question_id string | question string | question_images list | answer string | solution string | solution_images list | subject string | topic string | difficulty string | question_type string | has_image bool | exam string | source_paper string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
ADV-C01-Q20 | Consider the following volume-temperature (V-T) diagram
for the expansion of 5 moles of an ideal monoatomic gas.
[IMAGE] Considering on P-V work is involved, the total change in enthalpy (in
Joule) for the transformation of state in the sequence
$X \rightarrow Y \rightarrow Z$ is \_\_\_\_\_\_\_\_.
\[Use the given data:... | [
"images/image14.png"
] | (8120) | $X \rightarrow Y$ is an isothermal process and for ideal
bas
$$\mathrm{\Delta}H = 0$$
$Y \rightarrow Z$ is an isochoric process
$$\mathrm{\Delta}U = nC_{V,m}\left( T_{2} - T_{1} \right)$$
$$= 5 \times 12(415 - 335)$$
$$= 4800\ J$$
$$\mathrm{\Delta}H = \mathrm{\Delta}U + \mathrm{\Delta}(PV)$$
$$= \mathrm{\Delta}U + nR\m... | [] | Chemistry | Chemical Thermodynamics | subjective | true | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q6 | Consider a 70% efficient hydrogen-oxygen fuel cell
working under standard conditions at 1 bar and 298 K. Its cell reaction
is
$$H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l)$$
The work derived from the cell on the consumption of
$1.0 \times 10^{- 3}mol$ of $H_{2}(g)$ is used to compress $1.00\ mol$
of a monoatom... | [] | 13.32 | For give reaction
$H_{2}(g) + \frac{1}{2}O_{2}(g)\overset{\ \ \ 2e^{-}\ \ \ }{\rightarrow}H_{2}O(l)$
$E{^\circ} = 1.23V$
$$\mathrm{\Delta}G{^\circ} = - FE{^\circ}_{cell} = \lbrack - 2 \times 96500 \times 1.23\rbrack 1 \times 10^{- 3} \times 0.7 = - 166.173\ J$$
$$W = \frac{nR\mathrm{\Delta}T}{\gamma - 1}$$
$$166.173 = ... | [] | Chemistry | Electrochemistry | numerical | false | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q15 | The stoichiometric reaction of 516 g of
dimethyldichlorosilane with water results in a tetrameric cyclic product
X in 75% yield. The weight (in g) of X obtained is \_\_\_\_\_\_\_\_.
\[Use, molar mass
$\left( g\ mol^{- 1} \right):H = 1,\ C = 12,O = 16,Si = 28,Cl = 35.5$\] | [] | (222) | [IMAGE] $$\left( CH_{3} \right)_{2}SiCl_{2} = \frac{516}{129} = 4\ mol$$
On Applying POAC on Si
$$1 \times n_{\left( CH_{3} \right)_{2}SiCl_{2}} = 4 \times n_{tetramer}\ \ $$
$$n_{tetramer} = 1\ mol$$
$$\downarrow \times 296\frac{g}{mol}$$
$$296g$$
$$296 \times \frac{75}{100} = 222\ g$$ | [
"images/image10.png"
] | Chemistry | Stoichiometry | subjective | true | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q5 | Consider the following reversible reaction,
$$A(g) + B(g) \rightleftharpoons AB(g)$$
The activation energy of the backward reaction exceeds that of the
forward reaction by $2\ RT$ (in $J\ mol^{- 1}$).
If the pre-exponential factor of the forward reaction is 4 times that
of the reverse reaction, the absolute value of
$\... | [] | 8500 J/mole | $A(g) + B(g) \rightleftharpoons AB(g)$
$E_{a_{b}} = E_{a_{f}} + 2RT$ & $A_{f} = 4A_{b}$
Now, Rate constant of forward reaction $k_{f} = A_{f}e^{- Ea_{f}/RT}$
Rate constant of reverse reaction $K_{b} = A_{b}e^{E_{ab}/RT}$
Equilibrium constant
$$K_{eq} = \frac{K_{f}}{K_{b}} = \frac{A_{f}}{A_{b}}e^{- \left( E_{af} - E_{ab... | [] | Chemistry | Chemical Equilibrium | subjective | false | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q10 | One mole of an ideal gas at 900 K, undergoes two
reversible processes, I followed by II, as shown below. If the work done
by the gas in the two processes are same the value of
$\ln\frac{v_{3}}{v_{2}}$ is \_\_\_\_\_\_\_.
[IMAGE] (U: internal energy, S: entropy, p: pressure, V: volume, R: gas
constant)
(Given: molar heat... | [
"images/image6.png"
] | (10) | $\mathrm{\Delta}U_{1} = nC_{v}\mathrm{\Delta}T$
$$- 1800\ R = 1 \times \frac{5R}{2} \times \mathrm{\Delta}T$$
$$\mathrm{\Delta}T_{1} = - 720k$$
$$T_{2} - T_{1} = - 720\ \left\lbrack T_{1} = 900k \right\rbrack$$
$T_{2} = 180\ k$.
Process II is reversible isothermal
$$W_{II} = - nRT\ln\left( \frac{V_{3}}{V_{2}} \right)$$... | [] | Chemistry | Chemical Thermodynamics | subjective | true | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q14 | Consider the strong electrolytes
$Z_{m}X_{n},U_{m}Y_{p}$ and $V_{m}X_{n}$. Limiting molar conductivity
$(\Lambda{^\circ})$ of $U_{m}Y_{p}$ and $V_{m}X_{n}$ are 250 and
$440\ S\ cm^{2}\ mol^{- 1}$, respectively. The value of $(m + n + p)$ is
\_\_\_\_\_\_\_\_.
------------------------------------------------------------... | [
"images/image9.png"
] | (7) | $Z_{m}X_{n},$ $U_{m}Y_{p},$ $V_{m}X_{n}$
$\lambda_{M}^{\infty} = \ ?$ 250 $440\ S\ cm^{2}mol^{- 1}$
$$\lambda_{M}^{\infty}\left( U_{m}Y_{p} \right) = m \times 25 + p(100) = 250$$
$25m + 100p = 250$ ... (i)
$\lambda_{M}^{\infty}\left( V_{m}X_{m} \right) = m \times 100 + n \times 80 = 440$
... (ii)
For
$Z_{m}X_{n} \Right... | [] | Chemistry | Electrochemistry | subjective | true | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q16 | The plot of $\log k_{f}$ versus $\frac{1}{T}$ for a
reversible reaction $A(g) \rightleftharpoons P(g)$ is shown.
[IMAGE] Pre-exponential factors for the forward and backward reactions are
$10^{15}s^{- 1}$ and $10^{11}s^{- 1}$, respectively.
If the value of $\log K$ for the reaction at $500\ K$ is 6, the value
of $\left... | [
"images/image11.png"
] | (5) | $\frac{1}{T} = 0.002$
$$T = 500$$
$$\log k_{f} = 9$$
$$k_{f} = 10^{9}$$
$$\log K_{eq} = 6$$
$$K_{eq} = 10^{6} = \frac{k_{f}}{k_{b}} = \frac{10^{9}}{k_{b}}$$
$k_{b} = 10^{3}$ (at 500 K)
$$A_{b} = 10^{11}$$
$$\log k_{b} = \log A_{b} - \frac{(Ea)_{b}}{2.303R(500)}$$
$$3 = 11 - \frac{(Ea)_{b}}{2.303R(500)}$$
$$\frac{(Ea)_{... | [] | Chemistry | Chemical Equilibrium | subjective | true | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q19 | For $He^{+}$, a transition takes place from the orbit
of radius 105.8 pm to the orbit of radius 26.45 pm. The wavelength (in
nm) of the emitted photon during the transition is \_\_\_\_\_\_\_\_.
Bohr radius, a = 52.9 pm
Rydberg constant $R_{H} = 2.2 \times 10^{- 18}J$
Planck's constant, $h = 6.6 \times 10^{- 34}Js$
Spee... | [] | 30 | $\frac{1}{\lambda} = \frac{R_{H}Z^{2}}{h.c}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$$\frac{1}{\lambda} = \frac{2.2 \times 10^{- 18} \times Z^{2}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack$$
⇒
$\frac{2.2 \times 10^{- 18} \times ... | [] | Chemistry | Atomic Structure | numerical | false | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q7 | $_{92}^{238}U$ is known to undergo radioactive decay to
form $_{82}^{206}{Pb}$ by emitting alpha and beta particles. A rock
initially contained $68 \times 10^{- 6}g$ of $_{92}^{238}U$. If the
number of alpha particles that it would emit during its radioactive
decay of $_{92}^{238}U$ to $_{82}^{206}{Pb}$ in three half-l... | [] | $1.2 \times 10^{18}$ | $_{92}^{238}U \rightarrow_{82}^{206}{Pb} + 8_{2}^{4}{He} + 6_{- 1}^{0}\beta$
$$0.286 \times 10^{- 6}mole$$
$$= 0.286 \times 6.02 \times 10^{17}$$
$$= 1.72 \times 10^{17}$$
After 3 Half lives
$$N = \frac{No}{(2)^{3}} = \frac{1.72 \times 10^{17}}{8}$$
$$= 0.215 \times 10^{17}$$
So, no. of molecule or uranium decayed
$$= ... | [] | Chemistry | Nuclear Chemistry | subjective | false | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q13 | Concentration of $H_{2}SO_{4}$ and $N_{2}SO_{4}$ in a
solution is $1\ M$ and $1.8 \times 10^{- 2}\ M,$ respectively. Molar
solubility of $PbSO_{4}$ in the same solution is $X \times 10^{- Y}\ M$
(expressed in scientific notation). The value of Y is \_\_\_\_\_\_\_.
\[Given: Solubility product of
$PbSO_{4}\left( K_{sp} \... | [] | (6) | $\left\lbrack H_{2}SO_{4} \right\rbrack = 1M$ &
$\left\lbrack Na_{2}S_{4} \right\rbrack = 1.8 \times 10^{- 2}\ M$
$$H_{2}SO_{4} \rightarrow H^{+} + HSO_{4}^{1 -}$$
1M 1 M 1 M
$H^{+} + SO_{4}^{2 -}$ $\rightarrow$ $HSO_{4}^{1 -}$
$K = \frac{1}{K_{a_{2}}} = \frac{10^{2}}{1.2}$
1M $1.8 \times 10^{- 2}$ 1M
(1-x) $\left( 1.8... | [] | Chemistry | Ionic Equilibrium | subjective | false | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q18 | In a one-litre flask, 6 moles of A under goes the
reaction $A(g) \rightleftharpoons P(g)$. The progress of product
formation at two temperatures (in Kelvin), $T_{1}$ and $T_{2}$, is shown
in the figure:
[IMAGE] If $T_{1} = 2T_{2}$ and
$\left( \mathrm{\Delta}G_{2}^{\ominus} - \mathrm{\Delta}G_{1}^{\ominus} \right) = RT_... | [
"images/image13.png"
] | (8) | $\mathrm{\Delta}G_{1}^{\ominus} = - RT_{1}\ln\frac{4}{2}$
$$\mathrm{\Delta}G_{2}^{\ominus} = - RT_{2}\ln\frac{2}{4}$$
$$\mathrm{\Delta}G_{2}^{\ominus} - \mathrm{\Delta}G_{1}^{\ominus}$$
$$= - RT_{2}\ln{\left( \frac{1}{2} \right) + RT_{1}\ln 2}$$
$$= - RT_{2}\ln{\left( \frac{1}{2} \right) + R_{2}T_{2}\ln 2}$$
$$= RT_{2}... | [] | Chemistry | Stoichiometry | subjective | true | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q11 | 2 mole of $Hg(g)$ is combusted in a fixed volume bomb
calorimeter with excess of $O_{2}$ at $298\ K$ and 1 atm into $HgO(s)$.
During the reaction, temperature increases from $298.0\ K$ to
$312.8\ K$. If heat capacity of the bomb calorimeter and enthalpy of
formation of $Hg(g)$ are $20.00\ kJ\ K^{- 1}$ and $61.32\ kJ\ m... | [] | 90.39 | [IMAGE] $$q_{combustion} = \left( C_{T} \right)\mathrm{\Delta}T$$
$$= 20 \times \lbrack 312.8 - 298\rbrack$$
$$= (20 \times 14.8)$$
$$= 296\ kJ$$
$$\mathrm{\Delta}H = \mathrm{\Delta}U + \mathrm{\Delta}ngRT$$
$$= - 296 + \left( \frac{( - 3) \times 8.315 \times 298}{1000} \right)$$
$$= - 296 - 7.433$$
$$= - 303.43\ kJ$$
... | [
"images/image7.png",
"images/image8.png"
] | Chemistry | Chemical Thermodynamics | numerical | true | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q2 | Consider an ionic solid $MX$ with $NaCl$ structure.
Construct a new structure $(Z)$ whose unit cell is constructed from the
unit cell of $MX$ following the sequential instructions given below.
Neglect the charge balance.
\(i\) Remove all the anions $(X)$ except the central one
\(ii\) Replace all the face centered catio... | [] | 3 | As per given information cation form FCC lattice and anion
occpy all the octahedral void.
So $M^{+}$ $X^{-}$ & Formula MX
4 ion 4 ion
After step I 4 ion 1 ion
After step II 1 ion 4 ion
After step III 0 ion 4 ion
After step IV 1 ion 3 ion
So ratio of $\frac{No.\ of\ anion}{No.\ of\ cation} = \frac{3}{1}$ | [] | Chemistry | Solid State | numerical | false | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q12 | A solution is prepared by mixing 0.01 mole each of
$H_{2}CO_{3},NaHCO_{3},Na_{2}CO_{3}$ and $NaOH$ in 100 mL of water. pH
of the resulting solution is \_\_\_\_\_\_\_\_\_\_.
\[Given: $pKa_{1}$ and $pKa_{2}$ of $H_{2}CO_{3}$ are 6.37 and 10.32,
respectively; $\log 2 = 0.30$\] | [] | 10.02 | $n_{H_{2}CO_{3}} = n_{NaHCO_{3}} = n_{Na_{2}CO_{3}} = n_{NaOH} = 0.01\ mole$.
$H_{2}CO_{3}$ $+$ $NaOH$ $\rightarrow$ $NaHCO_{3}$
0.01 mole 0.01 mole 0.01 mole
0 0 0.02 mole
$NaHCO_{3}$ $+$ $Na_{2}co_{3}$
0.02 mole 0.01 mole
this act as acidic buffer solution.
$$pH = pKa_{2} + \log\frac{\left\lbrack Na_{2}CO_{3} \right\... | [] | Chemistry | Ionic Equilibrium | numerical | false | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q3 | The solubility of a salt of weak acid (AB) at pH 3 is
$Y \times 10^{- 3}\ mol\ L^{- 1}$. The value of $Y$ is
\_\_\_\_\_\_\_\_\_\_.
(Given that the value of solubility product of
$AB\ \left( K_{sp} \right)$ $= 2 \times 10^{- 10}$ and the value of
ionization constant of $HB\ \left( K_{a} \right) = 1 \times 10^{- 8}$) | [] | 4.47 | $AB \rightleftharpoons \ A^{+}\ \ + \ B^{-}$
$x$ $x$ $x - y$
$2 \times 10^{10} = x(x - y)$ ... (1)
$B^{-}\ \ \ \ + \ \ H^{+} \rightleftharpoons HB$
$x - y$ $10^{- 3}$ $y$
$$10^{8} = \frac{y}{(x - y)10^{- 3}}$$
$$\frac{y}{x - y} = 10^{5}$$
$x - y = 10^{- 5}y$ ... (2)
From (1) & (2)
$$2 \times 10^{- 10} = x^{2} - 2 \time... | [] | Chemistry | Ionic Equilibrium | numerical | false | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q17 | One mole of an ideal monoatomic gas undergoes two
reversible processes (A → B and B → C) as shown in the given figure:
[IMAGE] A → B is an adiabatic process. If the total heat absorbed in the
entire process (A → B and B → C) is
$RT_{2}\ln 10$, the value of $2\log V_{3}$ is
\_\_\_\_\_\_\_\_\_\_\_\_.
\[Use, molar heat ca... | [
"images/image12.png"
] | (7) | A -- B (Adiabatic)
Monoatomic gas $\gamma = 5/3$
$$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1}$$
$$600V_{1}^{5/3 - 1} = 60V_{2}^{5/3 - 1}$$
$$10.V_{1}^{2/3} = V_{2}^{2/3}$$
$$10(10)^{2/3} = V_{2}^{2/3}$$
$v_{2} = 10^{5/2}$ ... (i)
$$q_{Total} = q_{A - B} + q_{B - C}$$
$$RT_{2}\ln 10 = 0 + RT_{2}\ln\frac{V_{3}}{V_... | [] | Chemistry | Chemical Thermodynamics | subjective | true | JEE Advanced | Vijy_sir_Chemistry_21.docx | |
ADV-C01-Q8 | For the give close packed structure of a salt made of
cation X and anion Y shown below (ions of only one face are shown for
clarity), the packing fraction is
approximately packing fraction $= \frac{packing\ efficiency}{100}$
[IMAGE] | [
"images/image2.jpeg"
] | 0.63 | $2\left\lbrack r_{x} + r_{y} \right\rbrack = \sqrt{2}a$
$\left\{ a = 2r_{y};r_{y} = 0.59 \right\}$
$$2\left\lbrack r_{x} + r_{y} \right\rbrack = \sqrt{2}\left( 2r_{y} \right)$$
$$2r_{x} = 2\sqrt{2}r_{y} - 2r_{y}$$
$$r_{x} = r_{y}\left\lbrack \sqrt{2} - 1 \right\rbrack$$
$$r_{x} = 0.414,r_{y} = 0.207\ a$$
Packing fracti... | [] | Chemistry | Solid State | numerical | true | JEE Advanced | Vijy_sir_Chemistry_21.docx |
JEE Advanced — Question Bank
A structured dataset of JEE Advanced examination questions with full worked solutions and diagrams. JEE Advanced questions are more analytical than JEE Main — many are subjective, integer, or numerical-answer type with detailed multi-step solutions.
Subsets (PCM):
- Physics — 50 questions
- Chemistry — 21 questions
- Mathematics — 48 questions
Structure
Organised into subsets by subject and splits (train / test):
mathematics/ physics/ chemistry/
├── train.jsonl ├── train.jsonl ├── train.jsonl
├── test.jsonl ├── test.jsonl ├── test.jsonl
└── images/ └── images/ └── images/
Schema
| Field | Type | Description |
|---|---|---|
question_id |
string | Unique ID, e.g. ADV-M01-Q1, ADV-P01-Q1, ADV-C01-Q1. |
question |
string | Question text in LaTeX. [IMAGE] marks inline figures. |
question_images |
list[str] | Relative paths to images used in the question. |
answer |
string | The answer — integer, LaTeX expression, relation, or value(s). |
solution |
string | Full worked solution in LaTeX. [IMAGE] marks inline figures. |
solution_images |
list[str] | Relative paths to images used in the solution. |
subject |
string | Mathematics, Physics, or Chemistry. |
topic |
string | Auto-detected from content (e.g. Coordination Compounds). |
difficulty |
string | Reserved (not provided in source). |
question_type |
string | numerical / subjective / single_correct. |
has_image |
bool | Whether the question or solution contains a figure. |
exam |
string | JEE Advanced. |
source_paper |
string | Original source paper reference. |
Notes
- Unlike JEE Main (4-option MCQs), JEE Advanced questions here are largely
subjective / integer / numerical, so the schema uses a single
answerfield rather than four options. topicis auto-detected from the question and solution content, since the source papers do not carry explicit topic labels. Chemistry topics are detected at fine granularity (e.g. Coordination Compounds, Solid State, Nuclear Chemistry, Electrochemistry, Chemical Thermodynamics).- All figures are stored under each subject's
images/folder and referenced by relative path.
Loading
from datasets import load_dataset
math = load_dataset("eQOURSE/jee-advanced-questions", "mathematics")
phys = load_dataset("eQOURSE/jee-advanced-questions", "physics")
chem = load_dataset("eQOURSE/jee-advanced-questions", "chemistry")
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