JEE Mains Important Questions with Answers - To strategise for the JEE Mains exam properly, you must also focus on the JEE Mains important questions with answers. There is a set of questions that are most repeated and often carry a high weightage in the JEE Mains exam 2026. If you are aiming to score high marks with a good rank, you must practice these kinds of questions at least once. In this article, we have given you all the resources that will help you understand and solve these JEE Main questions, as you can also refer to the answer. Let’s begin by understanding the exam pattern.
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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.
Also Read:
The following table will give you the most weighted chapters. Practising from these might help you understand the concepts and types of questions asked from these chapters.
Chapter Name | Weightage |
Optics | 13.26% |
Electrostatics | 10.74% |
Properties of Solids and Liquids | 9.05% |
Physics and Measurement | 6.11% |
Rotational Motion | 6.53% |
Q: A convex lens of focal length 20 cm is placed 30 cm away from an object. Find the nature and position of the image formed.
Solution:
Using lens formula:
$$
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \
& \frac{1}{20}=\frac{1}{v}-\frac{1}{-30} \
& \frac{1}{20}=\frac{1}{v}+\frac{1}{30} \
& \frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60} \
& v=60 \mathrm{~cm} \text { (real, inverted image formed). }
\end{aligned}
$$
Q. 2 In an experiment for determination of the refractive index of glass of a prism by
Option 1:
1.5
Option 2:
1.6
Option 3:
1.7
Option 4:
1.8
Solution:
If
$$
\mu=\frac{\sin i}{\sin r}=\frac{\sin \left(A+\delta_m\right) / 2}{\sin A / 2} \cdots(i)
$$
where,
Given,
So, angle of deviation by a glass prism
$$
\delta=i+e-A \Rightarrow 40^{\circ}=35^{\circ}+79^{\circ}-A
$$
i.e.Angle of prism
Such that,
Let us put
$$
\begin{aligned}
1.5 & =\frac{\sin \left(\frac{A+\delta_{\text {min }}}{2}\right)}{\sin A / 2} \
\Rightarrow 1.5 & =\frac{\sin \left(\frac{74^{\circ}+\delta_{\text {min }}}{2}\right)}{\sin 37^{\circ}} \
\Rightarrow 0.9 & =\sin \left(37^{\circ}+\frac{\delta_{\text {min }}}{2}\right)\left(\because \sin 37^{\circ} \approx 0.6\right) \
\sin 64^{\circ} & =\sin \left(37^{\circ}+\frac{\delta_{\text {min }}}{2}\right)\left(\because \sin 64^{\circ}=0.9\right) \
\Rightarrow \delta_{\text {min }} & \approx 54^{\circ}
\end{aligned}
$$
This angle is greater than the
Hence, the closest answer will be 1.5.
Hence, the answer is option (1).
Q: Two equal point charges of
Solution:
At midpoint, fields due to both charges are equal and opposite
So, net
Q. A charge of +4 C is kept at a distance of 50 cm from a charge of -6 C . Find the two points where the potential is zero
Option 1:
Internal point lies at a distance of 20 cm from 4C charge and external point lies at a distance of 100 cm from 4C charge.
Option 2:
Internal point lies at a distance of 30 cm from 4C charge and external point lies at a distance of 100 cm from 4C charge
Option 3:
Potential is zero only at 20 cm from 4C charge between the two charges
Option 4:
Potential is zero only at 20 cm from - 6C charge between the two charges
Correct Answer:
Internal point lies at a distance of 20 cm from 4C charge and external point lies at a distance of 100 cm from 4C charge.
Solution:
The charges are given as,
$$
\begin{aligned}
& q_1=4 \mathrm{C} \
& q_2=-6 \mathrm{C}
\end{aligned}
$$
Let the distance of pt. P from 4 C charge be x , consequently the distance of charge -6 C from pt. P is (
The potential at P is given as,
$$
\begin{aligned}
& 0=k \frac{q_1}{r_1}+k \frac{9_2}{r_2} \
& 0=k \frac{4}{x}+k \frac{(-6)}{0.5-x} \
& 0=k \frac{4}{x}+k \frac{(-6)}{0.5-x} \
& x=0.2 \mathrm{~m}
\end{aligned}
$$
Thus, the potential is zero only at 20 cm from the 4C charge between the two charges.
Hence, the answer is the option (1).
Important Questions of Electrostatics
Q: A steel wire of length 2 m and area of cross-section
Solution:
$$
\begin{aligned}
& \Delta L=\frac{F L}{A Y} \
& =\frac{100 \times 2}{\left(1 \times 10^{-6}\right)\left(2 \times 10^{11}\right)} \
& =1 \times 10^{-3} \mathrm{~m}=1 \mathrm{~mm}
\end{aligned}
$$
Q. Two tubes of radii
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Rate of liquid flow through a tube
Since the rate of flow is the same across the two tubes.
$$
\begin{aligned}
& \therefore \frac{\pi P_1 r_1^4}{8 \eta l_1}=\frac{\pi P_2 r_2^4}{8 \eta l_2} \
& \quad \frac{P_1 r_1^4}{l_1}=\frac{P_2 r_2^4}{l_2} \
& \text { Since } P_1=4 P_1 \quad l_2=l_1 / 4 \
& \therefore r_2^4=\frac{r_1^2}{16} \
& \therefore r_2=\frac{r_1}{2}
\end{aligned}
$$
Hence, the answer is option 4.
Important Questions of Properties of Solids and Liquids
Q: The time period of a simple pendulum is given by
Solution:
Using the formula for percentage error propagation:
$$
\frac{\Delta T}{T}=\frac{1}{2}\left(\frac{\Delta L}{L}+\frac{\Delta g}{g}\right)
$$
Substitute the given values:
$$
\begin{aligned}
\frac{\Delta T}{T} & =\frac{1}{2}\left(\frac{2}{100}+\frac{10}{980}\right) \
\frac{\Delta T}{T} & =\frac{1}{2}(0.02+0.0102) \
\frac{\Delta T}{T} & =0.0151=1.51 \%
\end{aligned}
$$
Therefore, the maximum percentage error in the time period is
Q. A physical quantity
$$
P=a^{1 / 2} b^2 c^3 d^{-4}
$$
If the relative errors in the measurement of
Option 1:
8\%
Option 2:
12\%
Option 3:
32\%
Option 4:
25\%
Correct Answer:
32\%
Solution:
Given :
$$
\begin{aligned}
& \Rightarrow \frac{\Delta \mathrm{b}}{\mathrm{~b}} \times 100=1 \
& \Rightarrow \frac{\Delta \mathrm{c}}{\mathrm{c}} \times 100=3
\end{aligned}
$$
Now, given,
$$
\begin{aligned}
& \Rightarrow \frac{\Delta \mathrm{P}}{\mathrm{P}} \times 100=\frac{2}{2} \pm 2 \times 1 \pm 3 \times 3 \pm 4 \times 5 \
& \Rightarrow \frac{\Delta \mathrm{P}}{\mathrm{P}} \times 100=32 \%
\end{aligned}
$$
Hence, the answer is option 3
Important Questions of Physics and Measurement
Q: A solid sphere of mass 2 kg and radius 0.2 m rolls without slipping with a linear speed of
Solution:
KE
$$
=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2
$$
For sphere,
So KE
$$
=25+10=35 \mathrm{~J} .
$$
Q. A solid sphere of mass
Correct Answer:
140
Solution:
Now for solid sphere
Given
$$
\begin{aligned}
& I_2=\left(\frac{M}{8} r^2\right)(2 / 5)=\left(\frac{M}{8}\right)\left(\frac{R}{2}\right)^2 2 / 5=I_2 \
& I_1 / I_2=\frac{\frac{7 M}{8} \frac{(2 R)^2}{2}}{\frac{2}{5} \frac{M}{8}\left(\frac{R}{2}\right)^2}=140
\end{aligned}
$$
Hence, the answer is the option (2).
These chapters listed below have the highest chance of repeating again. The weightage given is the weightage of these chapters according to past trends. Therefore, the questions from these chapters are considered to be JEE Mains important questions and answers.
Chapter Name | Weightage |
8.21% | |
Chemical Thermodynamics | 6.95% |
Some basic concepts in chemistry | 7.79% |
Hydrocarbons | 5.26% |
Organic Compounds containing Oxygen | 5.26% |
Questions:
Q: The IUPAC name of
A) Pentaamminechloridocobalt(III) chloride
B) Pentamminecobalt(III) chloride
C) Pentaamminecobalt(II) chloride
D) Pentamminechloridocobalt(II) chloride
Solution:
- Central metal: Co
- Oxidation state of
- Correct name
Answer: A
Q. The total number of possible isomers for square-planar
Option 1:
8
Option 2:
12
Option 3:
16
Option 4:
24
Correct Answer:
12
Solution:
As we have learnt,
Now, the given square planar complex can show geometrical isomerism as well as linkage isomers.
The number of isomers possible is listed below:
\begin{tabular}{|l|l|}
\hline Compound & Number of Isomers \
\hline
\hline
\hline
\hline
\hline
\end{tabular}
Thus, the total number of Isomers is 12.
Hence, the answer is an option (2).
Important Questions of Co-ordination Compounds
Q: For a reaction,
Solution:
$$
\begin{aligned}
& \Delta G=\Delta H-T \Delta S \
& =-120-(300 \times-0.200) \
& =-120+60 \
& =-60 \mathrm{~kJ}
\end{aligned}
$$
Since
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,
$$
\begin{gathered}
\Delta H=n C_p \Delta T \
\Delta S=n \operatorname{Rln}\left(V_f / V_i\right) \geqslant 0
\end{gathered}
$$
Enthalpy remains constant but entropy increases.
Hence, the answer is an option (2).
Important Questions of Chemical Thermodynamics
Q: 0.5 g of an impure sample of
Solution:
Moles
Q. At 300 K and
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Volume of
Volume of
$$
\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2(g)+\frac{y}{2} \mathrm{H}_2 \mathrm{O}(l) \
& 15 \mathrm{ml} \quad 15\left(x+\frac{y}{4}\right) \
& 0 \quad 0 \quad 15 \mathrm{x}-
\end{aligned}
$$
After combustion total volume
$$
\begin{aligned}
& 330=V_{N_2}+V_{\mathrm{CO}_2} \
& 330=300+15 \mathrm{x} \
& \mathrm{x}=2
\end{aligned}
$$
Volume of
$$
\begin{aligned}
& 15\left(x+\frac{y}{4}\right)=75 \
& \left(x+\frac{y}{4}\right)=5 \
& y=12
\end{aligned}
$$
Important Questions of Some Basic Concepts in Chemistry
Q: Which of the following alkanes will give only one monochlorination product?
A) Butane
B) 2-Methylpropane
C) Neopentane
D) Pentane
Solution:
- Neopentane
- Only one product formed.
Answer: C
Q. Which hydrogen in compound (E) is easily replaceable during bromination reaction in presence of light ?
$$
\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\underset{\beta}{\mathrm{CH}}=\underset{\alpha}{\mathrm{CH}}
$$
(E)
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
\begin{aligned}
&\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}-\mathrm{CH}=\mathrm{CH}_2\
&\text { Hence, the answer is the option (2). }
\end{aligned}
Important Questions of Hydrocarbons
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 oxidized easily, giving positive Tollen's (silver mirror) and Fehling's/Benedict's (red precipitate).
- Ketones do not respond.
Answer: C
Q. For standardizing 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:
These chapters listed below have the highest chance of repeating. Therefore, you must practice the questions from the chapters listed below.
Chapter Name | weightage |
Co-ordinate geometry | 17.89% |
Integral Calculus | 10.74% |
Limit , continuity and differentiability | 8.84% |
Sets, Relations and Functions | 7.79% |
Complex numbers and quadratic equations | 6.95% |
You can also use: JEE Main 2026 - A complete preparation strategy.
Questions:
Q : The equation of a line passing through
Solution:
- Slope of given line
- Required line
- Equation:
$$
4 x+3 y-17=0
$$
Q. The area enclosed by the curves
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
Q: Evaluate
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:
- Second: let
$$
\int_0^\pi \sin x \cos ^2 x d x=-\int_1^{-1} u^2 d u=\int_{-1}^1 u^2 d u=\frac{2}{3}
$$
So result
Q.
The integral
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:
$$
\begin{aligned}
& I=\int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}} d x \
& \quad I=\int \frac{1}{(x-1)^2\left(\frac{x+2}{x-1}\right)^{\frac{5}{4}}} d x \
& \text { Let } \frac{x+2}{x-1}=t \Rightarrow \frac{(x-1)-(x+2)}{(x-1)^2} d x=d t
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow I=\frac{-1}{3} \int \frac{d t}{t^{\frac{5}{4}}}=\frac{-1}{3} \times \frac{t^{\frac{-5}{4}+1}}{-\frac{5}{4}+1}+c=\frac{4}{3} t^{\frac{-1}{4}}+c \
& =\frac{4}{3} \times\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+c
\end{aligned}
$$
option (2)
Important Questions of Co-ordinate Geometry
Q: Evaluate
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
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
Q: If
Solution:
- Total ordered pairs possible
- Each pair can be either in relation or not
Answer: 64 relations
Q. Consider the following two relations on the set
and
Then :
Option 1:
both
Option 2:
Option 3:
Option 4:
both
Correct Answer:
Solution:
As we learned
REFLEXIVE RELATION: A relation
SYMMETRIC RELATION: A relation
TRANSITIVE RELATION: A relation
Now,
Reflexive:
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.
Reflexive: since (
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
Q : If
Solution:
-
-
So,
Q. Let
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
$$
\begin{aligned}
z^3 & =(1+a i)^3=1-3 a^2+3 a i-a^3 i \
& =1-3 a^2+\left(3 a-a^3\right) i
\end{aligned}
$$
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,
$$
\begin{aligned}
& 1+z+z^2+\ldots \ldots+z^{11}=\frac{1\left(1-z^{12}\right)}{1-z^2} \
& \frac{1-2^{12}(\cos 4 \pi+i \sin 4 \pi)}{1-(1+i \sqrt{3})}=\frac{1-2^{12}}{-i \sqrt{3}} \
& =\frac{4095}{i \sqrt{3}}=-1365 \sqrt{3} i
\end{aligned}
$$
Hence, the answer is option 4.
Important Questions of Complex Numbers & Quadratic Equations
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.
On Question asked by student community
Hello,
The link to the question paper is attached here. Careers360 help students in their exam preparation. students can find all the required study materials and exam materials, exam materials, results and cutoffs on the website.
Thank you.
Hello,
Despite of the prior backlogs, you can appear for JEE Mains in 2026 because you passed class 12; that marks the basic eligibility.
Anyways, this would be regarded as the third JEE Mains attempt, because there's a limit that a candidate can attempt only a limited number of times;
You can get free mock tests for JEE 2026 on the Careers360 website. The mock tests are based on the latest exam pattern and help in understanding the level of questions and time management. You just need to register and start attempting the tests online.
Direct link for JEE mock
Hello,
If you passed 12th in 2025 and did not appear in JEE till now, then this is your eligibility:
JEE Main can be given for 3 consecutive years after 12th.
Each year has 2 sessions .
You are eligible in 2025, 2026 and 2027 .
Since you
Hello,
For JEE (for IT / engineering) :
10th percentage
There is no fixed minimum percentage needed in Class 10 to prepare for JEE.
You just need to pass Class 10 .
Important part is Class 12
You must study Physics, Chemistry, and Mathematics in Class 11 and 12 .
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