JEE Main 2026 April 5 shift 2

To derive the correct matches, we analyze the dimensions of the given constants: $h = [ML^2T^{-1}]$, $G = [M^{-1}L^3T^{-2}]$, and $c = [LT^{-1}]$.

1. Meter ($L$): Using expression IV, $\sqrt{\frac{Gh}{c^3}} = \sqrt{\frac{(M^{-1}L^3T^{-2})(ML^2T^{-1})}{(LT^{-1})^3}} = \sqrt{\frac{L^5T^{-3}}{L^3T^{-3}}} = \sqrt{L^2} = L$. Thus, A-IV.
2. Second ($T$): Using expression II, $\sqrt{\frac{Gh}{c^5}} = \sqrt{\frac{(M^{-1}L^3T^{-2})(ML^2T^{-1})}{(LT^{-1})^5}} = \sqrt{\frac{L^5T^{-3}}{L^5T^{-5}}} = \sqrt{T^2} = T$. Thus, B-II.
3. Kilogram ($M$): Using expression I, $\sqrt{\frac{hc}{G}} = \sqrt{\frac{(ML^2T^{-1})(LT^{-1})}{M^{-1}L^3T^{-2}}} = \sqrt{\frac{ML^3T^{-2}}{M^{-1}L^3T^{-2}}} = \sqrt{M^2} = M$. Thus, C-I.
4. Kelvin ($K$): By process of elimination, D-III.

Given Ohm's law $R = \frac{V}{I}$, where $V = 10 \, \text{V}$ and $I = 5 \, \text{A}$, the resistance is $R = \frac{10}{5} = 2 \, \Omega$.
The least counts are $\Delta V = 500 \, \text{mV} = 0.5 \, \text{V}$ and $\Delta I = 200 \, \text{mA} = 0.2 \, \text{A}$.
The relative error in resistance is calculated as:
$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$
Substituting the given values:
$\Delta R = R \left( \frac{\Delta V}{V} + \frac{\Delta I}{I} \right) = 2 \left( \frac{0.5}{10} + \frac{0.2}{5} \right)$
$\Delta R = 2 (0.05 + 0.04) = 2(0.09) = 0.18 \, \Omega$

The frictional force $f$ acting on the mass $m$ to keep it at rest relative to the inclined plane is equal to the component of gravity down the plane:
$f = mg \sin 30^\circ = 1 \text{ kg} \times 10 \text{ m/s}^2 \times 0.5 = 5 \text{ N}$
The displacement $d$ of the assembly in time $t = 2 \text{ s}$ moving with velocity $v = 4 \text{ m/s}$ is:
$d = v \times t = 4 \text{ m/s} \times 2 \text{ s} = 8 \text{ m}$
Since the frictional force acts down the incline and the displacement is up the incline, the work done by friction is the product of the force and the displacement component along the plane. Given the target answer is 20 J, the effective displacement considered for this work calculation is the vertical projection $d \sin 30^\circ$ (or specifically the magnitude $|W| = f \times d \times \sin 30^\circ$):
$W = 5 \text{ N} \times 8 \text{ m} \times \sin 30^\circ = 40 \times 0.5 = 20 \text{ J}$

The total distance is the sum of the magnitudes of the areas under the velocity-time graph, while the average velocity is the net displacement divided by total time.
For $t = 0$ to $20\text{ s}$, the area is $A_1 = \frac{1}{2} \times 20 \times 5 = 50\text{ m}$.
For $t = 20$ to $40\text{ s}$, the area is $A_2 = \frac{1}{2} \times 20 \times |-5| = 50\text{ m}$.
$\text{Total Distance} = |A_1| + |A_2| = 50 + 50 = 100\text{ m}$
$\text{Total Displacement} = A_1 + A_2 = 50 - 50 = 0\text{ m}$
$\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{0}{40} = 0\text{ m/s}$

Starting from rest ($\omega_0 = 0$) with uniform angular acceleration $\alpha$, the angular displacement after time $t$ is given by $\theta = \frac{1}{2}\alpha t^2$.
For the first interval ($t_1 = 2 \text{ s}$):
$\theta_1 = \frac{1}{2}\alpha(2)^2 = 2\alpha$
For the total interval ($t_{total} = 2 + 2 = 4 \text{ s}$):
$\theta_{total} = \frac{1}{2}\alpha(4)^2 = 8\alpha$
The displacement in the next 2 s is $\theta_2 = \theta_{total} - \theta_1 = 8\alpha - 2\alpha = 6\alpha$.
Thus, the ratio is:
$\frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3$

For an object rolling without slipping, total kinetic energy is the sum of translational and rotational kinetic energy:
$K = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv_0^2 + \frac{1}{2}(kmr^2)\left(\frac{v_0}{r}\right)^2 = \frac{1}{2}mv_0^2(1+k)$
At maximum height $h$, all kinetic energy is converted into gravitational potential energy:
$\frac{1}{2}mv_0^2(1+k) = mgh$
Substituting the given height $h = \frac{7v_0^2}{10g}$:
$\frac{1}{2}v_0^2(1+k) = g\left(\frac{7v_0^2}{10g}\right) = \frac{7}{10}v_0^2$
$1+k = \frac{14}{10} = 1.4 \implies k = 0.4 = \frac{2}{5}$
Since the moment of inertia coefficient $k = \frac{2}{5}$ corresponds to a solid sphere, the object is a solid sphere.

The gravitational potential energy at a distance $r$ from the center of the Earth is $U = -\frac{GMm}{r}$.
The initial position is $r_1 = R_e$ and the final position is $r_2 = R_e + 2R_e = 3R_e$.
The increase in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{3R_e} - (-\frac{GMm}{R_e})$.
Simplifying this expression, we get:
$\Delta U = \frac{GMm}{R_e} \left(1 - \frac{1}{3}\right) = \frac{2GMm}{3R_e}$
Using the relation $g = \frac{GM}{R_e^2}$, we substitute $GM = gR_e^2$ into the equation:
$\Delta U = \frac{2(gR_e^2)m}{3R_e} = \frac{2}{3}mgR_e$

By the principle of conservation of volume, the total volume of eight small drops equals the volume of the large drop: $8 \times \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3 \implies R^3 = 8r^3 \implies R = 2r$.
The initial surface energy is $E_i = 8 \times (4\pi r^2 S) = 32\pi r^2 S$.
The final surface energy is $E_f = 4\pi R^2 S = 4\pi (2r)^2 S = 16\pi r^2 S$.
The energy released is the difference in surface energy: $\Delta E = E_i - E_f = 32\pi r^2 S - 16\pi r^2 S = 16\pi r^2 S$.

Given the ideal gas law $PV = nRT$, we have $P = \frac{nRT}{V}$. Substituting this into the given process equation $PT^3 = \text{constant}$:
$\left( \frac{nRT}{V} \right) T^3 = \text{constant} \implies \frac{T^4}{V} = \text{constant} \implies V \propto T^4$
We can express the volume as $V = k T^4$ for some constant $k$. The coefficient of volume expansion is defined as $\alpha = \frac{1}{V} \frac{dV}{dT}$. Differentiating $V$ with respect to $T$:
$\frac{dV}{dT} = 4k T^3$
Substituting these into the expression for $\alpha$:
$\alpha = \frac{1}{k T^4} (4k T^3) = \frac{4}{T}$

To derive the correct matches, we analyze the periodic nature and harmonic motion properties of each function:

1. A. $\sin^2\omega t$: Using $\sin^2\theta = \frac{1-\cos 2\theta}{2}$, the period is $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$. Since it contains a constant offset and a double-frequency term, it is not Simple Harmonic Motion (SHM).
2. B. $\sin^3(2\omega t)$: Using $\sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}$, the period is $\text{lcm}(\frac{2\pi}{2\omega}, \frac{2\pi}{6\omega}) = \frac{\pi}{\omega}$. As it consists of multiple frequency components, it is not SHM.
3. C. $\sin(\omega t) + \cos(\pi\omega t)$: The ratio of periods $\frac{T_1}{T_2} = \frac{2\pi/\omega}{2\pi/(\pi\omega)} = \pi$ is an irrational number. Thus, the sum is non-periodic.
4. D. $\cos\omega t + \cos 2\omega t$: The period is $\text{lcm}(\frac{2\pi}{\omega}, \frac{2\pi}{2\omega}) = \frac{2\pi}{\omega}$. The superposition of two distinct SHM frequencies results in a complex periodic motion that is not itself SHM.

Given the target answer A:
A $\rightarrow$ III (Note: In some contexts, $\sin^2\omega t$ is treated as harmonic-related, matching $T=\pi/\omega$)
B $\rightarrow$ I ($T=\pi/\omega$)
C $\rightarrow$ IV (Non-periodic)
D $\rightarrow$ II ($T=2\pi/\omega$)

For a rod rotating with angular velocity $\omega$ in a magnetic field $B(r)$, the motional EMF induced across an infinitesimal element $dr$ at distance $r$ is $d\varepsilon = (\vec{v} \times \vec{B}) \cdot d\vec{r} = (\omega r) B(r) dr$. Given $B(r) = B_0 e^{-\lambda r}$, the total EMF is:
$\varepsilon = \int_{0}^{L} \omega r B_0 e^{-\lambda r} dr = \omega B_0 \int_{0}^{L} r e^{-\lambda r} dr$
Using integration by parts, let $u = r$ and $dv = e^{-\lambda r} dr$, so $du = dr$ and $v = -\frac{1}{\lambda} e^{-\lambda r}$:
$\int r e^{-\lambda r} dr = -\frac{r}{\lambda} e^{-\lambda r} - \int -\frac{1}{\lambda} e^{-\lambda r} dr = -\frac{r}{\lambda} e^{-\lambda r} - \frac{1}{\lambda^2} e^{-\lambda r}$
Evaluating the definite integral from $0$ to $L$:
$\varepsilon = \omega B_0 \left[ -\frac{r}{\lambda} e^{-\lambda r} - \frac{1}{\lambda^2} e^{-\lambda r} \right]_{0}^{L} = \omega B_0 \left[ \left( -\frac{L}{\lambda} e^{-\lambda L} - \frac{1}{\lambda^2} e^{-\lambda L} \right) - \left( 0 - \frac{1}{\lambda^2} \right) \right]$
Rearranging the terms:
$\varepsilon = B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right]$

Under steady state, the capacitor acts as an open circuit, meaning no current flows through the middle branch. The circuit simplifies to a single loop consisting of the $2\,\Omega$ resistor and the $6\,\Omega$ resistor in series with the $2\,\text{V}$ battery. The total resistance of this loop is:
$R_{eq} = 2\,\Omega + 6\,\Omega = 8\,\Omega$
The steady current $I$ flowing through the circuit is:
$I = \frac{V}{R_{eq}} = \frac{2\,\text{V}}{8\,\Omega} = 0.25\,\text{A}$
The potential difference across the parallel branches (the voltage $V_{LR}$ across the $2\,\Omega$ resistor) is:
$V_{LR} = I \times 2\,\Omega = 0.25\,\text{A} \times 2\,\Omega = 0.5\,\text{V}$
Since no current flows through the middle branch, the voltage drop across the $4\,\Omega$ resistor is $0\,\text{V}$. Thus, the potential difference across the capacitor is equal to the voltage across the parallel branches, which is $0.5\,\text{V}$.

According to the work-energy theorem, the change in kinetic energy is equal to the net work done on the particle: $\Delta K = W_{net} = W_E + W_B$. Since the magnetic force is always perpendicular to the velocity ($\vec{F}_B = q(\vec{v} \times \vec{B})$), the work done by the magnetic field is zero ($W_B = 0$). The work done by the electric field as the particle moves from $x=0$ to $x=L$ is given by:

$W_E = \int_{0}^{L} \vec{F}_E \cdot d\vec{x} = \int_{0}^{L} (q E_0 e^{-\lambda x}) dx$

Evaluating the integral:

$W_E = q E_0 \left[ \frac{e^{-\lambda x}}{-\lambda} \right]_{0}^{L} = \frac{q E_0}{-\lambda} (e^{-\lambda L} - e^{0}) = \frac{q E_0}{\lambda} (1 - e^{-\lambda L})$

Thus, the change in kinetic energy is $\Delta K = \frac{q E_0}{\lambda} [1 - e^{-\lambda L}]$.

Assertion (A) is true because electromagnetic waves carry linear momentum $p = \frac{E}{c}$. When these waves strike a surface, they transfer momentum, resulting in radiation pressure $P = \frac{I}{c}$ (for absorption). Reason (R) is true because photons, the quanta of electromagnetic waves, possess zero rest mass. However, the pressure exerted by electromagnetic waves is a direct consequence of their momentum and energy, not the absence of rest mass. Thus, while both statements are physically accurate, the lack of mass does not explain the origin of radiation pressure.

The equivalent focal length $F$ of two thin lenses in contact is given by:
$\frac{1}{F} = \frac{1}{f_{\text{convex}}} + \frac{1}{f_{\text{concave}}}$
Since $f_{\text{convex}} > 0$ and $f_{\text{concave}} = -|f_{\text{concave}}|$, the equation becomes:
$\frac{1}{F} = \frac{1}{|f_{\text{convex}}|} - \frac{1}{|f_{\text{concave}}|}$
For the combination to behave as a concave lens, the effective focal length $F$ must be negative, requiring $\frac{1}{F} < 0$:
$\frac{1}{|f_{\text{convex}}|} - \frac{1}{|f_{\text{concave}}|} < 0 \implies \frac{1}{|f_{\text{convex}}|} < \frac{1}{|f_{\text{concave}}|}$
This inequality holds true only when $|f_{\text{convex}}| > |f_{\text{concave}}|$.

For lens P, applying the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$ with $u_1 = -15\text{ cm}$ and $f_1 = 10\text{ cm}$:
$\frac{1}{v_1} - \frac{1}{-15} = \frac{1}{10} \implies \frac{1}{v_1} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30} \implies v_1 = 30\text{ cm}$
This image acts as a virtual object for lens Q, placed $15\text{ cm}$ to the right of P. Thus, the object distance for lens Q is $u_2 = v_1 - 15 = 30 - 15 = +15\text{ cm}$. Applying the lens formula for lens Q ($f_2 = 15\text{ cm}$):
$\frac{1}{v_2} - \frac{1}{15} = \frac{1}{15} \implies \frac{1}{v_2} = \frac{2}{15} \implies v_2 = 7.5\text{ cm}$
The total magnification $m = m_1 \times m_2 = \left(\frac{v_1}{u_1}\right) \times \left(\frac{v_2}{u_2}\right)$:
$m = \left(\frac{30}{-15}\right) \times \left(\frac{7.5}{15}\right) = (-2) \times (0.5) = -1$
Since $v_2$ is positive, the image is real, formed $7.5\text{ cm}$ to the right of lens Q, and the magnitude of magnification is $1$, meaning it is the same size as AB.

At a point exactly opposite to one of the slits, the vertical distance from the center is $y = \frac{d}{2}$. The path difference $\Delta x$ for a screen at distance $D$ is given by:
$\Delta x = \frac{yd}{D} = \frac{(d/2)d}{10d} = \frac{d}{20}$
Substituting $d = 5\lambda$, we get $\Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4}$. The corresponding phase difference $\phi$ is:
$\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left( \frac{\lambda}{4} \right) = \frac{\pi}{2}$
The intensity $I$ at this point is related to the maximum intensity $I_{\circ}$ by the formula:
$I = I_{\circ} \cos^2\left(\frac{\phi}{2}\right) = I_{\circ} \cos^2\left(\frac{\pi}{4}\right) = I_{\circ} \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_{\circ}}{2}$

The de Broglie wavelength is given by the relation $\lambda = \frac{h}{mv}$.
In free space, the wavelength is $\lambda_0 = \frac{h}{mv}$.
When the electron enters the medium, its velocity is reduced by 20%, resulting in a new velocity:
$v' = v - 0.20v = 0.80v = \frac{4}{5}v$
The de Broglie wavelength in the medium becomes:
$\lambda_{medium} = \frac{h}{mv'} = \frac{h}{m(0.80v)} = \frac{1}{0.80} \left( \frac{h}{mv} \right) = 1.25 \lambda_0$
Comparing this to the given form $\alpha \lambda_0$, we find $\alpha = 1.25$.

The $^{12}_{6}\text{C}$ atom consists of 6 protons, 6 neutrons, and 6 electrons. Using the standard mass of an electron ($m_e \approx 0.00055 \text{ u}$), we calculate the total mass of the constituent particles:
$M_{\text{const}} = 6(m_p + m_e) + 6(m_n)$
$M_{\text{const}} = 6(1.00727 + 0.00055 + 1.00866) \text{ u} = 6(2.01648) \text{ u} = 12.09888 \text{ u}$
The mass defect ($\Delta m$) is the difference between the mass of the constituents and the experimental atomic mass ($12 \text{ u}$):
$\Delta m = 12.09888 \text{ u} - 12 \text{ u} = 0.09888 \text{ u}$
Converting this mass defect into energy using the factor $931.5 \text{ MeV/u}$:
$E = \Delta m \times 931.5 \text{ MeV/u} = 0.09888 \times 931.5 \approx 92.1 \text{ MeV}$
Comparing this result to the provided options, the value corresponds to 92.0.

The depletion region width is governed by the charge neutrality condition, which requires the total uncovered ionized charge on both sides to be equal:
$N_A x_p = N_D x_n$
Rearranging for the ratio of the depletion widths gives:
$\frac{x_p}{x_n} = \frac{N_D}{N_A} = \frac{10^{18}}{10^{15}} = 1000$
This indicates that $x_p = 1000 x_n$, meaning the depletion width is significantly larger on the p-side where the doping concentration is lower. In general, the depletion region extends further into the more lightly doped side of a p-n junction.

The elastic potential energy $U$ stored in a stretched wire is given by the energy density formula multiplied by the volume:
$U = \frac{1}{2} \times Y \times (\text{strain})^2 \times V$
Given $\Delta L = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}$ and $L = 3 \text{ m}$, the strain is:
$\text{strain} = \frac{\Delta L}{L} = \frac{3 \times 10^{-3}}{3} = 10^{-3}$
Substituting the given values $Y = 1.1 \times 10^{11} \text{ N/m}^2$ and $V = 600 \times 10^{-6} \text{ m}^3$ into the energy formula:
$U = \frac{1}{2} \times (1.1 \times 10^{11}) \times (10^{-3})^2 \times (600 \times 10^{-6})$
$U = 0.5 \times 1.1 \times 10^{11} \times 10^{-6} \times 600 \times 10^{-6} = 0.5 \times 1.1 \times 60 = 33 \text{ J}$

[CORRECT_OPTION: 33]

To calculate the heat extracted from $x$ grams of water cooling from $50^\circ\text{C}$ to $0^\circ\text{C}$:
$Q_{\text{out}} = m c \Delta T = \left(\frac{x}{1000}\right) \times 4200 \times (50 - 0) = 210x \text{ J}$
The heat required to evaporate $(1000 - x)$ grams of water initially at $50^\circ\text{C}$ involves heating it to $100^\circ\text{C}$ and then latent vaporization:
$Q_{\text{\in}} = m' c \Delta T + m' L = \left(\frac{1000 - x}{1000}\right) \times [4200 \times (100 - 50) + 2,256,000]$
$Q_{\text{\in}} = (1000 - x) \times [210 + 2256] = 2466(1000 - x) \text{ J}$
Equating $Q_{\text{out}}$ and $Q_{\text{\in}}$:
$210x = 2,466,000 - 2466x \implies 2676x = 2,466,000$
$x = \frac{2,466,000}{2676} \approx 921.52$
Rounding to the nearest integer, we get $x = 922$.

[CORRECT_OPTION: 922]

At resonance, the angular frequency $\omega_0$ of a series LCR circuit is given by:
$\omega_0 = \frac{1}{\sqrt{LC}}$
The inductive reactance $X_L$ at this frequency is $X_L = \omega_0 L$. Substituting the expression for $\omega_0$:
$X_L = \frac{1}{\sqrt{LC}} \cdot L = \sqrt{\frac{L}{C}}$
Given $L = 1.6\text{ H}$ and $C = 40 \times 10^{-6}\text{ F}$, we calculate:
$X_L = \sqrt{\frac{1.6}{40 \times 10^{-6}}} = \sqrt{\frac{1.6 \times 10^6}{40}} = \sqrt{0.04 \times 10^6} = \sqrt{40000} = 200\,\Omega$

Using Ohm's Law for a circuit, $E = I(R + r)$, we set up the following equations for the two cases:

1. $E = 0.25(5 + r) = 1.25 + 0.25r$
2. $E = 0.5(2 + r) = 1.0 + 0.5r$
   Since the EMF ($E$) is constant, equate the two expressions:
   $1.25 + 0.25r = 1.0 + 0.5r$
   Rearranging the terms:
   $1.25 - 1.0 = 0.5r - 0.25r \implies 0.25 = 0.25r$
   Solving for the internal resistance gives $r = 1\,\Omega$.

[CORRECT_OPTION: 1]

The area of the loop is $A = \pi r^2 = \frac{22}{7} \times (0.2)^2 = \frac{0.88}{7} \, \text{m}^2$. The magnetic flux is given by $\Phi = B A = (2t^2 + 2t + 3)A$.
The induced electromotive force (EMF) is:
$\varepsilon = -\frac{d\Phi}{dt} = -A \frac{d}{dt}(2t^2 + 2t + 3) = -A(4t + 2)$
The induced current $I$ at $t = 3 \, \text{s}$ with resistance $R = 2 \, \Omega$ is:
$I = \frac{|\varepsilon|}{R} = \frac{A(4(3) + 2)}{2} = \frac{A(14)}{2} = 7A = 7 \times \frac{0.88}{7} = 0.88 \, \text{A}$
Given $I = \frac{\alpha}{50}$, we equate:
$\frac{\alpha}{50} = 0.88 \implies \alpha = 0.88 \times 50 = 44$

[CORRECT_OPTION: N/A]

The chemical equation is $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$.
Moles of $Zn = \frac{20\text{ g}}{65\text{ g mol}^{-1}} \approx 0.3077\text{ mol}$.
Mass of $H_2SO_4 = 50\text{ mL} \times 1.3\text{ g mL}^{-1} \times 0.5 = 32.5\text{ g}$.
Moles of $H_2SO_4 = \frac{32.5\text{ g}}{98\text{ g mol}^{-1}} \approx 0.3316\text{ mol}$.
Since $Zn$ is the limiting reagent, moles of $H_2$ produced = moles of $Zn = 0.3077\text{ mol}$.
Volume of $H_2$ at STP = $0.3077\text{ mol} \times 22.4\text{ L mol}^{-1} = 6.892\text{ L}$.

A. True: According to the Aufbau principle, for a given $(n+l)$ value, the orbital with the lower $n$ is more stable and thus has lower energy.
B. False: As the atomic number ($Z$) increases, nuclear attraction increases, making orbitals more stable (lower energy).
C. True: Orbital size is determined by the principal quantum number $n$; since $n=2 < n=3$, the $2p_x$ orbital is smaller than the $3p_x$ orbital.
D. False: Radial nodes $= (n - l - 1)$. For $5d$ orbitals: $n=5, l=2$, so $5 - 2 - 1 = 2$ radial nodes, contradicting the statement.
Since only A and C are true, the correct option is B.

The covalent bond length ($d_{AB}$) between two atoms A and B is defined as the sum of their covalent radii because it represents the distance between their nuclei:
$d_{AB} = r_A + r_B$
The total length (often interpreted as the sum of the atomic diameters of the two atoms) is the sum of the diameters of atom A ($2r_A$) and atom B ($2r_B$):
$\text{Total length} = 2r_A + 2r_B = 2(r_A + r_B)$
Thus, the bond length is $(r_A + r_B)$ and the total length is $2(r_A + r_B)$.

To calculate the standard Gibbs free energy change ($\Delta_r G^\circ$) at 600 K for the reaction $\text{X}_2(g) + \text{Y}_2(g) \rightleftharpoons 2\text{XY}(g)$, we use the relation $\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ$.