Contents
The JEE Main 2027 Sample Paper serves as a crucial study resource for the Joint Entrance Examination Main. If you are participating in the JEE Main conducted by the National Testing Agency (NTA), you have the option to download the PDF of the JEE Main model question paper for preparation. This sample paper includes practice questions aligned with the JEE Main syllabus, offering insight into the test’s expectations, difficulty level, and aiding in the enhancement of your overall preparation.
JEE Mains 2027: Sample Paper
What is JEE Main 2027 Sample Paper?
The JEE Main 2026 Sample Paper is a set of practice questions that mirrors the syllabus of the Joint Entrance Examination Main. It includes model questions designed to provide candidates with a comprehensive understanding of the topics covered in the exam. Additionally, the sample paper outlines the exam’s format and structure, allowing candidates to acquaint themselves with the scheme of the Engineering entrance exam conducted by the National Testing Agency (NTA).
JEE Main 2027 Sample Paper by NTA (PDF)
- JEE Main Physics Sample Paper
- JEE Main Chemistry Sample Paper
- JEE Main Mathematics Sample Paper
- JEE Main B.Planning Sample Paper
A block of mass \(m\) is attached to a spring of spring constant \(k\) and suspended vertically. If the mass is released from rest when the spring is in its natural length, the maximum extension produced in the spring is:
Solution: By Work-Energy Theorem: \(mgx = \frac{1}{2}kx^2 \implies x = \frac{2mg}{k}\).
A disk of mass \(M\) and radius \(R\) rolls without slipping down an inclined plane of angle \(\theta\). The acceleration of the center of mass of the disk is:
Solution: Acceleration \(a = \frac{g \sin\theta}{1 + I/(MR^2)}\). For a disk, \(I = \frac{1}{2}MR^2 \implies a = \frac{2}{3}g\sin\theta\).
Two charges \(+q\) and \(-q\) are placed at a distance \(2a\) apart. The electric potential at a point on the equatorial axis at a distance \(r\) (\(r \gg a\)) from the center of the dipole is:
Solution: Equatorial potential of an electric dipole is zero because distances to \(+q\) and \(-q\) are identical.
A Carnot engine operating between temperatures \(T_1 = 500\text{ K}\) and \(T_2 = 300\text{ K}\) absorbs \(600\text{ J}\) of heat per cycle. The work done per cycle is:
Solution: Efficiency \(\eta = 1 – \frac{300}{500} = 0.40\). Work \(W = \eta \times Q_1 = 0.40 \times 600 = 240\text{ J}\).
A wire of length \(L\) carries a current \(I\) along the positive x-axis in a magnetic field \(\vec{B} = B_0(\hat{i} + 2\hat{j} – \hat{k})\). The magnitude of force on the wire is:
Solution: \(\vec{F} = I(L\hat{i} \times B_0(\hat{i} + 2\hat{j} – \hat{k})) = ILB_0(2\hat{k} + \hat{j})\). Magnitude \(|\vec{F}| = ILB_0\sqrt{2^2+1^2} = ILB_0\sqrt{5}\).
In Young’s double slit experiment, if the distance between slits is halved and distance to screen is doubled, the fringe width becomes:
Solution: \(\beta = \frac{\lambda D}{d}\). New fringe width \(\beta’ = \frac{\lambda (2D)}{(d/2)} = 4\beta\).
The de Broglie wavelength of an electron accelerated through a potential difference of \(V\) volts is approximately:
Solution: \(\lambda = \sqrt{\frac{150}{V}}\ \text{\AA} = \frac{12.27}{\sqrt{V}}\ \text{\AA}\).
A nucleus \(X\) decays as: \(X \xrightarrow{\alpha} X_1 \xrightarrow{\beta^-} X_2 \xrightarrow{\alpha} X_3\). If \(X\) has atomic number 92 and mass number 238, those of \(X_3\) are:
Solution: Mass number \(= 238 – 4 – 0 – 4 = 230\). Atomic number \(= 92 – 2 + 1 – 2 = 89\).
A uniform rod of length \(1\text{ m}\) and mass \(2\text{ kg}\) is pivoted at one end. A force of \(6\text{ N}\) is applied perpendicular to the rod at its free end. The angular acceleration produced is ________ \(\text{rad/s}^2\).
Solution: Torque \(\tau = F \times L = 6\). Moment of inertia \(I = \frac{1}{3}mL^2 = \frac{2}{3}\). Angular acceleration \(\alpha = \tau/I = 6 / (2/3) = 9\text{ rad/s}^2\).
In a series LCR circuit with \(R = 10\ \Omega\), \(L = 0.1\text{ H}\), and \(C = 10\ \mu\text{F}\), the resonant angular frequency \(\omega_0\) is ________ \(\text{rad/s}\).
Solution: \(\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 10^{-5}}} = \frac{1}{10^{-3}} = 1000\text{ rad/s}\).
Which of the following molecules has a square planar shape?
Solution: \(\text{XeF}_4\) has 4 bond pairs and 2 lone pairs (\(sp^3d^2\) hybridization), adopting a square planar geometry.
The set of quantum numbers not allowed for an electron in an atom is:
Solution: For \(n=2\), maximum allowed value for \(l\) is \(n-1 = 1\). Thus \(l=2\) is invalid.
An aqueous solution of which salt will have \(\text{pH} > 7\) at \(25^\circ\text{C}\)?
Solution: \(\text{CH}_3\text{COONa}\) is a salt of a weak acid and strong base, undergoing anionic hydrolysis to make the solution basic (\(\text{pH} > 7\)).
When ethyl bromide reacts with alcoholic \(\text{KOH}\), the primary organic product obtained is:
Solution: Alcoholic \(\text{KOH}\) causes dehydrohalogenation via an E2 mechanism to yield ethene.
The major product formed when phenol reacts with concentrated \(\text{HNO}_3\) in the presence of concentrated \(\text{H}_2\text{SO}_4\) is:
Solution: Nitration of phenol using concentrated nitrating mixture produces 2,4,6-trinitrophenol (Picric acid).
Which of the following transition metal ions is diamagnetic?
Solution: \(\text{Zn}^{2+}\) has fully filled d-orbitals (\(3d^{10}\)), meaning no unpaired electrons (diamagnetic).
In a zero-order reaction \(A \rightarrow B\), the half-life (\(t_{1/2}\)) is directly proportional to:
Solution: For a zero-order reaction, \(t_{1/2} = \frac{[A]_0}{2k} \propto [A]_0\).
The IUPAC name of \(\text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CHO}\) is:
Solution: Numbering starts from the aldehyde carbon: C1 is -CHO, C3 has the -OH group. Hence, 3-hydroxybutanal.
The molarity of a solution prepared by dissolving \(4\text{ g}\) of \(\text{NaOH}\) in enough water to form \(250\text{ mL}\) of solution is ________ \(\text{M}\).
Solution: Moles of \(\text{NaOH} = \frac{4}{40} = 0.1\). Molarity \(= \frac{0.1}{0.250} = 0.4\text{ M}\).
The standard potential \(E^\circ\) for \(\text{Zn}^{2+}/\text{Zn}\) is \(-0.76\text{ V}\) and for \(\text{Cu}^{2+}/\text{Cu}\) is \(+0.34\text{ V}\). The standard EMF of the cell \(\text{Zn}|\text{Zn}^{2+} || \text{Cu}^{2+}|\text{Cu}\) is ________ \(\text{V}\).
Solution: \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}} = +0.34 – (-0.76) = 1.10\text{ V}\).
The maximum value of the function \(f(x) = 3\sin x + 4\cos x\) is:
Solution: Maximum value of \(a\sin x + b\cos x = \sqrt{a^2+b^2} = \sqrt{3^2+4^2} = 5\).
If \(\lim_{x \to 0} \frac{\sin(ax)}{bx} = 2\), where \(a, b \neq 0\), then the value of \(\frac{a}{b}\) is:
Solution: \(\lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b} \lim_{x \to 0} \frac{\sin(ax)}{ax} = \frac{a}{b} = 2\).
The derivative of \(\tan^{-1} x\) with respect to \(\cot^{-1} x\) is:
Solution: \(\frac{d(\tan^{-1}x)}{dx} = \frac{1}{1+x^2}\) and \(\frac{d(\cot^{-1}x)}{dx} = -\frac{1}{1+x^2}\). Dividing yields \(-1\).
The value of the determinant \(\begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix}\) is:
Solution: Apply \(C_3 \rightarrow C_3 + C_2\). \(C_3\) becomes \((a+b+c)[1, 1, 1]^T\), making \(C_1\) and \(C_3\) proportional. Hence, value is 0.
The area bounded by the curve \(y = x^2\) and the line \(y = 4\) is:
Solution: Area \(= 2 \int_{0}^{4} \sqrt{y}\, dy = 2 \left[ \frac{2}{3} y^{3/2} \right]_{0}^{4} = \frac{32}{3}\).
If two dice are thrown simultaneously, the probability of getting a sum of 8 is:
Solution: Favorable outcomes: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 cases. Total cases = 36. \(P = 5/36\).
The general solution of the differential equation \(\frac{dy}{dx} = e^{x-y}\) is:
Solution: Separating variables: \(e^y dy = e^x dx \implies e^y = e^x + C\).
The vector equation of a line passing through \((1, 2, 3)\) and parallel to \(3\hat{i} + 2\hat{j} – 2\hat{k}\) is:
Solution: Equation of line: \(\vec{r} = \vec{a} + \lambda\vec{b}\). Here \(\vec{a} = \hat{i}+2\hat{j}+3\hat{k}\) and \(\vec{b} = 3\hat{i}+2\hat{j}-2\hat{k}\).
The sum of the roots of the quadratic equation \(2x^2 – 8x + 5 = 0\) is ________.
Solution: Sum of roots \(= -b/a = -(-8)/2 = 4\).
Evaluate the definite integral \(\int_{0}^{2} (3x^2 + 2x + 1) \, dx\). The numerical answer is ________.
Solution: \(\left[ x^3 + x^2 + x \right]_0^2 = (2^3 + 2^2 + 2) – 0 = 8 + 4 + 2 = 14\).
JEE Main 2026 Sample Paper
- JEE Main Sample Paper – Set A
- JEE Main Sample Paper – Set B
- JEE Main Sample Paper – Set C
- JEE Main Sample Paper – Set D
JEE Main 2026 Sample Paper Full Length (PDF)
- Practice Paper 1
- Practice Paper 2
- Practice Paper 3
- Practice Paper 4
- Practice Paper for B.Arch Set A
- Practice Paper for B.Arch Set B
- B.Plan Practice Paper
- Drawing Test Practice Paper
JEE Main 2026 Sample Paper PDF
The complete model question paper is as follows.
JEE Main
JEE Main (Joint Entrance Examination Main) is a national-level entrance examination in India for admission to undergraduate engineering and architecture programs. It’s a highly competitive exam, and the scores are used for admission to top engineering colleges, including the National Institutes of Technology (NITs) and the Indian Institutes of Information Technology (IIITs).
Quick Links:
- JEE Mains syllabus
- JEE Mains Question Papers
- JEE Mains Marks, Percentile, and Rank
- JEE Main Exam Pattern
- Sample Paper
- Login
- Application Form
- Application Form Correction
- City Allotment
- Response Sheet
- Admit Card
- Answer Key
- Result
- IIT JEE Rank Predictor
- IIT JEE Study Material and Books
JEE Main Sample Paper – An Overview
| Aspects | Details |
|---|---|
| Examination | JEE Main |
| JEE Main Full Form | Joint Entrance Examination Main |
| Educational Resource | Model Paper of JEE Main |
| More About The Exam | JEE Main |
| Similar Exams | Engineering Entrance Exams |
| Official Body | NTA |
| NTA Full Form | National Testing Agency |
| Scale of Exam | National Level |
| Official Website | jeemain.nta.ac.in |
| Programs | BE, B.Tech, B.Arch, B.Planning |
| Score Accepting Institutes | Engineering Colleges across India |
| Regions Of These Institutes | India |
