JEE Mains Physics PYQ Chapterwise PDF - Free Download

JEE Mains Physics PYQ Chapterwise PDF: Weightage

In this section, we will be seeing some of the high-weightage chapters and questions from those chapters. First, let's understand the weightage of the chapters in JEE Mains Physics.

JEE Main Physics Chapter-wise Questions with Solutions: Optics

You can access the pdf download here: Optics PYQs PDF

Some important questions asked in Optics are given below:

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 :

Option 1: 3 mm

Option 2: 9 mm

Option 3: 4.5 mm

Option 4: 1.5 mm

Correct Answer: 9 mm

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

Option 1: 1.5

Option 2: 1.6

Option 3: 1.7

Option 4: 1.8

Correct Answer: 1.5

Solution:

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

μ=sin⁡isin⁡r=sin⁡(A+δm)/2sin⁡A/2⋯(i)

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

Given, i=35^∘,δ=40^∘,e=79^∘

So, the angle of deviation by a glass prism

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

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

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

Let us put $\mu = 1.5$ in eq. (i), we get

$1.5 = \dfrac{\sin\left(\dfrac{A + \delta_{\min}}{2}\right)}{\sin A/2}$

$\Rightarrow 1.5 = \dfrac{\sin\left(\dfrac{74^\circ + \delta_{\min}}{2}\right)}{\sin 37^\circ}$

$\Rightarrow 0.9 = \sin\left(37^\circ + \dfrac{\delta_{\min}}{2}\right) \quad \left(\because \sin 37^\circ \approx 0.6\right)$

$\sin 64^\circ = \sin\left(37^\circ + \dfrac{\delta_{\min}}{2}\right) \quad \left(\because \sin 64^\circ = 0.9\right)$

$37^\circ + \dfrac{\delta_{\min}}{2} = 64^\circ \Rightarrow \delta_{\min} \approx 54^\circ$

This angle is greater than the $40^\circ$ deviation angle already given. For greater $\mu$,
The deviation will be even higher. Hence, $\mu$ of the given prism should be less than $1.5$.

Hence, the closest answer will be $1.5$.

Hence, the answer is option (1).

JEE Mains Physics PYQ chapterwise: Electrostatics

You can access the pdf download here: Electrostatics PYQS PDF

Some questions asked from this chapter are given below:

Q. 1 Three charges Q,+q, and +q are placed at the vertices of a right-angled isosceles triangle as shown below. The net electrostatic energy of the configuration is zero if the value of Q is:

Option 1: $\frac{-q}{1+\sqrt{2}}$

Option 2: $\frac{-\sqrt{2} q}{\sqrt{2}+1}$

Option 3: +q

Option 4: -2q

Correct Answer:

$\frac{-\sqrt{2} q}{\sqrt{2}+1}$

Solution:

Potential Energy of a System of n Charges:

$U = K\left(\dfrac{Q_1 Q_2}{r_{12}} + \dfrac{Q_2 Q_3}{r_{23}} + \dfrac{Q_1 Q_3}{r_{13}}\right)$

wherein, for a system of 3 charges:

$U = K\left[\dfrac{q^2}{d} + \dfrac{Qq}{d} + \dfrac{Qq}{d\sqrt{2}}\right] = 0$

$\Rightarrow Q = \dfrac{-q\sqrt{2}}{\sqrt{2}+1}$

Q. 2 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 Δ, V, then:

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

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

Option 3: $\rho(r) \propto \dfrac{1}{r}$

Option 4: $\rho(r) \propto \dfrac{1}{r^2}$

Correct Answer: $\rho(r) \propto \dfrac{1}{r}$

Solution:

As we learned, the relation between field and potential:

$E = \dfrac{-dV}{dr}$

wherein $\dfrac{dV}{dr}$ is the potential gradient.

If P lies inside:

$E_{in} = \dfrac{1}{4\pi\epsilon_0} \dfrac{Qr}{R^3}$

$V_{in} = \dfrac{Q}{4\pi\epsilon_0} \dfrac{3R^2 - r^2}{2R^3}$

$E_{in} = \dfrac{\rho r}{3\epsilon_0}$

$V_{in} = \dfrac{\rho(3R^2 - r^2)}{6\epsilon_0}$

We know $E = \dfrac{-dV}{dr}$, here $\Delta V$ and $\Delta r$ are the same for any pair of surfaces.

Therefore, $E = \text{constant}$

Now, the electric field inside the spherical charge distribution:

$E = \dfrac{\rho}{3\epsilon_0} r$

$E$ would be constant if $\rho r = \text{constant}$

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

Hence, the answer is Option (3).

JEE Mains Physics PYQ chapterwise: Rotational Motion

You can access the pdf download here: Rotational Motion PYQs PDF

Some important questions to practice from this chapter are given below:

Q. 1 A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of the wheel is 10 kg, and the radius is 10 cm, and it can freely rotate without any friction. Initially, the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m, would be :

Option 1: $20 \ \text{rad/s}$

Option 2: $30 \ \text{rad/s}$

Option 3: $10 \ \text{rad/s}$

Option 4: $0 \ \text{rad/s}$

Correct Answer: $20 \ \text{rad/s}$

Solution:

$W_F = 20 \times 1 = 20 \ \text{J}$

$\therefore \quad \Delta KE = 20 \ \text{J} = \dfrac{1}{2} I\omega^2$

$I = MR^2 = 10 \times (0.1)^2 = 0.1 \ \text{kg m}^2$

$\therefore \quad 20 = \dfrac{1}{2} \times 0.1 \times \omega^2$

$\Rightarrow \omega = 20 \ \text{rad/s}$

Hence, the answer is Option (1).

Q. 2 In a physical balance working on the principle of moments, when a 5 mg weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct?

Option 1: The left arm is longer than the right arm

Option 2: Both arms are of the same length

Option 3: The left arm is shorter than the right arm

Option 4: Every object that is weighed using this balance appears lighter than its actual weight.

Correct Answer: The left arm is shorter than the right arm

Solution:

Point of application of force: The point at which, if the net force is assumed to be acting, then it will produce the same effect of both translation and rotation.

For balance:

Anticlockwise Moment $=$ Clockwise Moment

i.e. $\text{Load} \times \text{load arm} = \text{effort} \times \text{effort arm}$

$F_1 d_1 = F_2 d_2$

When $5mg$ weight is placed on the left side:

$(5mg + mg)d_1 = mg \cdot d_2$

$6mg \cdot d_1 = mg \cdot d_2$

$d_2 = 6d_1$

The load arm shifts to the left side, hence the left arm becomes shorter than the right arm.

Hence, the answer is Option (3).

JEE Main Physics Chapter-wise Questions with Solutions: Magnetic Effects of Current and Magnetism

You can access the pdf download here: Magnetic Effects of Current and Magnetism PYQs PDF

Some of the important questions asked in this chapter are given below:

Q.1 A magnetic dipole in a constant magnetic field has:

Option 1: maximum potential energy when the torque is maximum.

Option 2: zero potential energy when the torque is minimum.

Option 3: zero potential energy when the torque is maximum.

Option 4: minimum potential energy when the torque is maximum.

Correct Answer: zero potential energy when the torque is maximum.

Solution:

As we learned, Torque is given by:

$\vec{T} = \vec{M} \times \vec{B}, \quad M = NiA$

$T = MB\sin\theta = NBiA\sin\theta$

where $M$ is the magnetic moment.

Work done by current carrying coil:

$W = MB(1 - \cos\theta)$

For maximum torque, $\theta = 90^\circ$, at this value of $\theta$, potential energy will be zero.

Hence, the answer is Option (3).

Q.2 When a current of $5 \ \text{mA}$ is passed through a galvanometer having a coil of resistance $15 \ \Omega$, it shows full-scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range $0-10 \ \text{V}$ is:

Option 1: $1.985 \times 10^3 \ \Omega$

Option 2: $2.045 \times 10^3 \ \Omega$

Option 3: $2.535 \times 10^3 \ \Omega$

Option 4: $4.005 \times 10^3 \ \Omega$

Correct Answer: $1.985 \times 10^3 \ \Omega$

Solution:

Given:

$I_g$ (Galvanometer Current) $= 5 \ \text{mA}$

$G$ (Resistance of Galvanometer) $= 15 \ \Omega$

$R$ = resistance to be put in series with the galvanometer.

We know:

$V = I_g(R + G)$

$10 = 5 \times 10^{-3}(R + 15)$

$2000 = R + 15$

$\Rightarrow R = 1985 \ \Omega = 1.985 \times 10^3 \ \Omega$

Hence, the answer is Option (1).

JEE Mains Physics PYQ chapterwise: Physics and Measurement

You can access the pdf download here: Physics and Measurement PYQs PDF

Some questions from this chapter are given below:

Q.1 If time $(t)$, velocity $(v)$, and angular momentum $(l)$ are taken as the fundamental units. Then the dimension of mass $(m)$ in terms of $t$, $v$ and $l$ is:

Option 1: $\left[t^{-1} v^{1} l^{-2}\right]$

Option 2: $\left[t^{1} v^{2} l^{-1}\right]$

Option 3: $\left[t^{-2} v^{-1} l^{1}\right]$

Option 4: $\left[t^{-1} v^{-2} l^{1}\right]$

Correct Answer: $\left[t^{-1} v^{-2} l^{1}\right]$

Q.2 A, B, C, and D are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation $AD = C\ln(BD)$ holds. Then, which of the combinations is not a meaningful quantity?

Option 1: $A^2 - B^2C^2$

Option 2: $\dfrac{(A-C)}{D}$

Option 3: $\dfrac{A}{B} - C$

Option 4: $\dfrac{C}{BD} - \dfrac{(AD)^2}{C}$

Correct Answer: $\dfrac{(A-C)}{D}$

Solution:

Dimension: The power to which fundamental quantities must be raised to express the given physical quantities.

Given:

$A \cdot D = C\ln(B \cdot D)$

Hence $[B \cdot D]$ is dimensionless, which means:

$[B] = \left[\dfrac{1}{D}\right]$

and $[A \cdot D] = [C]$

In the expression $\dfrac{A-C}{D}$, the dimensions of $A$ are not the same as the dimensions of $C$, so this is not a meaningful quantity.

Hence, the answer is Option (2).

JEE Mains Physics PYQ chapterwise: Kinematics

You can access the pdf download here: Kinematics PYQs PDF

For some questions asked in this chapter, refer to the following:

Q.1 A ball having kinetic energy $KE$ is projected at an angle of $60^\circ$ from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?

Option 1: $\dfrac{KE}{8}$

Option 2: $\dfrac{KE}{4}$

Option 3: $\dfrac{KE}{16}$

Option 4: $\dfrac{KE}{2}$

Correct Answer: $\dfrac{KE}{4}$

Solution:

Initial K.E:

$KE = \dfrac{1}{2}mu^2$

Speed at the highest point:

$v = u\cos 60^\circ = \dfrac{u}{2}$

$KE_f = \dfrac{1}{2}m\left(\dfrac{u}{2}\right)^2$

$= \dfrac{1}{4} \times \dfrac{1}{2}mu^2$

$= \dfrac{KE}{4}$

Hence, the answer is Option (2).

Q.3 The magnetic susceptibility of a material of a rod is $499$. Permeability in vacuum is $4\pi \times 10^{-7} \ \text{H/m}$. The absolute permeability of the material of the rod is:

Option 1: $4\pi \times 10^{-4} \ \text{H/m}$

Option 2: $2\pi \times 10^{-4} \ \text{H/m}$

Option 3: $3\pi \times 10^{-4} \ \text{H/m}$

Option 4: $\pi \times 10^{-4} \ \text{H/m}$

Correct Answer: $2\pi \times 10^{-4} \ \text{H/m}$

JEE Mains Physics PYQ chapterwise: Atoms And Nuclei

For pdf download, visit this page: Atoms And Nuclei

For some questions asked in this chapter, refer to the following:

Q.1 The acceleration of an electron in the first orbit of the hydrogen atom $(n = 1)$ is:

Option 1: $\dfrac{h^2}{\pi^2 m^2 r^3}$

Option 2: $\dfrac{h^2}{8\pi^2 m^2 r^3}$

Option 3: $\dfrac{h^2}{4\pi^2 m^2 r^3}$

Option 4: $\dfrac{h^2}{4\pi m^2 r^3}$

Correct Answer: $\dfrac{h^2}{4\pi^2 m^2 r^3}$

Solution:

Bohr quantisation principle: The angular momentum of an electron in a stationary orbit is quantised.

$mvr = \dfrac{nh}{2\pi}$

For $n = 1$:

$mvr = \dfrac{h}{2\pi} \Rightarrow v = \dfrac{h}{2\pi mr}$

$\therefore a = \dfrac{v^2}{r} = \left(\dfrac{h}{2\pi mr}\right)^2 \cdot \dfrac{1}{r} = \dfrac{h^2}{4\pi^2 m^2 r^3}$

Hence, the answer is Option (3).

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Q.2 Imagine that a reactor converts all given mass into energy and that it operates at a power level of $10^9\ \text{W}$. The mass of the fuel consumed per hour in the reactor will be:
(velocity of light, $c = 3 \times 10^8\ \text{m/s}$)

Option 1: $0.96\ \text{gm}$

Option 2: $0.8\ \text{gm}$

Option 3: $4 \times 10^{-2}\ \text{gm}$

Option 4: $6.6 \times 10^{-5}\ \text{gm}$

Correct Answer: $4 \times 10^{-2}\ \text{gm}$

Solution:

Energy mass equivalence:

$\Delta E = \Delta m \cdot c^2$

where $\Delta m$ is the mass defect and $\Delta E$ is the energy released.

Amount of energy produced in 1 hour:

$E = 3600 \times 10^9\ \text{J} = 3.6 \times 10^{12}\ \text{J}$

This energy is produced by converting mass into energy, so
$\Delta m \cdot c^2 = 3.6 \times 10^{12}$:

$\Delta m = \dfrac{3.6 \times 10^{12}}{9 \times 10^{16}} = 0.4 \times 10^{-4}\ \text{kg}$

$\Delta m = 4 \times 10^{-2}\ \text{gm}$

Hence, the answer is Option (3).

JEE Mains Physics PYQ chapterwise: Thermodynamics

You can access the pdf download here: Thermodynamics PYQs PDF

For some questions asked in this chapter, refer to the following:

Q.1 $N$ moles of a diatomic gas in a cylinder are at a temperature $T$. Heat is supplied to the cylinder such that the temperature remains constant,t, but $n$ moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy $(nRT)$
of the gas?

Correct Answer: $0.5$

Solution:

Number of moles of diatomic gas $= N$

Hence, the number of moles of monoatomic gas $= 2N$

Since $T =$ constant and volume is also constant:

Work done $= 0$

Change in internal energy $=$ Heat supplied

Change in total kinetic energy $= n_f C_{vf} T - n_i C_{vi} T$

$n_i = N, \quad C_{vi} = \dfrac{5R}{2}$

$n_f = 2N, \quad C_{vf} = \dfrac{3R}{2}$

$\therefore \Delta K = 3nRT - \dfrac{5}{2}nRT = \dfrac{1}{2}nRT$

Hence, the answer is $0.5$.

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Q.2 In a process, the temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation $VT = K$, where $K$ is a constant. In this process, the temperature of the gas is increased by $\Delta T$. If the amount of heat absorbed by gas is $x \times R\Delta T$, then what will be the value of $x$ ($R$ isthe as constant)?

Correct Answer: $0.50$

Solution:

Given $VT = K$

$V\left[\dfrac{PV}{nR}\right] = K$

$PV^2 = nRK$

Here the polytropic index is $x = 2$, since $PV^2 = \text{constant}$

For a polytropic process:

$C = \dfrac{R}{1-x} + C_v$

$C = \dfrac{R}{1-2} + \dfrac{3R}{2} = -R + \dfrac{3R}{2} = \dfrac{R}{2}$

$\therefore \Delta Q = nC\Delta T = \dfrac{R}{2}\Delta T$

Therefore $x = 0.5$

Hence, the answer is $0.5$.

JEE Mains Physics PYQ chapterwise: Current Electricity

You can access the pdf download here: Current Electricity PYQs PDF

For some questions asked in this chapter, refer to the following:

Q.1 Two identical cells, when connected either in parallel or in series gives same current in an external resistance $5\ \Omega$. The internal resistance of each cell will be ____$\Omega$.

Correct Answer: $5$

Solution:

For parallel combination:

$r_{eq_1} = \dfrac{r}{2}, \quad E_{eq_1} = \dfrac{r}{2}\left(\dfrac{E}{r} + \dfrac{E}{r}\right) = E$

For series combination:

$r_{eq_2} = 2r, \quad E_{eq_2} = 2E$

Currents in each case:

$I_1 = \dfrac{E}{5 + \dfrac{r}{2}}, \quad I_2 = \dfrac{2E}{2r + 5}$

Since $I_1 = I_2$:

$2r + 5 = 2\left(5 + \dfrac{r}{2}\right)$

$2r + 5 = 10 + r$

$\therefore r = 5\ \Omega$

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Q.2 The resistance of an electric toaster has a temperature dependence given by $R(T) = R_0[1 + \alpha(T - T_0)]$ in its range of operation. At $T_0 = 300\ \text{K}$, $R = 100\ \Omega$ and at $T = 500\ \text{K}$, $R = 120\ \Omega$. The toaster is connected to a voltage source at $200\ \text{V}$, and its temperature is raised at a constant rate from $300$ to $500\ \text{K}$ in $30\ \text{s}$. The total work done in raising the temperature is:

Option 1: $400\ln\dfrac{1.5}{1.3}\ \text{J}$

Option 2: $200\ln\dfrac{2}{3}\ \text{J}$

Option 3: $400\ln\dfrac{5}{6}\ \text{J}$

Option 4: $300\ \text{J}$

Correct Answer: $400\ln\dfrac{5}{6}\ \text{J}$

Solution:

Given: $R(T) = R_0(1 + \alpha(T - T_0))$

Applying boundary conditions:

$120 = 100(1 + 200\alpha)$

$\therefore \alpha = 10^{-3}\ \text{K}^{-1}$

Since temperature increases at a constant rate from $300\ \text{K}$ to $500\ \text{K}$ in $30\ \text{s}$:

$T(t) = 300 + \dfrac{20t}{3}$

By Joule's Law, heat dissipated in a resistor is given by:

$W = \displaystyle\int_0^{30} \dfrac{V^2}{R}\, dt
= \displaystyle\int_0^{30} \dfrac{V^2}{R_0\left(1 + \alpha(T - T_0)\right)}\, dt
= \dfrac{V^2}{R_0}\displaystyle\int_0^{30} \dfrac{1}{1 + \dfrac{20\alpha t}{3}}\, dt$

Solving:

$W = 400\ln\left(\dfrac{6}{5}\right)$

Work done on resistor $= -W = 400\ln\dfrac{5}{6}\ \text{J}$

Hence, the answer is Option (3).

For the rest of the chapter-wise pyq, refer to the table below. You will find the entire pdf download for each of the chapters as mentioned below, along with their weightage:

Benefits of JEE Mains Physics PYQ Chapterwise Practice

Solving Physics PYQ chapterwise offers multiple advantages:

1. Identifying the easiest type or high-scoping chapters and concepts.

2. Identify the repetitive concepts in questions and study them properly.

3. Understanding how the NCERT theory is included in questions.

4. Improves knowledge and understanding of concepts. Plus, you must also refer to the JEE Main Latest Syllabus 2026.

5. Helps become faster and solve questions quickly.

6. Solve and practice JEE Main 2026 Important Formulas for Physics.

7. Boosts confidence for a real-time exam.