NEET 2026 Answer Key

Time taken by light: $t = 6 \text{ min } 40 \text{ s} = 360 + 40 = 400 \text{ s}$

Since the speed of light in vacuum is taken as unity (i.e., $v = 1$), the distance in the new unit is:

$d = v \times t = 1 \times 400 = 400$

• A. Young's Modulus = $\dfrac{\text{Stress}}{\text{Strain}} = \dfrac{FL}{A \Delta L}$ → II
• B. Compressibility = $\dfrac{1}{\text{Bulk Modulus}} = -\dfrac{1}{\Delta P}\left(\dfrac{\Delta V}{V}\right)$ → III
• C. Bulk Modulus = $-P\left(\dfrac{V}{\Delta V}\right)$ → IV
• D. Poisson's Ratio = $\dfrac{\text{Lateral strain}}{\text{Longitudinal strain}} = \dfrac{\Delta d}{\Delta L}\left(\dfrac{L}{d}\right)$ → I

Hence the correct match is A-II, B-III, C-IV, D-I.

For an ideal diode: forward resistance = 0 and reverse biased resistance = $\infty$.

Analysing the circuit, the forward-biased diodes short-circuit their respective branches, leaving only the 4 Ω and 2 Ω resistors in parallel across the 10 V battery:

Circuit diagram with diodes and resistors

$I = \frac{10}{2} + \frac{10}{4} = 5 + 2.5 = \frac{30}{4} = \frac{15}{2} \text{ A}$

Angular acceleration:

$\alpha = \frac{\omega_2 - \omega_1}{\Delta t} = \left(\frac{1200 - 600}{10}\right)\frac{2\pi}{60} = 2\pi \text{ rad/s}^2$

Convert angular velocities:

$\omega_1 = 600 \times \frac{2\pi}{60} = 20\pi \text{ rad/s}, \quad \omega_2 = 1200 \times \frac{2\pi}{60} = 40\pi \text{ rad/s}$

Using $\omega_2^2 = \omega_1^2 + 2\alpha\theta$:

$(40\pi)^2 = (20\pi)^2 + 2 \times 2\pi\theta$

$1200\pi^2 = 4\pi\theta \implies \theta = 300\pi \text{ radian}$

$\text{Number of revolutions} = \frac{\theta}{2\pi} = \frac{300\pi}{2\pi} = 150$

Kinetic energy of a simple pendulum:

$K = \frac{1}{2}mv^2 = \frac{1}{2}mA^2\omega^2\cos^2(\omega t + \phi)$

$\therefore K \propto \cos^2(\omega t + \phi)$

Since $\cos^2$ is always non-negative and oscillates with a period of $T/2$, the K.E. vs time graph is always above the time axis, with peaks at $t = 0, T/2, T, \ldots$ and zeros at $t = T/4, 3T/4, \ldots$ This matches option (3).

Circuit can be draw as,

Circuit

Terminal voltage of the battery:

$V = E - ir = 12 - (0.6 \times 2) = 12 - 1.2 = 10.8 \text{ V}$

The rms speed is given by:

$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$

For the same temperature, $v_{\text{rms}} \propto \dfrac{1}{\sqrt{M}}$. Therefore:

$\frac{V^{\text{Ar}}_{\text{rms}}}{V^{\text{Cl}}_{\text{rms}}} = \sqrt{\frac{M_{\text{Cl}}}{M_{\text{Ar}}}} = \sqrt{\frac{70}{40}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}$

Equilateral prism diagram

Since QR is parallel to the base BC, the ray travels symmetrically through the prism, meaning the angle of emergence $e = i = 50°$.

Using the prism equation:

$i + e = A + \delta$

$50° + 50° = 60° + \delta$

$\delta = 100° - 60° = 40°$

• A. $E = h\nu$ is the energy of a photon → IV
• B. Diffraction and Interference confirm the wave nature of light → III
• C. $\lambda = h/p$ is the de Broglie wavelength of a particle → I
• D. Compton effect confirms the particle nature of light → II

Hence the correct match is A-IV, B-III, C-I, D-II.

The potential energy equals twice the total energy in magnitude (by the virial theorem), so:

$\frac{KQ^2}{2r} = 3.4 \text{ eV}$

$r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times 3.4 \times 1.6 \times 10^{-19}}$

$r = 2.1176 \times 10^{-10} \approx 2.1 \times 10^{-10} \text{ m}$

Forces diagram

The maximum static friction provides the net horizontal force on the box:

$f_{\max} = \mu mg = ma_{\max}$

$a_{\max} = \mu g = 0.12 \times 10 = 1.2 \text{ m/s}^2$

Circuit diagram

From the circuit, $C_1$, $C_2$, $C_3$, $C_4$ (10 µF each) are in series in pairs, giving two series combinations of 5 µF each, which are in parallel giving 10 µF — but this is in series with $C_5 = 2.5$ µF equivalent branch. After full simplification:

$C_{\text{eq}} = 2.5 + 2.5 = 5\ \mu\text{F}$

Charge on each capacitor:

$q_1 = q_2 = q_3 = q_4 = 2.5 \times 50 = 125\ \mu\text{C}$

$q_5 = 2.5 \times 50 = 125\ \mu\text{C}$

Hence all five capacitors carry 125 µC each.

$W = U_2 - U_1 = -\frac{GMm}{R+R} - \left(-\frac{GMm}{R}\right)$

$= -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{GMm}{2R} = \frac{mgR}{2}$

$\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{5.580 \text{ kg}}{(9.0 \text{ cm})^3} = \frac{5.580}{729} \times 10^6 \text{ kg/m}^3$

$= 0.007654321 \times 10^6 = 7.654 \times 10^3 \text{ kg/m}^3$

Since the side measurement (9.0 cm) has only 2 significant figures, the result must be rounded to 2 significant figures:

$\text{Density} = 7.7 \times 10^3 \text{ kg/m}^3 \implies X = 7.7$

During the entire journey (upward and downward), the acceleration due to gravity acts vertically downward (constant). Therefore, the slope of the velocity vs. time curve must be negative and constant throughout the journey.

Diagram

The ball starts with a positive velocity (upward), decelerates to zero at the highest point, then accelerates downward (negative velocity). This gives a straight line with a constant negative slope crossing the time axis — which matches plot C.

At the equilibrium position, the potential energy is zero, so all the total mechanical energy is kinetic:

$\frac{1}{2}mv^2 = 0.02 \text{ J}$

$\frac{1}{2} \times 20 \times 10^{-3} \times v^2 = 2 \times 10^{-2}$

$v^2 = 2 \implies v = \sqrt{2} \approx 1.41 \text{ m/s}$

The intensity formula is:

$I = I_0 \cos^2\frac{\Delta\phi}{2}, \quad \text{where } \Delta\phi = \frac{2\pi}{\lambda} \cdot \Delta x$

For path difference $= \lambda$:

$K = I_0 \cos^2\left(\frac{2\pi}{\lambda} \cdot \frac{\lambda}{2}\right) = I_0\cos^2(\pi) = I_0 \implies K = I_0$

For path difference $= \dfrac{\lambda}{3}$:

$K_1 = I_0 \cos^2\left(\frac{2\pi}{\lambda} \cdot \frac{\lambda}{6}\right) = I_0\cos^2\left(\frac{\pi}{3}\right) = I_0 \cdot \left(\frac{1}{2}\right)^2 = \frac{I_0}{4} = \frac{K}{4}$

Diode circuit with AC source and resistor

The voltage drop appears across the diode only when it is in reverse bias. In the positive half cycle, the diode is reverse biased — the full positive half of the AC input appears across it. In the negative half cycle, the diode is forward biased (zero resistance), so no voltage drops across it. Thus, $V_D$ shows only the positive half-cycle of the input waveform, which corresponds to option (4).

Diode circuit with AC source and resistor

The resonance frequency of a series LC circuit:

$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{1 \times 10^{-3} \times 0.1 \times 10^{-6}}}$

$= \frac{1}{2\pi\sqrt{10^{-10}}} = \frac{1}{2\pi \times 10^{-5}} \approx 15.9 \text{ kHz}$

• Statement A is TRUE: In both interference and diffraction, there is no net gain or loss of energy — energy is merely redistributed from dark fringes to bright fringes, consistent with the conservation of energy.
• Statement B is FALSE: Diffraction and interference are wave phenomena exhibited by all types of waves, including sound waves, not just light waves.

Comparing with the standard form, the wave number is:

$k = 2\pi \times 8 \times 10^{-3} \text{ rad/cm}$

The phase difference between two points separated by $\Delta x = 0.5 \text{ m} = 50 \text{ cm}$ is:

$\Delta\phi = k \cdot \Delta x = 2\pi \times 8 \times 10^{-3} \times 50 = 2\pi \times 8 \times 10^{-3} \times \frac{100}{2}$

$= 8\pi \times 10^{-1} = 0.8\pi \text{ rad}$

Solution diagram

Net force:

$F_{\text{net}} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ N}$

Acceleration:

$a = \frac{F_{\text{net}}}{m} = \frac{10}{5} = 2 \text{ m/s}^2$

Direction with respect to the 8 N force:

$\tan\theta = \frac{6}{8} = \frac{3}{4} \implies \theta = \tan^{-1}\!\left(\frac{3}{4}\right) \text{ from the 8 N force}$

Energy lost when a charged capacitor is connected to an identical uncharged capacitor:

$\Delta E = \frac{1}{2} \cdot \frac{C_1 C_2}{C_1 + C_2} \cdot V^2$

$= \frac{1}{2} \times \frac{200 \times 200}{400} \times 10^{-12} \times (100)^2$

$= \frac{1}{2} \times 100 \times 10^{-12} \times 10^4 = \frac{1}{2} \times 10^{-6} = 0.5 \times 10^{-6} \text{ J}$

$P = \frac{\text{Work}}{\text{Time}} = \frac{mgh}{t} = \frac{10^3 \times 9.8 \times 20}{10} = 19.6 \text{ kW}$

$\text{L.C.} = 1 \text{ MSD} - 1 \text{ VSD}$

Since $20 \text{ VSD} = 16 \text{ MSD}$:

$1 \text{ VSD} = \frac{16}{20} \text{ MSD}$

$\text{L.C.} = 1 \text{ MSD} - \frac{16}{20} \text{ MSD} = \frac{4}{20} \text{ MSD} = \frac{1}{5} \text{ mm} = 0.2 \text{ mm} = 0.02 \text{ cm}$

The distance fallen is $s = \frac{1}{2}gt^2$, which is directly proportional to the square of reaction time. Therefore, greater the reaction time, greater the distance fallen.

Descending order of reaction times: $t_B (0.22) > t_E (0.21) > t_A (0.20) > t_D (0.19) > t_C (0.18)$

Hence descending order of distance: $S_B > S_E > S_A > S_D > S_C$

Square loop Wheatstone bridge circuit

• The total wire resistance is 4 Ω, so each of the four sides has resistance 1 Ω.
• The circuit forms a balanced Wheatstone bridge (each arm = 1 Ω), so no current flows through the 2 Ω resistance between B and D.
• The effective resistance between A and C is two parallel paths of 2 Ω each:

$R_{\text{eff}} = \frac{2 \times 2}{2 + 2} = 1\ \Omega$

$I = \frac{E}{R_{\text{eff}}} = \frac{2}{1} = 2 \text{ A}$

The resistance of the heater is fixed. Using $P = V^2/R$:

$\frac{P_c}{P_0} = \frac{V^2}{V_0^2}$

$P_c = P_0 \times \left(\frac{V}{V_0}\right)^2 = 400 \times \left(\frac{200}{220}\right)^2 \approx 331 \text{ W}$

Magnetic field at centre of a circular coil:

$B_0 = \frac{\mu_0 N i}{2R}$

$i = \frac{2RB_0}{\mu_0 N} = \frac{2 \times 5 \times 10^{-2} \times 3.14 \times 10^{-3}}{4\pi \times 10^{-7} \times 100} = 2.5 \text{ A}$

Magnetic moment:

$M = NiA = Ni(\pi R^2) = 100 \times 2.5 \times 3.14 \times (5 \times 10^{-2})^2 = 2 \text{ A m}^2$

The loop moves normal to the shorter side (3 cm), so the effective cutting length is the shorter side:

$E_{\text{induced}} = Bvl = 0.3 \times 2 \times 10^{-2} \times 3 \times 10^{-2} = 1.8 \times 10^{-4} \text{ V}$

Nuclear radius: $r = r_0 A^{1/3}$, so $r^3 = r_0^3 \cdot A$, hence $V = \frac{4}{3}\pi r^3 \propto A$. So A is wrong, B is correct.

Mass defect is defined as the difference between the mass of the individual nucleons (protons + neutrons) and the actual mass of the assembled nucleus — not the difference between atom and nucleus. So C is wrong, D is correct.

Using $R = R_0 A^{1/3}$ and $V = \dfrac{4}{3}\pi R^3 = \dfrac{M}{\rho}$:

$\frac{4}{3}\pi R_0^3 A = \frac{M}{\rho}$

$A = \frac{M}{\rho} \times \frac{3}{4\pi R_0^3} = \frac{19.926 \times 10^{-27} \times 3}{2.29 \times 10^{17} \times 12.56 \times (1.2 \times 10^{-15})^3} \approx 12$

Time for 30 oscillations = 60 s, so time period:

$T = \frac{60}{30} = 2 \text{ s}$

Using $T = 2\pi\sqrt{\dfrac{l}{g}}$:

$l = \frac{gT^2}{4\pi^2} = \frac{9.8 \times 2 \times 2}{4 \times 9.8} = \frac{9.8 \times 4}{4 \times 9.8} = 1 \text{ m}$

By the first law of thermodynamics:

$\frac{dQ}{dt} = \frac{d(\Delta U)}{dt} + \frac{dW}{dt}$

$100 = \frac{d(\Delta U)}{dt} + 75$

$\frac{d(\Delta U)}{dt} = 25 \text{ W}$

Total mass of the wire: $M = mL$

The wire forms a circular ring of circumference $L$, so radius:

$r = \frac{L}{2\pi}$

Moment of inertia about a diameter (CM axis along $xx'$): $I_{CM} = \dfrac{Mr^2}{2}$

Using the parallel axis theorem, the axis $yy'$ passes through the rim (tangent to the circle at one point), at distance $r$ from the CM:

$I_{yy'} = I_{CM} + Mr^2 = \frac{Mr^2}{2} + Mr^2 = \frac{3}{2}Mr^2$

$= \frac{3}{2} \times (mL) \times \left(\frac{L}{2\pi}\right)^2 = \frac{3mL^3}{8\pi^2}$

circuit

$i_G = 1\ \text{mA} = 0.001\ \text{A}$, $i_S \approx 10\ \text{A}$

Since the shunt and galvanometer are in parallel:

$i_S r_S = i_G R_G$

$10 \times r_S = 0.001 \times 100$

$r_S = 0.01\ \Omega$

• The position of the null (balance) point does not change when $G$ and $E$ are interchanged — this is a property of the Wheatstone bridge.
• At the balance point, there will be no deflection in the galvanometer.
• At all other positions of the jockey (unbalanced condition), the galvanometer will show deflection, which can be either left-sided or right-sided depending on the position relative to the null point.

$i = i_{\text{peak}}\sin(\omega t), \quad \omega = 2\pi f = 2\pi \times 60 = 120\pi \text{ rad/s}$

For current to reach its peak value:

$\sin(120\pi t) = 1 \implies 120\pi t = \frac{\pi}{2}$

$t = \frac{1}{240} \text{ s}$

For a long straight solid wire with uniformly distributed current:

• Inside the wire ($r < a$): $B = \dfrac{\mu_0 I r}{2\pi a^2} \implies B \propto r$ (linear increase)
• Outside the wire ($r > a$): $B = \dfrac{\mu_0 I}{2\pi r} \implies B \propto \dfrac{1}{r}$ (hyperbolic decrease)

The field increases linearly from the axis, reaches a maximum at $r = a$, then decreases as $1/r$. This matches option (1).

Solution

• Statement A is TRUE: In the V–I characteristic of a forward-biased p-n junction diode, once the forward bias exceeds the threshold (knee) voltage (~0.7 V for Si), the diode current rises sharply.
• Statement B is FALSE: This large forward current is simply called the forward current, not the reverse saturation current. Reverse saturation current is the very small current that flows under reverse bias conditions.

• A — TRUE: The electrostatic field inside a conductor is zero in the static situation.
• B — FALSE: The electric field at the surface is $E = \dfrac{\sigma}{\varepsilon_0}$, which clearly depends on surface charge density $\sigma$.
• C — TRUE: Any excess charge resides only on the surface; the interior has no excess charge in electrostatic equilibrium.
• D — TRUE: The electrostatic field at the surface must be perpendicular (normal) to the surface; a tangential component would cause charge to flow.
• E — FALSE: The electrostatic potential inside a conductor is constant (equal to the surface potential), but it is not necessarily zero.

Hence A, C and D are correct.

For photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function of the metal. If $\frac{hc}{\lambda} < W_0$, the photoelectric effect does not occur, which means $\lambda > \frac{hc}{W_0}$.

Calculating the threshold wavelength:

$\lambda_0 = \frac{hc}{W_0} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 1.6 \times 10^{-19}} = \frac{3 \times 10^{-7}}{1.6} = \frac{300}{1.6} \text{ nm} \approx 187.5 \text{ nm}$

Any wavelength greater than 187.5 nm will NOT produce the photoelectric effect. Among the options, only 200 nm > 187.5 nm, so option (3) is correct.

Solution

Note: This question has been marked (2*) indicating a contested/ambiguous answer. The official key awards (2), though technically the ray appears to diverge from the second principal focus (F₂) of a concave lens. The explanation below reflects the official reasoning.

For a concave (diverging) lens:

• F₂ (Second principal focus): The virtual image position for an object at infinity. A ray parallel to the principal axis, after refraction, appears to diverge from F₂ on the same side as the object.

Solution

• F₁ (First principal focus): The virtual object position for which the refracted ray emerges parallel to the principal axis.

A ray parallel to the principal axis, after passing through a concave lens, diverges such that its extension backward appears to come from the second principal focus. The best available option from the given choices is (2), as the paper intended "first principal focus" to mean the focus on the same side as the incoming ray.

Using the hydrostatic pressure formula:

$P = P_0 + \rho g h$

$100 \times 10^5 = 10^5 + 10^3 \times 10 \times h$

$10^7 = 10^5 + 10^4 h$

$10^3 = 10 + h$

$h = 1000 - 10 = \boxed{990 \text{ m}}$

The correct matching is:

• Microwave (A–IV): Produced by Klystron valve or magnetron valve.
• Visible light (B–I): Produced when electrons in atoms transition from a higher to a lower energy level.
• Gamma rays (C–II): Produced by radioactive decay of the nucleus.
• Infra-red rays (D–III): Produced by vibration of atoms and molecules.

The reagents that reduce nitriles ($\text{R-CN}$) to primary amines ($\text{R-CH}_2\text{-NH}_2$) are:

• A. LiAlH₄ / H₂O: $\text{R-CN} \xrightarrow{\text{(i) LiAlH}_4,\text{ (ii) H}_2\text{O}} \text{R-CH}_2\text{-NH}_2$ ✓
• C. H₂/Ni: $\text{R-CN} \xrightarrow{\text{H}_2/\text{Ni}} \text{R-CH}_2\text{-NH}_2$ ✓
• D. Na(Hg)/C₂H₅OH: $\text{R-CN} \xrightarrow{\text{Na(Hg)/C}_2\text{H}_5\text{OH}} \text{R-CH}_2\text{-NH}_2$ ✓

Reagents that do NOT give primary amines from nitriles:

• B. Sn + HCl: Reduces $\text{R-CN}$ to an aldehyde ($\text{R-CHO}$), not a primary amine.
• E. Br₂/aq. NaOH (Hofmann bromamide reaction): This works on amides, not nitriles, to give a primary amine with one less carbon.

• V₂O₅ (A–III): Catalyses the oxidation of SO₂ to SO₃ in the Contact Process for manufacture of H₂SO₄.
• Fe (B–I): Acts as catalyst in the Haber process for preparation of ammonia from N₂/H₂ mixture.
• PdCl₂ (C–IV): Catalyses the Wacker oxidation — oxidation of ethyne to ethanal.
• Ni complex (D–II): Catalyses polymerisation of alkynes.

For the reaction $2A(g) + B(g) \rightarrow 2D(g)$:

$\Delta n_g = 2 - (2+1) = -1$

First, calculate $\Delta H^\ominus$ from $\Delta U^\ominus$:

$\Delta H^\ominus = \Delta U^\ominus + \Delta n_g RT = -10 - \frac{1 \times 298 \times 8.31}{1000} = -10 - 2.48 = -12.48\ \text{kJ mol}^{-1}$

Then, calculate $\Delta G^\ominus$:

$\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus = -12.48 - \frac{298 \times (-44)}{1000}$

$= -12.48 + 13.112 = +0.632\ \text{kJ mol}^{-1}$

Since $\Delta G^\ominus > 0$, the reaction is non-spontaneous under standard conditions. Option (3) is closest to the calculated value.

The azimuthal quantum number $l$ identifies the subshell: $l=0 \rightarrow s$, $l=1 \rightarrow p$, $l=2 \rightarrow d$, $l=3 \rightarrow f$. The orbital name is formed by combining $n$ with the subshell letter:

• A: $n=2,\ l=1$ → 2p (II)
• B: $n=4,\ l=0$ → 4s (III)
• C: $n=5,\ l=3$ → 5f (IV)
• D: $n=3,\ l=2$ → 3d (I)

Hence: A–II, B–III, C–IV, D–I → Option (3).

Let the solubility be $s$. Since BiO(OH) is a pure solid, its activity is 1:

$K = [\text{BiO}^+][\text{OH}^-] = s \times s = s^2$

$s = \sqrt{K} = \sqrt{4 \times 10^{-10}} = 2 \times 10^{-5}\ \text{M}$

So $[\text{OH}^-] = 2 \times 10^{-5}$ M.

$[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{10^{-14}}{2 \times 10^{-5}} = \frac{1}{2} \times 10^{-9}$

$\text{pH} = -\log[\text{H}^+] = 9 + \log 2 = 9 + 0.3010 = \boxed{9.301}$

• DNA has a double-stranded helix secondary structure (Watson-Crick model) consisting of two polynucleotide chains wound around each other. Its four nitrogenous bases are Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).
• RNA is typically single-stranded and contains Uracil (U) instead of thymine. It does not have thymine.

Options (1), (3), and (4) are incorrect because they either assign the wrong base or the wrong strand structure to DNA/RNA.

Metamerism is a type of structural isomerism that arises due to different alkyl groups on either side of the same functional group in compounds belonging to the same homologous series.

• Option (1): These are position isomers (both alcohols, –OH at different positions), not metamers.
• Option (2): These are chain isomers (both alkanes with different carbon skeletons).
• Option (3): These are functional group isomers (a ketone and an aldehyde).
• Option (4): $\text{CH}_3\text{OCH}_2\text{CH}_2\text{CH}_3$ (methyl propyl ether) and $\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3$ (diethyl ether) — both are ethers with the same molecular formula but different alkyl groups on either side of the –O– group. These are metamers.

Solution

• [Pt(NH₃)₂Cl₂] (A–III): Exists as cis and trans forms → Geometrical isomerism.
• [Co(en)₃]³⁺ (B–I): The tris(ethylenediamine) cobalt(III) complex is non-superimposable on its mirror image → Optical isomerism.
• [Co(NH₃)₅NO₂]Cl₂ (C–IV): The nitrite ligand (NO₂⁻) can coordinate through N (nitro) or O (nitrito) → Linkage isomerism. Its linkage isomer is [Co(NH₃)₅(ONO)]Cl₂.
• [Cr(H₂O)₆]Cl₃ (D–II): Can exist as [Cr(H₂O)₅Cl]Cl₂·H₂O where a water molecule is replaced by Cl in the coordination sphere → Solvate (ionisation) isomerism.

The unit of rate constant $k$ for an $n^\text{th}$ order reaction is:

$\left(\frac{\text{mol}}{\text{L}}\right)^{1-n} \text{s}^{-1}$

• Zero order ($n=0$): $\text{mol L}^{-1}\ \text{s}^{-1}$ → (IV)
• First order ($n=1$): $\text{s}^{-1}$ → (III)
• Second order ($n=2$): $\text{mol}^{-1}\ \text{L}\ \text{s}^{-1}$ → (I)
• Third order ($n=3$): $\text{mol}^{-2}\ \text{L}^2\ \text{s}^{-1}$ → (II)

Hence: A–IV, B–III, C–I, D–II → Option (2).

Solution.

Numbering from the end that gives lowest locants to substituents:

Substituents are listed in alphabetical order (ethyl before methyl):

IUPAC name: 3-ethyl-5-methylheptane

Total energy consumed per second = $150\ \text{W} \times 1\ \text{s} = 150\ \text{J}$

Energy converted to light:

$E_\text{light} = \frac{150 \times 8}{100} = 12\ \text{J}$

Number of photons emitted:

$n = \frac{E_\text{light}}{E_\text{photon}} = \frac{12}{4.42 \times 10^{-19}} \approx 2.715 \times 10^{19}$

Hence, approximately $\mathbf{2.71 \times 10^{19}}$ photons are emitted per second.

The steam reforming reaction of methane:

$\text{CH}_4(g) + \text{H}_2\text{O}(g) \xrightarrow{\text{Ni},\ 1273\text{ K}} \text{CO}(g) + 3\text{H}_2(g)$

This reaction is used industrially for the preparation of dihydrogen (syngas). The products are carbon monoxide and hydrogen gas.

Degree of unsaturation $= \frac{2(8)+2-8}{2} = \frac{10}{2} = 5$

Five degrees of unsaturation suggest an aromatic ring (4 degrees) plus one additional degree (C=O).

Solution.