JEE Mains Important Questions with Answers PDF - – Physics, Chemistry, Maths

JEE Mains Important Questions with Answers: Chapter-Wise Important Topics for JEE Main

In this section, we will understand the high weighted chapters. Along with these chapters, we will see JEE mains important questions with answers. Some JEE practice questions PDF with answers are also given below so that you can get a reference of what type of questions are there. Additionally, we have provided you the entire JEE mains important questions with answers pdf which you can download and practice accordingly. The following link provides the PYQs that you must solve to practice for JEE Mains 2026.

JEE Mains Important Questions with Answers PDF: Physics

The following table will give you the most weighted chapters of JEE Main January Session. Practising from these might help you understand the concepts and types of questions asked from these chapters.

JEE Mains Physics Important Questions with Answers:
1. Optics

Q. 1 A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000\AA and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :

1) 3 mm

2) 9 mm (Correct)

3) 4.5 mm

4) 1.5 mm

Solution:

Fraunhofer Diffraction -

$b \sin \theta=n \lambda$

- wherein

Condition of nth minima.

b= slit width

θ= angle of deviation

$a=0.1 \mathrm{~mm}=10^{-4} \mathrm{~cm}$

$lambda=6000 AA=6 \times 10^{-7} \mathrm{~cm}$

$D=0.5 \mathrm{~m}$

for 3rd dark bond, asin⁡θ=3λ

$sin \theta=\frac{3 \lambda}{a}=\frac{x}{D} $

$therefore$ $x=\frac{3 \lambda D}{a}$

Q. 2 In an experiment for determination of the refractive index of glass of a prism by i−δ, plot, it was found that a ray incident at angle 35∘, suffers a deviation of 40∘ and that it emerges at angle 790. In that case which of the following is closest to the maximum possible value of the refractive index?

1) 1.5

2) 1.6

3) 1.7

4) 1.8

Correct Answer:

1.5

Solution:

If μ is refractive index of material of prism, then from Snell's law

μ=sin⁡isin⁡r=sin⁡(A+δm)/2sin⁡A/2⋯(i)

where, A is angle of prism and δ¯m is minimum deviation through prism.

Given, $\mathrm{i}=35^{\circ}, \delta=40^{\circ}, \mathrm{e}=79^{\circ}$

So, angle of deviation by a glass prism

$\delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \Rightarrow 40^{\circ}=35^{\circ}+79^{\circ}-\mathrm{A}$

i.e.Angle of prism $\Rightarrow A=74^{\circ}$

Such that, $r_1+r_2=A=74^{\circ}$

Important Question of Optics

2. Electrostatics

Q. 1 Within a spherical charge distribution of charge density ρ(r),N equipotential surfaces of potential V₀, V₀+V, V₀+2 V,………..V₀+NΔV(ΔV>0), are drawn and have increasing radii r₀,r₁, r₂,……….r_N, respectively. If the difference in the radii of the surfaces is constant for all values of V₀ and ΔV then:

1) $\rho(r) \propto r$

2) $\rho(r)=\text { constant }$

3) $\rho(r) \propto \frac{1}{r}$

4) $\rho(r) \propto \frac{1}{r^2}$

Correct Answer:

$\rho(r) \propto \frac{1}{r}$

Solution:

As we know, the relation between electric field and potential is

$E=-\frac{d V}{d r}$

Important Questions of Electrostatics

3. Properties of Solids and Liquids

Q. 1 A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains the same, the stress in the leg will change by a factor of :

Solution:

As we learned in

Stress -

$\begin{aligned}
& \text { Stress }=\frac{\text { force }}{\text { Area }}=\frac{F}{A} \\
& F=\text { applied force } \\
& A=\text { area } \\
& \text { Stress(S) }=\frac{F}{A}=\frac{m g}{A} \\
& \text { Stress }=\frac{\text { Volume × density } \times g}{L^2}
\end{aligned}$

$\begin{aligned}
& \text { Stress }=\frac{L^3 \rho g}{L^2}=\text { Stress } \alpha L \\
& \text { Stress }\left(\mathrm{S}^{\prime}\right)=\mathrm{ma} /\left.9\right|^* 9 \mathrm{~b}=\rho \mathrm{Va} /\left.8\right|^* 8 \mathrm{~b} \\
& \text { therefore } \mathrm{S}^{\prime} / \mathrm{S}=\left(9^3 \mathrm{~V} / 81^* \mathrm{lb}\right)^* \mathrm{lb} / \mathrm{V} \\
& \mathrm{~S}^{\prime}=9 \mathrm{~S}
\end{aligned}$

Hence stress in the leg will change by a factor of 9 .

Hence, the answer is the option (1).

Important Questions of Properties of Solids and Liquids

4. Physics and Measurement

Q. 1 Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

Option 1:

[M]=[T⁻¹C⁻² h]

Option 2:

[M]=[T⁻¹C² h]

Option 3:

[M]=[T⁻¹C⁻² h⁻¹]

Option 4:

[M]=[TC⁻² h]

Correct Answer:

[M]=[T⁻¹C⁻² h]

Solution:

Dimension of length [L]=[CT]

Dimension of mass =[ Angular Momentum ][ Velocity ][ Length ]

=[h][C][CT]=[C⁻²T⁻¹h]

Hence, the answer is option 1.

Q. 2 The diameter and height of a cylinder are measured by a meter scale to be 12.6±0.1 cm and 34.2±0.1 cm, respectively. What will be the value of its volume in the appropriate significant figure?

Option 1:

$4300 \pm 80 \mathrm{~cm}^3$

Option 2:

$4264.4 \pm 81.0 \mathrm{~cm}^3$

Option 3:

$4264 \pm 81 \mathrm{~cm}^3$

Option 4:

$4260 \pm 80 \mathrm{~cm}^3$

Correct Answer:

$4260 \pm 80 \mathrm{~cm}^3$

Solution:

$\mathrm{v}=\frac{\pi \mathrm{d}^2}{4} \mathrm{~h}=4260 \mathrm{~cm}^3$

$\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{2 \Delta \mathrm{~d}}{\mathrm{~d}}+\frac{\Delta \mathrm{h}}{\mathrm{~h}}$

$\Delta \mathrm{v}=2 \times \frac{0.1 \mathrm{v}}{12.6}+\frac{0.1 \mathrm{v}}{34.2}=\frac{0.2}{12.6} \times 4260+\frac{0.1 \times 4260}{34.2}=80$

$\therefore \text { Volume }=4260 \pm 80 \mathrm{~cm}^3$

Hence, the answer is the option is (4).

Important Questions of Physics and Measurement

5. Rotational Motion

Q. 1 The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N , and the distance of the particle from the origin is 5 m , the angle between the force and the position vector is (in radians) :

Correct Answer:

0.52

Solution:

$T=\vec{r} \times \vec{F}$

$T=|\vec{r}| \cdot|\vec{F}| \cdot \sin \theta----(1)$

$T=2.5 \mathrm{Nm}$

$|\vec{r}|=5 \mathrm{~m}$

$|\vec{F}|=1 \mathrm{~N} \cdots-\text { put in (1) }$

$T=2.5=1 \times 5 \times \sin \theta$

$ \sin \theta=0.5=\frac{1}{2}$

$\theta=\frac{\pi}{6}$

$\theta=0.52 \text { radian }$

Hence, the answer is 0.52 .

Important Questions of Rotational Motion

JEE Practice Questions PDF with Answers: Chemistry

These chapters listed below have the highest chance of repeating again. The weightage given is the weightage of these chapters according to January Session of JEE Main 2026. Therefore, the questions from these chapters are considered to be JEE Mains important questions and answers.

Questions:

1. Co-ordination Compounds

Q: The IUPAC name of [Co(NH₃)₅Cl]Cl₂ is:

A) Pentaamminechloridocobalt(III) chloride

B) Pentamminecobalt(III) chloride

C) Pentaamminecobalt(II) chloride

D) Pentamminechloridocobalt(II) chloride

Solution:

- Central metal: Co

- Oxidation state of Co:x+0( from NH₃)+(−1)=+2. But total charge =+1 (complex ion is +2 , balanced by 2Cl⁻). So Co=+3.

- Correct name = Pentaamminechloridocobalt(III) chloride.

Answer: A

Q. 2 The pair of compounds having metals in their highest oxidation state is :

Option 1:
$\mathrm{MnO}_2$ and $\mathrm{CrO}_2 \mathrm{Cl}_2$

Option 2:
$\left[\mathrm{NiCl}_4\right]^{2-}$ and $\left[\mathrm{CoCl}_4\right]^{2-}$

Option 3:
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ and $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$
Option 4:
[ $\left.\mathrm{FeCl}_4\right]^{-}$and $\mathrm{Co}_2 \mathrm{O}_3$

Solution:

The oxidation state of metals in respective compounds are:

(1) $\mathrm{MnO}_2$ Oxidation State $=+4$
$\mathrm{CrO}_2 \mathrm{Cl}_2$ Oxidation State $=+6$
(2) $\left[\mathrm{NiCl}_4\right]^{2-}$ Oxidation State $=+2$
$\left[\mathrm{CoCl}_4\right]^{2-}$ Oxidation State $=+2$
(3) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ Oxidation State $=+3$
$\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$ Oxidation State $=+2$
(4) $\left[\mathrm{FeCl}_4\right]^{-}$Oxidation State $=+3$
$\mathrm{Co}_2 \mathrm{O}_3$ Oxidation State $=+3$

Observing all the figures we can say that the first pair has the highest oxidation state Hence, the answer is an option (1).

Important Questions of Co-ordination Compounds

2. Chemical Thermodynamics

Q: For a reaction, ΔH=−120 kJ and ΔS=−200 J/K at 300 K . Predict spontaneity.

Solution:

$\Delta G=\Delta H-T \Delta S$

=-120-($300 \times -0.200$)

=-120+60

$ =-60 \mathrm{~kJ}$

Since ΔG<0→ reaction is spontaneous.

Q. An ideal gas undergoes isothermal expansion at constant pressure. During the process :

Option 1:

enthalpy increases but entropy decreases.

Option 2:

enthalpy remains constant but entropy increases.

Option 3:

enthalpy decreases but entropy increases.

Option 4:

Both enthalpy and entropy remain constant.

Correct Answer:

enthalpy remains constant but entropy increases.

Solution:

In the equation,

$\Delta H=n C_p \Delta T$

$\Delta S=n R \ln \left(\frac{V_f}{V_i}\right) \geq 0$

Enthalpy remains constant but entropy increases.

Hence, the answer is an option (2).

Important Questions of Chemical Thermodynamics

3. Some Basic Concepts in Chemistry

Q: 0.5 g of an impure sample of CaCO₃ is treated with excess HCl , releasing 120 mL of CO₂ at STP. Calculate % purity of sample.

Solution:

1 mol CaCO₃→1 mol CO₂=22.4 L

120 mL=0.120 L→ moles CO2=0.120÷22.4=0.00536 mol

Moles CaCO3=0.00536 mol→ mass =0.00536×100=0.536 g

% purity =(0.536÷0.5)×100=107% (practically, would round ∼100%; excess due to measurement)

Q. At 300 K and 1 atm,15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O₂ by volume for complete combustion. After combustion, the gases occupy 330 mL . Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :

Option 1:

C₄H₈

Option 2:

C₄H₁₀

Option 3:

C₃H₆

Option 4:

C₃H₈

Correct Answer:

C₃H₈

Solution:

Volume of N₂ in air =375×0.8=300ml

Volume of O₂ in air =375×0.2=75ml

$\begin{aligned} & \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \longrightarrow x \mathrm{CO}_2(g)+\frac{y}{2} \mathrm{H}_2 \mathrm{O}(l) \\ & 15 \mathrm{~mL} \quad 15\left(x+\frac{y}{4}\right) \\ & 0 \quad 0 \quad 15 x\end{aligned}$

After combustion total volume

$\begin{aligned} & 330=V_{N_2}+V_{\mathrm{CO}_2} \\ & 330=300+15 x \\ & x=2\end{aligned}$

Volume of O₂ used

$\begin{aligned} & 15\left(x+\frac{y}{4}\right)=75 \\ & \left(x+\frac{y}{4}\right)=5 \\ & y=12\end{aligned}$

Important Questions of Some Basic Concepts in Chemistry

4. Hydrocarbons

Q: Which of the following alkanes will give only one monochlorination product?

A) Butane

B) 2-Methylpropane

C) Neopentane

D) Pentane

Solution:

- Neopentane (C(CH₃)₄) has all equivalent hydrogens.

- Only one product formed.

Answer: C

Q. Which hydrogen in compound (E) is easily replaceable during the bromination reaction in the presence of light?

$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\underset{\beta}{\mathrm{CH}}=\underset{\alpha}{\mathrm{CH}}$

(E)

Option 1:

α− hydrogen

Option 2:

γ-hydrogen

Option 3:

δ-hydrogen

Option 4:

β - hydrogen

Correct Answer:

γ-hydrogen

Solution:

γ-Hydrogen is easily replaceable during bromination reaction in the presence of light because, in the reaction, a free radical is formed as an intermediate at the gamma position which is very stable due to allylic conjugation.

$\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}-\mathrm{CH}=\mathrm{CH}_2$

Hence, the answer is option (2).

Important Questions of Hydrocarbons

5. Organic Compounds Containing Oxygen

Q: Which test is used to distinguish between aldehydes and ketones?

A) Tollen's Test

B) Benedict's Test

C) Both A and B

D) Fehling's Test

Solution:

- Aldehydes are oxidised easily, giving positive Tollen's (silver mirror) and Fehling's/Benedict's (red precipitate).

- Ketones do not respond.

Answer: C

Q. For standardising NaOH solution, which of the following is used as a primary standard?

Option 1:

Ferrous Ammonium Sulfate

Option 2:

dil. HCl

Option 3:

Oxalic acid

Option 4:

Sodium tetraborate

Correct Answer:

Oxalic acid

Solution:

As we have learned

Primary Standard -

The primary standard is a substance of known high purity that may be dissolved in a known volume of solvent to give a primary standard solution. It is a reference chemical used to measure an unknown concentration of another known chemical.

Important Questions of Organic Compounds Containing Oxygen

Also Read:

• JEE Main 2026 Chemistry Important Notes and Formulas

• JEE Main 2026 Important Notes and Formulas for Physics

• JEE Main 2026 Important Notes and Formulas for Maths

JEE Practice Questions with Answers: Maths

These chapters listed below have the highest chance of repeating. Therefore, you must practice the questions from the chapters listed below.

You can also use: JEE Main 2026 - A complete preparation strategy.

Questions:

1. Co-ordinate Geometry

Q : The equation of a line passing through (2,3) and perpendicular to the line 3x−4y+5=0 is:

Solution:

- Slope of given line =34.

- Required line ⊥ this, so slope =−43.

- Equation: y−3=−43(x−2).

$4 x+3 y-17=0$

Q. The area enclosed by the curves xy+4y=16 and x+y=6 is equal to :

Option 1:

$28-30 \log _e 2$

Option 2:

$30-28 \log _{\mathrm{e}} 2$

Option 3:

$30-32 \log _e 2$

Option 4:

$32-30 \log _{\mathrm{e}} 2$

Correct Answer:

$30-32 \log _e 2$

Solution:

$y(x+4)=16 \& y=6-x$

solve both curves,

$(6-x)(x+4)=16$

$6 x-x^2+24-4 x=16$

$x^2-2 x-8=0 \Rightarrow x=-2,4$

Important Questions of

2. Integral Calculus

Q: Evaluate ∫0πsin3⁡xdx.

Solution:

$\sin ^3 x=\sin x\left(1-\cos ^2 x\right)$

$\int_0^\pi \sin ^3 x d x=\int_0^\pi \sin x d x-\int_0^\pi \sin x \cos ^2 x d x$

- First integral: ∫0πsin⁡xdx=2.

- Second: let u=cos⁡x,du=−sin⁡xdx.

$\int_0^\pi \sin x \cos ^2 x d x=-\int_1^{-1} u^2 d u=\int_{-1}^1 u^2 d u=\frac{2}{3}$

So result =2−23=43.

Q.

The integral ∫1(x−1)3(x+2)54 dx is equal to : (where C is a constant of integration)

Option 1:

$\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{5}{4}}+C$

Option 2:

$\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$

Option 3:

$\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{5}{4}}+C$

Option 4:

$\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{1}{4}}+C$

Correct Answer:

$\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$

Solution:

$\begin{aligned} & I=\int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}} d x \\ & I=\int \frac{1}{(x-1)^2\left(\frac{x+2}{x-1}\right)^{\frac{5}{4}}} d x \\ & \text { Let } \frac{x+2}{x-1}=t \Rightarrow \frac{(x-1)-(x+2)}{(x-1)^2} d x=d t\end{aligned}$

$\begin{aligned} & \Rightarrow I=-\frac{1}{3} \int \frac{d t}{t^{\frac{5}{4}}}=-\frac{1}{3} \times \frac{t^{-\frac{5}{4}+1}}{-\frac{5}{4}+1}+c=\frac{4}{3} t^{-\frac{1}{4}}+c \\ & =\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+c\end{aligned}$

option (2)

Important Questions of Co-ordinate Geometry

3. Limit, Continuity & Differentiability

Q: Evaluate limx→0sin⁡5x.

Solution:

$\lim _{x \rightarrow 0} \frac{\sin 5 x}{x}=\lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x} \cdot 5=1 \cdot 5=5$

Q. If the maximum value of a , for which the function fa(x)=tan−1⁡2x−3ax+7 is nondecreasing in (−π6,π6), is a―, then fa―(π8) is equal to

Option 1:

$8-\frac{9 \pi}{4\left(9+\pi^2\right)}$

Option 2:

$8-\frac{4 \pi}{9\left(4+\pi^2\right)}$

Option 3:

$8\left(\frac{1+\pi^2}{9+\pi^2}\right)$

Option 4:

$8-\frac{\pi}{4}$

Correct Answer:

$8-\frac{9 \pi}{4\left(9+\pi^2\right)}$

Important Questions of Limit, Continuity & Differentiability

4. Sets, Relations and Functions

Q: If A={1,2,3},B={a,b}, find the number of relations from A to B.

Solution:

- Total ordered pairs possible =|A|×|B|=3×2=6.

- Each pair can be either in relation or not → total =26=64.

Answer: 64 relations

Q. Consider the following two relations on the set A={a,b,c},

R₁={(c,a),(b,b),(a,c),(c,c),(b,c),(a,a)}

and R₂={(a,b),(b,a),(c,c),(c,a),(a,a),(b,b),(a,c)}.

Then :

Option 1:

Both R₁ and R₂ is symmetric.

Option 2:

R₁ is not symmetric but it is transitive.

Option 3:

R2 is symmetric but it is not transitive.

Option 4:

both R1 and R₂ are transitive.

Correct Answer:

R₂ is symmetric, but it is not transitive.

Solution:

As we learned

REFLEXIVE RELATION: A relation R in A is said to be reflexive if aRa,∀a∈A

SYMMETRIC RELATION:
A relation $R$ on a set $A$ is said to be symmetric if

$$
a R b \Rightarrow b R a, \quad \forall a, b \in A .
$$

TRANSITIVE RELATION:
A relation $R$ on a set $A$ is said to be transitive if

$$
a R b \text { and } b R c \Rightarrow a R c, \quad \forall a, b, c \in A \text {. }
$$

Now,

R1:{(c,a)(b,b)(a,c)(c,c),(b,c)(a,a)}

Reflexive: R1 is reflexive since ( a,a), ( b,b ) and ( c,c ) are present.

Symmetric: (b,c) is present, but (c,b) is not, so not symmetric

Transitive: (b,c) and (c, a) belong to the relation, but (b, a) does not. So, it is not transitive.

R₂={(a,b),(b,a),(c,c),(c,a),(a,a),(b,b),(a,c)

Reflexive: since ( a,a ), ( b,b ) and ( c,c ) are present, it is reflexive

Symmetric: (a,b) and (b, a) both are present. Also (c, a) and (a,c) are present. So, it is symmetric

Transitive: (b, a), (a,c) are present but (b,c) is not present, so not transitive

Hence, the answer is the option 3.

Important Questions of Sets, Relations and Functions

5. Complex Numbers & Quadratic Equations

Q : If z=3+4i, find |z| and arg⁡(z).

Solution:

$\begin{gathered}|z|=\sqrt{3^2+4^2}=\sqrt{9+16}=5 \\ \arg (z)=\tan ^{-1}\left(\frac{4}{3}\right)\end{gathered}$

So, |z|=5,arg⁡(z)=tan−1⁡(43).

Q. Let z=1+ai be a complex number, a>0, such that z3 is a real number. Then the sum $1+z+z^2+\cdots+z^{11}$ is equal to :

Option 1:

$-1250 \sqrt{3} i$

Option 2:

$1250 \sqrt{3} i$

Option 3:

$1365 \sqrt{3} i$

Option 4:

$-1365 \sqrt{3} i$

Correct Answer:

$-1365 \sqrt{3} i$

Solution:

$z=1+a i, a>0$

given that z₃ is a real number.

$\begin{aligned} z^3 & =(1+a i)^3 \\ & =1-3 a^2+3 a i-a^3 i \\ & =1-3 a^2+\left(3 a-a^3\right) i\end{aligned}$

since, z is real number, so imaginary part is 0

$\begin{aligned} & \Rightarrow 3 a-a^3=0 \\ & \Rightarrow a=\sqrt{3}, \quad a>0 \\ & \Rightarrow z=1+i \sqrt{3} \\ & \Rightarrow z=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)\end{aligned}$

Now,

$\begin{aligned} & 1+z+z^2+\cdots+z^{11}=\frac{1-z^{12}}{1-z} \\ & \frac{1-2^{12}(\cos 4 \pi+i \sin 4 \pi)}{1-(1+i \sqrt{3})}=\frac{1-2^{12}}{-i \sqrt{3}} \\ & =\frac{4095}{i \sqrt{3}}=-1365 i \sqrt{3}\end{aligned}$

Hence, the answer is option 4.

Important Questions of Complex Numbers & Quadratic Equations

JEE Mains Important Questions with Answers FAQ:

Q1. How can I get JEE Mains important questions with answers?
You can download the chapterwise JEE Mains important questions with detailed answers in the PDF link provided in this article above. Prepare the resource properly to get a good rank in the exam.

Q.2 How often these questions are asked in the examination?

Almost every year, there will be questions that would relate to the same concepts, especially on important chapters high in weightage.

Q.3 Can I study through PDFs only for revision?

PDFs are great for final revision; however, for best results, revise them along with your short notes and formula sheets too.

Q.4 Are the solutions in the PDF explained well?

Yes, all the solutions are explained step-by-step and are aligned with the latest JEE Main marking scheme.