JEE MAIN 2026 ONLINE 22ND JANUARY MORNING SHIFT

$$\text { Match the LIST-I with LIST-II }$$

Dissociation of a gas $\mathrm{A}_2$ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K .

$$\mathrm{A}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g})$$

The standard Gibbs energy of formation of the involved substances has been provided below:

$$\begin{array}{|c|c|} \hline \text { Substance } & \Delta \mathrm{G}_{\mathrm{f}}^{\circ} / \mathrm{kJ} \mathrm{~mol}^{-1} \\ \hline \hline \mathrm{~A}_2 & -100.00 \\ \hline \mathrm{~A} & -50.832 \\ \hline \end{array}$$

The degree of dissociation of $\mathrm{A}_2(\mathrm{~g})$ is given by $\left(x \times 10^{-2}\right)^{1 / 2}$ where $x=$

$\_\_\_\_$ . (Nearest integer).

[ Given : $\mathrm{R}=8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 2=0.3010, \log 3=0.48$ ]

Assume degree of dissociation is not negligible.

Let A be a $3 \times 3$ matrix such that $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\mathrm{O}$. If $\mathrm{A}\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{l}3 \\ 3 \\ 2\end{array}\right], \mathrm{A}^2\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}-3 \\ 19 \\ -24\end{array}\right]$ and $\operatorname{det}(\operatorname{adj}(2 \operatorname{adj}(\mathrm{~A}+\mathrm{I})))=(2)^\alpha \cdot(3)^\beta \cdot(11)^\gamma, \alpha, \beta, \gamma$ are non-negative integers, then $\alpha+\beta+\gamma$ is equal to $\_\_\_\_$

A circular disc has radius $R_1$ and thickness $T_1$. Another circular disc made of the same material has radius $R_2$ and thickness $T_2$. If the moment of inertia of both discs are same and $\frac{R_1}{R_2}=2$ then $\frac{T_1}{T_2}=\frac{1}{\alpha}$. The value of $\alpha$ is $\_\_\_\_$ .

Consider the following electrochemical cell at 298 K

$\mathrm{Pt}\left|\mathrm{HSnO}_2^{-}(\mathrm{aq})\right| \mathrm{Sn}(\mathrm{OH})_6{ }^{2-}(\mathrm{aq})\left|\mathrm{OH}^{-}(\mathrm{aq})\right| \mathrm{Bi}_2 \mathrm{O}_3(\mathrm{~s}) \mid \mathrm{Bi}(\mathrm{s})$.

If the reaction quotient at a given time is $10^6$, then the cell EMF $\left(\mathrm{E}_{\text {cell }}\right)$ is

$\_\_\_\_$ $\times 10^{-1} \mathrm{~V}$ (Nearest integer).

Given the standard half-cell reduction potential as

$$\mathrm{E}_{\mathrm{Bi}_2 \mathrm{O}_3 / \mathrm{Bi}, \mathrm{OH}^{-}}^{\circ}=-0.44 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{Sn}(\mathrm{OH})_6^{2-} / \mathrm{HSnO}_2^{-}, \mathrm{OH}^{-}}^{\circ}=-0.90 \mathrm{~V}$$

Let ABC be a triangle. Consider four points $\mathrm{p}_1, \mathrm{p}_2, \mathrm{p}_3, \mathrm{p}_4$ on the side AB , five points $p_5, p_6, p_7, p_8, p_9$ on the side $B C$, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC . None of these points is a vertex of the triangle ABC . Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, \ldots, p_{13}$, is $\_\_\_\_$