flows through $C_1$ to $C_4$) and $C_5$ also holds $2.5 \times 50 = 125\ \mu C$. Every capacitor therefore stores 125 $\mu C$, option (2).

The work done equals the change in gravitational potential energy: $W = U_2 - U_1 = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right) = \frac{GMm}{2R}$. Using $g = \frac{GM}{R^2}$, this becomes $\frac{mgR}{2}$, option (4). It is not the naive $mgR$ because gravity weakens with height, so less work is needed than if g stayed constant.

Density = mass/volume = $\dfrac{5.580}{(9.0)^3}$ which numerically gives about $7.654 \times 10^3\ \text{kg m}^{-3}$. The side is measured to only two significant figures (9.0 cm), so the final answer must be rounded to two significant figures, giving $7.7 \times 10^3$. Hence X = 7.7 (option (4)); the other values retain more significant figures than the data justify.

Throughout the flight the only acceleration is g directed downward, so $dv/dt$ stays a constant negative value. The velocity therefore decreases linearly, reaches zero at the highest point and turns negative on the way down, tracing a single straight line of constant negative slope, which is plot C. Hence option (1). Plots with a change of slope, a V-shape, or a positive slope contradict the constant downward acceleration.

At the equilibrium (lowest) point the bob's energy is entirely kinetic, so $\tfrac{1}{2}mv^2$ equals the total energy 0.02 J. With m = 20 g = 0.02 kg, $v^2 = \dfrac{2(0.02)}{0.02} = 2$, giving $v = \sqrt{2} \approx 1.41\ \text{m/s}$ (option (2)). Forgetting to divide by the mass gives $\sqrt{2(0.02)} = 0.2\ \text{m/s}$ (option (1)).

The interference intensity follows $I = I_0\cos^2\left(\dfrac{\Delta\phi}{2}\right)$. A path difference of $\lambda$ gives $\Delta\phi = 2\pi$, so $I = I_0 = K$, i.e. $I_0 = K$. For a path difference $\lambda/3$, $\Delta\phi = \dfrac{2\pi}{3}$, giving $I = I_0\cos^2\left(\dfrac{\pi}{3}\right) = \dfrac{I_0}{4} = \dfrac{K}{4}$ (option (1)). The other choices ignore the $\cos^2$ dependence on phase.

The applied alternating current (ac) voltage drives current through the diode and R in series. During the half-cycle that reverse-biases the diode, almost the entire source voltage appears across the non-conducting diode, whereas in the forward-biased half-cycle the diode drop is nearly zero. So $v_D$ consists of one-sided humps with flat (near-zero) gaps, matching option (4). A full sine or a double-hump trace would wrongly imply the diode never blocks the current.

Series resonance occurs at $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$, which is independent of the resistance. With $L = 1\ \text{mH} = 10^{-3}\ \text{H}$ and $C = 0.1\ \mu\text{F} = 10^{-7}\ \text{F}$, $f_0 = \dfrac{1}{2\pi\sqrt{10^{-10}}} = \dfrac{1}{2\pi \times 10^{-5}} \approx 15.9\ \text{kHz}$ (option (4)). The resistance value is a distractor since it does not enter the resonance formula.

Interference and diffraction only redistribute light energy, so darkening of one region is compensated by brightening of another, keeping total energy unchanged. This makes statement I consistent with conservation of energy, i.e. Statement I is true. However, interference and diffraction are general wave phenomena shown by sound, water and other waves, not only light, so statement II is false. Hence 'I true, II false' (option (3)).

The spatial part of the phase is $2\pi(0.0080)x$ with x in cm, so the wave number term gives phase difference $\Delta\phi = 2\pi(0.0080)\,\Delta x$. Converting the separation to centimetres, $\Delta x = 0.5\ \text{m} = 50\ \text{cm}$, so $\Delta\phi = 2\pi \times 0.0080 \times 50 = 0.8\pi$ rad (option (2)). The wrong answers arise from failing to convert 0.5 m into 50 cm.

The two perpendicular forces give a resultant $F = \sqrt{8^2 + 6^2} = 10\ \text{N}$, so $a = F/m = 10/5 = 2\ \text{m s}^{-2}$. The resultant makes an angle $\theta$ with the 8 N force where $\tan\theta = 6/8 = 3/4$, i.e. $\tan^{-1}(3/4)$ measured from the 8 N force (option (4)). Option (1) forgets to divide by the mass, and the others attach the angle to the wrong force or invert the ratio.

When a charged capacitor shares its charge with an identical uncharged one, the energy dissipated is $\dfrac{1}{2}\dfrac{C_1 C_2}{C_1 + C_2}V^2$. With $C_1 = C_2 = 200\ \text{pF}$ and $V = 100\ \text{V}$, this equals $\dfrac{1}{2}(100\ \text{pF})(100)^2 = 0.5 \times 10^{-6}\ \text{J}$ (option (1)). The loss is real and independent of the connecting-wire resistance; answers in whole joules mishandle the pico-farad factor.

The useful power is work done against gravity divided by time, $P = \dfrac{mgh}{t}$. Substituting, $P = \dfrac{1000 \times 9.8 \times 20}{10} = 19600\ \text{W} = 19.6\ \text{kW}$ (option (3)). The 19.6 W answer drops a factor of 1000, and the 39.2 values double-count g or the height.

Least count = 1 main scale division (MSD) - 1 VSD. Since 20 VSD = 16 MSD, one VSD equals $\dfrac{16}{20}$ MSD, so L.C. = $\left(1 - \dfrac{16}{20}\right)$ MSD = $\dfrac{4}{20}$ MSD = $0.2 \times 1$ mm = 0.2 mm = 0.02 cm (option (3)). Forgetting to convert millimetres to centimetres gives the distractor 0.2 cm.

The ruler falls freely, so distance $s = \frac{1}{2}gt^2$ increases monotonically with reaction time. Ranking the reaction times from largest to smallest gives B(0.22) > E(0.21) > A(0.20) > D(0.19) > C(0.18), which is the same order as the distances. Option (4) is the exact reverse (smallest first). Option (2) is also smallest first and swaps persons B and E as well. Option (1) swaps persons D and C.

The 4 $\Omega$ wire splits equally, so each side of the square is 1 $\Omega$. With the 2 $\Omega$ across diagonal BD and the 2 V battery across A and C, the four 1 $\Omega$ arms form a balanced Wheatstone bridge, so no current flows through the 2 $\Omega$ branch. The effective resistance across AC is 1 $\Omega$, giving $I = E/R = 2/1 = 2$ A.

Why the other options are wrong:

- (2) Assumes far too low a resistance.
- (3) Arbitrary value from mis-combining resistances.
- (4) Would need 0.5 $\Omega$, as if each path from A to C had one 1 $\Omega$ side instead of two.

The heater resistance is fixed, so $P = V^2/R$ gives $P_c = \left(\frac{V}{V_0}\right)^2 P_0 = \left(\frac{200}{220}\right)^2 \times 400 \approx 331$ W. Power scales as the square of voltage, not linearly, so option (2) (unchanged) and option (1) (linear scaling) are wrong, and option (4) over-reduces.

From $B = \frac{\mu_0 N i}{2R}$, the current is $i = \frac{2RB}{\mu_0 N} = \frac{2\times5\times10^{-2}\times3.14\times10^{-3}}{4\pi\times10^{-7}\times100} = 2.5$ A. The magnetic moment $M = NiA = Ni\pi R^2 = 100\times2.5\times\pi\times(0.05)^2 \approx 2$ A m$^2$. Only option (4) gives both correct values.

The motional electromotive force (emf) is $E = Bvl$, where $l$ is the side perpendicular to the velocity. Since the loop moves normal to the shorter side, the effective length is the shorter side (3 cm): $E = 0.3 \times (2\times10^{-2}) \times (3\times10^{-2}) = 1.8\times10^{-4}$ V. Option (1) wrongly uses the 8 cm side.

Nuclear radius $r = r_0 A^{1/3}$, so volume $\propto r^3 \propto A$ (statement II correct, I wrong). The mass defect is the difference between the total mass of the constituent nucleons and the actual mass of the nucleus (statement IV correct, III wrong). Hence only II and IV are true.

Using density = mass/volume with volume $= \frac{4}{3}\pi R^3$ and $R = R_0 A^{1/3}$, we get $\frac{4}{3}\pi R_0^3 A = \frac{m}{\rho}$. Solving, $A = \frac{m}{\rho}\cdot\frac{3}{4\pi R_0^3} = \frac{19.926\times10^{-27}\times3}{2.29\times10^{17}\times12.56\times(1.2\times10^{-15})^3} \approx 12$.

The time period is $T = 60/30 = 2$ s. From $T = 2\pi\sqrt{L/g}$, the length is $L = \frac{gT^2}{4\pi^2} = \frac{9.8\times4}{4\times9.8} = 1$ m. The other options result from arithmetic or period errors.

By the first law of thermodynamics $Q = \Delta U + W$. Differentiating with respect to time, $\frac{dQ}{dt} = \frac{d(\Delta U)}{dt} + \frac{dW}{dt}$, so $\frac{d(\Delta U)}{dt} = 100 - 75 = 25$ W. The distractors correspond to the given work rate (1), heat rate (2), or the incorrect sum (3).

The ring's mass is $M = mL$ and its radius $r = L/2\pi$ (since $L = 2\pi r$). The axis $yy'$ is tangential to the ring in its plane, so by the parallel axis theorem $I_{yy'} = I_{CM} + Mr^2 = \tfrac{1}{2}Mr^2 + Mr^2 = \tfrac{3}{2}Mr^2 = \tfrac{3}{2}(mL)\left(\tfrac{L}{2\pi}\right)^2 = \dfrac{3mL^3}{8\pi^2}$. The correct expression has $L^3$ and $\pi^2$, eliminating the other options.

Why the other options are wrong:

- (1) Uses $\pi$ instead of $\pi$$^{2}$ in the denominator.
- (3) Wrong power of L and $\pi$.
- (4) Wrong power of L.

The shunt is in parallel with the galvanometer, so $i_s r_s = i_G R_G$, where $i_s \approx 10$ A and $i_G = 1$ mA. Thus $10\times r_s = 10^{-3}\times100$, giving $r_s = 0.01$ $\Omega$. The other values come from misplacing the factor of ten in the current ratio.

Why the other options are wrong:

- (2) Off by a factor of 10.
- (3) Off by a factor of 100.
- (4) Off by a factor of 10 (too small).

A metre bridge is a Wheatstone network, whose balance (null) condition is symmetric with respect to interchanging the cell and the galvanometer, so the balance point does not move. Away from that point the bridge is unbalanced and the galvanometer deflects; the direction of deflection depends on which side of balance the jockey lies, and it reads zero exactly at the balance point.

Why the other options are wrong:

- (1) Deflection is not restricted to one side; it flips about the null point.
- (2) No deflection occurs only at balance, not everywhere.
- (3) Same error as (1) but for the opposite side.

With $i = i_{peak}\sin(\omega t)$ and $\omega = 2\pi f = 120\pi\ \text{rad/s}$, the current first equals its peak when $\sin(\omega t) = 1$, i.e. $\omega t = \pi/2$. This gives $t = \dfrac{1}{4f} = \dfrac{1}{4 \times 60} = \dfrac{1}{240}\ \text{s}$ (a quarter of the period).

By Ampere's law, inside the wire ($r<a$) only the enclosed fraction of current contributes, giving $B = \dfrac{\mu_0 I r}{2\pi a^2}$, so $B \propto r$ (a straight line rising to a maximum at $r=a$). Outside ($r>a$) the full current is enclosed, giving $B = \dfrac{\mu_0 I}{2\pi r}$, so $B \propto 1/r$. Only graph (1) shows this linear rise followed by a $1/r$ decay.

Statement A is correct: beyond the threshold (knee) voltage in forward bias, the diode current rises steeply, as seen in the forward V-I characteristic. Statement B is wrong: this rapidly increasing current is the forward current, whereas reverse saturation current is the very small, nearly constant current that flows under reverse bias.

In electrostatic equilibrium, the electric field inside a conductor is zero, and any excess charge lies on its surface. The field just outside is normal to the surface and depends on the local surface charge density, $E = \dfrac{\sigma}{\varepsilon_0}\hat{n}$. The potential is constant inside but need not be zero. Therefore internal statements I, III and IV hold; II and V do not.

Photoemission requires photon energy $\dfrac{hc}{\lambda} \geq W_0$, i.e. wavelength $\lambda \leq \dfrac{hc}{W_0}$. Here $\dfrac{hc}{W_0} = \dfrac{6.6\times10^{-34}\times3\times10^{8}}{6.6\times1.6\times10^{-19}} \approx 187.5\ \text{nm}$. Any wavelength longer than this cannot eject electrons; $200\ \text{nm} > 187.5\ \text{nm}$, so it produces no photoelectric effect.

A concave lens is diverging, so a ray incident parallel to the principal axis bends away from the axis and travels as though it originated from the focal point on the same (incident) side of the lens. Traced backwards, the emergent ray appears to diverge from that near-side focus. Strictly, this near-side point is the second principal focus ($F_2$ in the figure), but option (2) is still the best answer, because it is the only option where the ray appears to diverge from a focus.

Why the other options are wrong:

- (1) A parallel ray relates to the focus, not to $2F$; this describes a different ray.
- (3) It does not stay parallel; a diverging lens spreads a parallel beam.
- (4) The emergent ray only appears to come from the second principal focus $F_2$ on the incident side; it moves away from that point and never passes through it.

Absolute pressure at depth $h$ is $P = P_0 + \rho g h$. So $100\times10^{5} = 1\times10^{5} + 1000\times10\times h$, giving $99\times10^{5} = 10^{4}\,h$, hence $h = 990\ \text{m}$.

Microwaves are generated by klystron or magnetron valves (P-IV). Visible light is emitted when atomic electrons drop from higher to lower energy levels (Q-I). Gamma rays are produced in radioactive decay of the nucleus (R-II). Infra-red radiation arises from the vibration of atoms and molecules (S-III). This matches option (3).

Nitriles $\mathrm{R}-\mathrm{CN}$ are reduced to primary amines $\mathrm{R}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$ by $\mathrm{LiAlH}_{4}$ followed by hydrolysis, by catalytic hydrogenation $\mathrm{H}_{2}/\mathrm{Ni}$, and by $\mathrm{Na}(\mathrm{Hg})/\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}$ (sodium amalgam in ethanol). So Statements I, III and IV are correct. $\mathrm{Sn} + \mathrm{HCl}$ (Stephen reduction) stops at the aldehyde, and $\mathrm{Br}_{2}$/aq. NaOH is the Hoffmann bromamide reaction on amides, not a nitrile reduction.

$\mathrm{V}_{2}\mathrm{O}_{5}$ is the contact-process catalyst linked with $\mathrm{H}_{2}\mathrm{SO}_{4}$/$\mathrm{SO}_{2}$ chemistry (III); Fe is the Haber-process catalyst for ammonia from $\mathrm{N}_{2}$/$\mathrm{H}_{2}$ (I); $\mathrm{PdCl}_{2}$ catalyses Wacker oxidation of ethyne/ethene to ethanal (IV); and Ni complexes catalyse polymerisation of alkynes (II). Hence P-III, Q-I, R-IV, S-II.

With $\Delta n_g = 2 - 3 = -1$, $\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT = -10 - \frac{298\times8.31}{1000} = -12.48\ \text{kJ}$. Then $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} = -12.48 - \frac{298\times(-44)}{1000} = -12.48 + 13.112 \approx +0.636\ \text{kJ mol}^{-1}$. A positive $\Delta G^{\ominus}$ means the reaction is non-spontaneous.

The orbital is named by n and the subshell letter for l (l = 0,1,2,3 -> s,p,d,f). So n=2,l=1 is 2p (II); n=4,l=0 is 4s (III); n=5,l=3 is 5f (IV); n=3,l=2 is 3d (I). This gives P-II, Q-III, R-IV, S-I.

For the dissolution, $K = s^2$ since a solid gives equal amounts of $\mathrm{BiO}^{+}$ and $\mathrm{OH}^{-}$. So $s = \sqrt{4\times10^{-10}} = 2\times10^{-5}\ \text{M} = [\mathrm{OH}^{-}]$. Then $\text{pOH} = 5 - \log 2 = 4.699$ and $\text{pH} = 14 - 4.699 = 9.301$.

DNA has a double-stranded helical secondary structure (two antiparallel polynucleotide chains) and its four bases are adenine, guanine, cytosine and thymine. RNA is generally single-stranded and contains uracil in place of thymine, so only the DNA statement pairing double-strand + thymine is correct.

Metamers have the same molecular formula and same functional group but differ in the alkyl groups attached on either side of that group. Methyl n-propyl ether $\mathrm{CH}_{3}\mathrm{OCH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}$ and diethyl ether $\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OCH}_{2}\mathrm{CH}_{3}$ are both $\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}$ ethers differing only in the distribution of carbons about the oxygen, so they are metamers.

Why the other options are wrong:

- (1) 1-propanol and 2-propanol are position isomers, not metamers.
- (2) n-Pentane and isopentane are chain isomers.
- (3) Propanone and propanal are functional-group isomers.

$[\mathrm{Pt}(\mathrm{NH}_{3})_{2}\mathrm{Cl}_{2}]$ shows cis/trans (geometrical) isomerism (III); $[\mathrm{Co}(\mathrm{en})_{3}]^{3+}$ is a tris-chelate that is chiral, showing optical isomerism (I); $[\mathrm{Co}(\mathrm{NH}_{3})_{5}\mathrm{NO}_{2}]\mathrm{Cl}_{2}$ shows linkage isomerism through the ambidentate $\mathrm{NO}_{2}^{-}$/$\mathrm{ONO}^{-}$ ligand (IV); and $[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]\mathrm{Cl}_{3}$ shows solvate (hydrate) isomerism (II). Hence A-III, B-I, C-IV, D-II.

The unit of the rate constant of an nth-order reaction is $(\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}$. For n=0 this is $\text{mol L}^{-1}\text{s}^{-1}$ (IV); n=1 gives $\text{s}^{-1}$ (III); n=2 gives $\text{mol}^{-1}\text{L s}^{-1}$ (I); n=3 gives $\text{mol}^{-2}\text{L}^2\text{s}^{-1}$ (II). Hence P-IV, Q-III, R-I, S-II.