Re-NEET 2026, on the exam screen.

The condition $\mu_k(L) = 0$ fixes $\alpha = \mu_0/L$, so $\mu_k(x) = \mu_0\left(1 - \frac{x}{L}\right)$ falls linearly from $\mu_0$ to zero. Because the force varies with position, the work must be integrated: $|W| = \int_0^L \mu_k(x) Mg\,dx = \mu_0 Mg\left[L - \frac{L}{2}\right] = \frac{1}{2}\mu_0 MgL$. Comparing with $n\mu_0 MgL$ gives $n = 1/2$, which is simply the average of the initial and final friction coefficients since the variation is linear. NCERT Class 11, Work, Energy and Power (work done by a variable force).

Mean free path is $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$, so $\lambda \propto \frac{1}{d^2 n}$. Taking the ratio, $\frac{\lambda_A}{\lambda_B} = \frac{d_B^2 n_B}{d_A^2 n_A}$. Substituting $\lambda_A/\lambda_B = \frac{1}{2}$ and $d_A = 2d_B$ gives $\frac{1}{2} = \frac{n_B}{4n_A}$, hence $n_A = \frac{1}{2}n_B$. NCERT Class 11, Kinetic Theory (mean free path).

In configuration $P$ the two inductors are in series and the segment between $a$ and $b$ spans just ONE of them, carrying the full current, so $E_P = L\frac{dI}{dt}$. In configuration $Q$ the points $a$ and $b$ straddle the parallel pair, whose equivalent inductance is $\frac{L}{2}$, so $E_Q = \frac{L}{2}\frac{dI}{dt}$. The ratio is therefore 2. NCERT Class 12, Electromagnetic Induction (self-inductance).

Why the other options are wrong:

- (1) takes $E_P$ across the series pair as $2L$ and inverts the ratio
- (2) inverts the correct answer by swapping which configuration sits in the numerator
- (3) assumes identical inductors must produce identical electromotive forces (emfs) regardless of how $a$ and $b$ are placed

An open-open pipe has antinodes at both ends, and its $p$-th harmonic contains exactly $p$ nodes, so the 5th harmonic gives $n = 5$. A closed pipe supports only odd harmonics, with a node at the closed end and an antinode at the open end; its $q$-th harmonic ($q$ odd) contains $\frac{q+1}{2}$ nodes, so the 9th harmonic gives $m = \frac{10}{2} = 5$. The ratio is therefore 1. NCERT Class 11, Waves (standing waves in air columns).

For a long solenoid $L = \mu_0 n^2 A l$, where $n$ is turns per unit length and $A$ the cross-sectional area. Substituting $A = \pi r^2$ gives $L = \mu_0\pi n^2 r^2 l$. NCERT Class 12, Electromagnetic Induction (self-inductance).

Average speed uses total PATH LENGTH, not displacement, so the turning point must be located first. Here $v = \frac{ds}{dt} = 2t - 6$, which vanishes at $t = 3$ s. Positions are $s(0) = 5\,\text{m}$, $s(3) = -4\,\text{m}$, $s(6) = 5\,\text{m}$, giving 9 m travelled in each leg and 18 m in total. Hence average speed $= 18/6 = 3\ \text{ms}^{-1}$. NCERT Class 11, Motion in a Straight Line.

The Q value is the mass defect converted to energy: $\Delta m = 238.050 - (234.043 + 4.003) = 0.004\ \text{u}$. Multiplying by the conversion factor, $Q = 0.004 \times 931.5 = 3.726\ \text{MeV} = 3726\ \text{keV}$. The positive value confirms the decay is spontaneous. NCERT Class 12, Nuclei (mass-energy and nuclear reactions).

Photocurrent is set by the NUMBER of photons striking the surface per second, so it is proportional to the source power and falls linearly with $t$. Stopping potential, by contrast, depends only on the maximum kinetic energy of the emitted electrons, $eV_0 = h\nu - \phi_0$, which is fixed by the FREQUENCY and the work function - neither of which changes here. So $|V|$ stays flat while $I$ decays. NCERT Class 12, Dual Nature of Radiation and Matter.

Why the other options are wrong:

- (2) treats stopping potential as intensity-dependent, the single most common photoelectric misconception
- (3) swaps the roles entirely, tying current to frequency and stopping potential to intensity
- (4) assumes neither quantity responds to the source, ignoring that current tracks photon flux

$P$ and $Q$ form a series branch of equivalent capacitance $C/2$, and that branch sits in parallel with $S$ across the battery, so $C_{eq} = \frac{C}{2} + C = \frac{3C}{2}$ and $U_T = \frac{1}{2}\left(\frac{3C}{2}\right)V^2 = \frac{3}{4}CV^2$. Identical capacitors in series split the voltage equally, so $P$ holds $V/2$ and $U_P = \frac{1}{2}C\left(\frac{V}{2}\right)^2 = \frac{CV^2}{8}$. The ratio is $\frac{CV^2/8}{3CV^2/4} = \frac{1}{6}$. NCERT Class 12, Electrostatic Potential and Capacitance.

Why the other options are wrong:

- (1) treats all three capacitors as being in parallel, so each stores an equal third and $P$ plus $Q$ gives 2/3
- (2) assumes the three capacitors share the energy equally, ignoring that $P$ sees only half the battery voltage
- (3) computes the ratio of $U_P$ to the energy of the series branch alone, forgetting to include $S$ in the total

Continuity gives $A_P v_P = A_Q v_Q$, so halving the area doubles the speed: $v_Q = 2v_P$. For a horizontal pipe Bernoulli's equation reduces to $\Delta P = \frac{1}{2}\rho(v_Q^2 - v_P^2) = \frac{3}{2}\rho v_P^2$. Substituting $15 = 1500\,v_P^2$ gives $v_P = 0.1\ \text{ms}^{-1} = 10\ \text{cm s}^{-1}$, and the volume flow rate is $A_P v_P = 40 \times 10 = 400\ \text{cm}^3\text{s}^{-1}$. NCERT Class 11, Mechanical Properties of Fluids.

Why the other options are wrong:

- (1) uses the narrow area $A_Q$ with the wide-section speed $v_P$, mismatching the pair
- (2) drops the factor 3 from $v_Q^2 - v_P^2$, effectively using $\Delta P = \frac{1}{2}\rho v_Q^2$ alone
- (3) mixes SI and centimetre-gram-second (CGS) units partway through, most often by leaving $\rho$ in $\text{kg m}^{-3}$ alongside areas in $\text{cm}^2$

The current entering at $A$ divides between the two arcs in INVERSE proportion to their resistances, so with $R_{ABC} = \frac{1}{2}R_{ADC}$ the currents are $\frac{2}{3}I_0$ and $\frac{1}{3}I_0$. Each arc is a semicircle ($A$ at the top, $C$ at the bottom), contributing $\frac{\mu_0 I}{4r}$ but in OPPOSITE directions at $O$. The net field is $\frac{\mu_0}{4r}\left(\frac{2}{3}I_0 - \frac{1}{3}I_0\right) = \frac{\mu_0 I_0}{12r}$. NCERT Class 12, Moving Charges and Magnetism (Biot-Savart law).

Why the other options are wrong:

- (2) sends the whole current $I_0$ round one semicircle and ignores the opposing contribution from the other arc
- (3) uses the full-loop result $\frac{\mu_0 I}{2r}$ without splitting the current or accounting for cancellation
- (4) applies the formula for a long straight wire, $\frac{\mu_0 I}{2\pi r}$, instead of an arc

Stokes' law gives $v = \frac{2r^2 g}{9\eta}(\sigma - \rho)$, which can be rewritten as $v = \frac{2r^2 g\rho}{9\eta}\left(\frac{\sigma}{\rho} - 1\right)$. With $r$ fixed this is a STRAIGHT LINE in $\sigma/\rho$ whose intercept on the horizontal axis is exactly at $\sigma/\rho = 1$. Below that value the ball is lighter than the liquid and does not sink at all, so the downward terminal velocity stays zero. NCERT Class 11, Mechanical Properties of Fluids (Stokes' law).

Why the other options are wrong:

- (2) takes $v \propto \sigma$ and lets the line pass through the origin, ignoring the buoyant term $-\rho$
- (3) places the zero-velocity crossing at some arbitrary density ratio rather than at the neutral-buoyancy point $\sigma = \rho$
- (4) treats terminal velocity as a fixed property of the ball once it sinks, missing its linear dependence on excess density

Escape speed is $v_e = \sqrt{\frac{2GM}{R}}$. With the masses equal, $v_e \propto R^{-1/2}$, so $\frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} = \sqrt{2}$. The denser, smaller planet is harder to escape from, which is the physical sense check. NCERT Class 11, Gravitation (escape speed).

Kepler's third law states $T^2 \propto R^3$ for orbits about a common central mass, so $T \propto R^{3/2}$. It also follows directly from equating gravitational and centripetal force: $\frac{GMm}{R^2} = m\omega^2 R$ with $\omega = 2\pi/T$. NCERT Class 11, Gravitation (Kepler's laws).

Parallel currents in the SAME direction attract, so wire $B$ pulls wire $A$ upwards with force per unit length $\frac{\mu_0 (I)(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}$. Wire $A$ stays on the floor as long as this does not exceed its weight per unit length $\lambda g$, so $\frac{\mu_0 I^2}{\pi h} \le \lambda g$ and $h_{min} = \frac{\mu_0 I^2}{\pi\lambda g}$. NCERT Class 12, Moving Charges and Magnetism (force between two parallel currents).

Once an ideal Zener diode is driven past its breakdown voltage it CLAMPS the voltage across itself at 3 V and behaves like a constant-voltage source. The remaining $5 - 3 = 2\ \text{V}$ of the supply must therefore appear across the series resistor, which is the element between $B$ and $A$. NCERT Class 12, Semiconductor Electronics (Zener diode as a voltage regulator).

Why the other options are wrong:

- (1) reports the voltage across the ZENER rather than across the resistor between $B$ and $A$
- (3) subtracts twice, or assumes the diode drops an extra 1 V forward-bias contribution
- (4) assumes the reverse-biased diode blocks all current, so no drop appears across the resistor - but breakdown means current does flow

In an adiabatic process $Q = 0$, so the work done by the gas comes entirely out of its internal energy: $W = -\Delta U = nC_V(T_1 - T_2)$. For a monatomic gas $C_V = \frac{3}{2}R$, giving $W = 1 \times \frac{3}{2}(8.3)(10) = 124.5\ \text{J}$. The same follows from $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$ with $\gamma - 1 = \frac{2}{3}$. NCERT Class 11, Thermodynamics.

Emission occurs only when the incident wavelength is SHORTER than the threshold wavelength. Cell 1 has $\lambda_1 < \lambda$, so the photons are too weak and $V_1 = 0$. For the other two, $eV_0 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_{th}}\right)$; a LARGER threshold wavelength means a smaller work function and hence a larger stopping potential. Since $\lambda_3 \gg \lambda_2$, the term $1/\lambda_3$ is nearly zero and $V_3 > V_2$. NCERT Class 12, Dual Nature of Radiation and Matter (Einstein's photoelectric equation).

A photon's momentum is $p = \frac{E}{c} = \frac{20 \times 1.6 \times 10^{-19}}{3 \times 10^8} \approx 1.07 \times 10^{-26}\ \text{kg m s}^{-1}$. A non-relativistic electron of the same kinetic energy has $p = \sqrt{2mE} = \sqrt{2(9 \times 10^{-31})(3.2 \times 10^{-18})} = 2.4 \times 10^{-24}\ \text{kg m s}^{-1}$. The ratio is therefore about 225 - a massive particle carries far more momentum than a photon of the same energy, because $p \propto \sqrt{E}$ for matter but $p \propto E$ for light. NCERT Class 12, Dual Nature of Radiation and Matter.

On an optical bench the positions read on the scale are those of the UPRIGHTS, not of the optical centres of the lens, mirror or object pin. The fixed offset between an upright's index mark and the actual optical element is removed by the index (or bench) correction, applied to every measured object and image distance before computing the focal length. NCERT Class 12 practical work, Ray Optics and Optical Instruments.

Work done in moving a charge slowly is path-independent and depends only on the endpoint potentials, so the detour through the tube is a red herring. Both $A$ and $B$ lie OUTSIDE the sphere, where $V(r) = \frac{Q}{4\pi\varepsilon_0 r}$ with $Q = \frac{4}{3}\pi R^3\rho$. Then $|W| = |V_B - V_A| = \frac{Q}{4\pi\varepsilon_0 R}\left|\frac{1}{3} - \frac{1}{2}\right| = \frac{\rho R^2}{18\varepsilon_0}$, giving $n = 18$. NCERT Class 12, Electrostatic Potential and Capacitance.

Why the other options are wrong:

- (1) uses only the $\frac{1}{2}$ from position $A$ and drops the $\frac{4}{3}\pi R^3$ volume factor
- (2) computes $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ but forgets the $\frac{1}{3}$ carried by the sphere's total charge
- (3) integrates the internal field along the tube as if the endpoints were inside the sphere

For near-normal viewing across a plane boundary, apparent depth $=$ real depth $\times \frac{n_{\text{observer}}}{n_{\text{object}}}$, and the object sits 2.5 cm from each face. Viewed from $P$: $h_1 = 2.5 \times \frac{1}{1.25} = 2\ \text{cm}$. Viewed from $R$: $h_2 = 2.5 \times \frac{1.5}{1.25} = 3\ \text{cm}$. Hence $|h_1 - h_2| = 1\ \text{cm}$. Looking in from the DENSER side makes the object appear farther away, which is the sense check here. NCERT Class 12, Ray Optics and Optical Instruments (refraction at plane surfaces).

Why the other options are wrong:

- (1) assumes the symmetric placement of the object makes the two apparent depths equal, ignoring the different observer media
- (3) uses the full 5 cm thickness instead of the 2.5 cm from the object to each face
- (4) computes $h_2$ alone and reports it as the difference

By symmetry the two particles sit at mirror-image angles about the vertical, so their tangential velocities have equal and opposite horizontal components while their vertical components add. Writing the particles at $\theta$ and $\pi - \theta$, the resultant works out to $P = 2m\omega r|\cos\theta|$, which is zero at the start ($\theta = \pi/2$), peaks when the particles are diametrically opposite the meeting point ($\theta = 0$), and returns to zero as they collide at $B$. That single smooth hump is option (3). NCERT Class 11, System of Particles and Rotational Motion (motion of the centre of mass).

Why the other options are wrong:

- (1) assumes momentum is conserved as a constant because the SPEEDS are constant, forgetting that momentum is a vector whose direction keeps changing
- (2) inverts the phase - takes $P \propto |\sin\theta|$, i.e. maximum where the particles start rather than where they are farthest apart
- (4) assumes $P$ falls off linearly with angle instead of following the cosine of the half-separation

For small expansions the coefficient of volume expansion is $\gamma = 3\alpha$. Hence $\Delta V = V\gamma\Delta T = \frac{4}{3}\pi R^3 \times 3\alpha \times \Delta T = 4\pi R^3\alpha\Delta T$. Equivalently, differentiating $V = \frac{4}{3}\pi R^3$ gives $\Delta V = 4\pi R^2 \Delta R$ with $\Delta R = R\alpha\Delta T$. NCERT Class 11, Thermal Properties of Matter.

Depressing the cork by $x$ creates an extra buoyant force $A\rho g x$ directed back towards equilibrium, so the effective spring constant is $k = A\rho g$ and $T = 2\pi\sqrt{\frac{m}{A\rho g}}$. With $m$ and $A$ unchanged, $T \propto \rho^{-1/2}$. Doubling the period therefore requires the density to fall by a factor of four: $\frac{\rho_2}{\rho_1} = \frac{1}{4}$. NCERT Class 11, Oscillations (simple harmonic motion (SHM) examples).

NTA dropped this question because no printed option matched the correct answer; option (4) is corrected here to 1.06 cm. Working: least count $= \frac{1\,\text{MSD}}{N} = \frac{1\,\text{mm}}{10} = 0.1\,\text{mm}$. The Vernier zero lies to the left of the main-scale zero, so the zero error is negative. With the 4th Vernier division coinciding, zero error $= -(N-n)\times LC = -(10-4)(0.1) = -0.6\,\text{mm}$. Therefore true length $= \text{observed reading} - \text{zero error} = 10.0 - (-0.6) = 10.6\,\text{mm} = 1.06\,\text{cm}$. NCERT Class 11, Units and Measurements (least count and instrumental error).

The spheres touch, so their centres are separated by $d = R + r$. About the axis through $A$: $I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m d^2$; about the axis through $B$: $I_B = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + M d^2$. The two spin terms are common to both and cancel, leaving $I_A - I_B = (m - M)(R + r)^2$, a negative quantity as expected since the heavier sphere sits off-axis in the second case. NCERT Class 11, System of Particles and Rotational Motion (parallel axis theorem).

Why the other options are wrong:

- (1) gets the separation right but flips the sign - attaches the parallel-axis term to the wrong body in each case
- (3) uses $R - r$ as the centre separation, as if the small sphere were embedded in the large one rather than attached externally
- (4) assumes the moment of inertia is independent of the choice of axis, ignoring the parallel axis theorem altogether

Conservation of energy gives $\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = E$, which rearranges to $\frac{v^2}{2E/m} + \frac{x^2}{2E/k} = 1$ - an ellipse in the $x\!-\!v$ plane. An ellipse becomes a CIRCLE only when the two semi-axes are equal, i.e. $\frac{2E}{m} = \frac{2E}{k}$, which requires $k = m$ numerically. NCERT Class 11, Oscillations (energy in simple harmonic motion).

For the convex lens, $u = -30\ \text{cm}$ and $f = +10\ \text{cm}$ give $\frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}$, so its image lies 15 cm to its right. The lenses are 3 cm apart, so that image sits 12 cm BEYOND the concave lens and acts as a VIRTUAL OBJECT with $u = +12\ \text{cm}$. Then $\frac{1}{v} = -\frac{1}{10} + \frac{1}{12} = -\frac{1}{60}$, giving $v = -60\ \text{cm}$: 60 cm to the left of the concave lens. NCERT Class 12, Ray Optics and Optical Instruments (combination of thin lenses).

Why the other options are wrong:

- (1) uses the thin-lens-in-contact formula $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$, which fails for a 3 cm separation
- (3) keeps the sign of $v$ positive, missing that a negative $v$ places a virtual image on the incoming side
- (4) gets the magnitude right but assigns the image to the wrong side of the concave lens

With $\omega = 2 \times 10^3\ \text{rad s}^{-1}$, the reactances are $X_L = \omega L = 20\ \Omega$ and $X_C = \frac{1}{\omega C} = 20\ \Omega$. Since $X_L = X_C$ the circuit is at RESONANCE, the reactances cancel and the impedance reduces to $Z = R = 100\ \Omega$. The current amplitude is then $I_0 = \frac{V_0}{Z} = \frac{220}{100} = 2.2\ \text{A}$. NCERT Class 12, Alternating Current (LCR series circuit and resonance).

The angular momentum of a rigid body about any point splits as $\vec{L} = \vec{L}_{cm} + \vec{r} \times M\vec{v}_{cm}$. Here the centre is FIXED, so $\vec{v}_{cm} = 0$ and the orbital term vanishes for every reference point. Only the spin term $I_{cm}\omega$ survives, which is the same whether you compute about $A$ or $B$, so the ratio is 1 regardless of the distances. NCERT Class 11, System of Particles and Rotational Motion (angular momentum of a rotating rigid body).

Why the other options are wrong:

- (1) assumes $L \propto 1/r^2$, importing an inverse-square intuition that does not apply
- (2) assumes $L$ scales inversely with the distance to the reference point
- (4) assumes $L$ grows in proportion to the distance, as it would for a single particle moving in a straight line

Faraday's law gives $\varepsilon = -\frac{d\Phi}{dt} = -BA_0\cos t$, so the dissipated power is $P = \frac{\varepsilon^2}{R} = \frac{B^2A_0^2}{R}\cos^2 t$. Because $\cos^2 t$ is MAXIMUM at $t = 0$ and touches zero at $t = \pi/2$, the correct graph starts at a peak and dips to zero, unlike a $\sin^2 t$ curve which would start at zero. $P$ is never negative because the electromotive force (emf) is squared. NCERT Class 12, Electromagnetic Induction (Faraday's law).

Why the other options are wrong:

- (1) differentiates $\sin t$ to $\sin t$ rather than $\cos t$, producing a $\sin^2 t$ curve that starts at zero
- (3) assumes the constant field means a constant emf, forgetting that it is the CHANGING area that drives induction
- (4) confuses the growing area with growing power, ignoring that the flux change is periodic and bounded

Point $A$ lies in the empty cavity, not in the conducting material, so the field of $Q$ is fully present there and $E_A \neq 0$. The induced $-Q$ on the cavity wall and $+Q$ on the outer surface arrange themselves so that the outer surface charge is UNIFORM regardless of where the cavity sits, so outside the sphere the field is exactly that of a point charge $Q$ at the centre. Since $B$ and $C$ are equidistant from that centre, $E_B = E_C$. NCERT Class 12, Electric Charges and Fields (electrostatics of conductors).

Why the other options are wrong:

- (1) confuses the cavity with the conducting material - the field vanishes inside the METAL, not inside the hollow
- (3) combines the cavity error with the idea that the off-centre cavity biases the external field towards $B$
- (4) correctly sees $E_A \neq 0$ but wrongly assumes the external field is skewed away from the cavity side

Outside the sphere the potential energy of the attracting pair is $U(r) = -\frac{Qq}{4\pi\varepsilon_0 r}$. Energy conservation from $r = 3R$ (at rest) to $r = R$ gives $\frac{1}{2}mv^2 = \frac{Qq}{4\pi\varepsilon_0 R}\left(1 - \frac{1}{3}\right) = \frac{Qq}{6\pi\varepsilon_0 R}$. Solving, $v = \sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$. NCERT Class 12, Electrostatic Potential and Capacitance.

Rutherford's scattering formula gives $N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$, so the count is enormous for small deflections and drops by several orders of magnitude as $\theta$ increases - which is exactly why the plots are drawn on a logarithmic vertical axis. The experimental finding was that the overwhelming majority of $\alpha$-particles pass almost undeviated, with only a tiny fraction turned through large angles by the nucleus. NCERT Class 12, Atoms (alpha-particle scattering).

Why the other options are wrong:

- (1) reverses the physics, implying most particles suffer large deflections and the atom is uniformly dense
- (2) captures the falling trend but far too weakly - it misses that the drop spans many orders of magnitude
- (4) invents a preferred scattering angle, as if the nucleus deflected particles by a characteristic amount

In circuit (A) the resistors are in SERIES, so $R_A = R_0(1-\alpha\Delta T) + R_0(1+\alpha\Delta T) = 2R_0$ - the two temperature terms cancel exactly and $I_A$ is unchanged. In circuit (B) they are in PARALLEL, giving $R_B = \frac{R_0(1-\alpha^2\Delta T^2)}{2}$, which decreases as $\Delta T$ grows because of the $-\alpha^2\Delta T^2$ term, so $I_B$ increases. NCERT Class 12, Current Electricity (temperature dependence of resistivity).

Why the other options are wrong:

- (2) assumes the series combination must heat up and resist more, missing the exact cancellation between $+\alpha$ and $-\alpha$
- (3) swaps the two circuits, applying the parallel result to (A) and the series result to (B)
- (4) spots the cancellation in series and wrongly assumes the same cancellation holds for reciprocals in parallel

Take each constant from its defining equation. Stefan's law $E = \sigma_s T^4$ makes $\sigma_s$ a power per unit area per $K^4$, so $[\sigma_s] = [MT^{-3}K^{-4}]$. Boltzmann's constant is an energy per kelvin, $[k_B] = [ML^2T^{-2}K^{-1}]$, and Wien's constant is a wavelength times a temperature, $[b] = [LK]$. Multiplying, $[MT^{-3}K^{-4}][M^{-1}L^{-2}T^{2}K][LK] = [L^{-1}T^{-1}K^{-2}]$. NCERT Class 11, Units and Measurements; the constants themselves come from Thermal Properties of Matter.