Top 10 Most Repeated Topics in Physics for JEE Mains - Boost Your Score in JEE 2026 Exam

Most Repeated Physics Topics for JEE Mains 2026 as Per Previous Years

In this section, we will be diving into the top 10 most repeated topics in physics for JEE Mains. Our experts have collected this data by analyzing the last 10 years question papers and JEE Main 2026 Physics syllabus. Along with the topic name, the total number of questions asked in the span of 10 years is also given in the table below. This data about the most repeated physics topics for JEE Mains is going to help students strategize for their exam. Students must avoid skipping the following topics as they are the top 10 most repeated topics in physics for JEE mains from 2016 to 2025.

As seen, the Projectile Motion is the most repeated topic so far with 43 questions. The 10th most important topic is Rotational Motion.

• JEE Main- Top 30 Most Repeated Questions & Topics

Previous Year Questions on Most Important Chapters for JEE Mains Physics 2026

In this section, we shall look at some of the topics questions asked in previous years to help you get an idea of what kind of questions are asked. We shall go topic wise from the above table of topics:

Question: A projectile is thrown with a velocity $u_0$ at an angle $\theta$ with the horizontal. The ratio of the rate of change of speed w.r.t. time at the highest point to that at the point of projection is

(1) $g \sin \theta$

(2) $-\mathrm{g} \sin \theta$

(3) zero

(4) g

Solution:

$\frac{d v}{d t}(\text { at the top })=0$

As tangential acceleration at the top $=0$

$\frac{d v}{d t}(\text { at the starting point })=-g \sin \theta$

So, required ratio = zero.

Hence, the correct answer is option (3)

Question: In which case the wire has maximum expansion, if the same force is applied to each wire

(1) $\mathrm{L}=500 \mathrm{~cm}, \mathrm{~d}=0.05 \mathrm{~mm}$

(2) $\mathrm{L}=200 \mathrm{~cm}, \mathrm{~d}=0.02 \mathrm{~mm}$

(3) $\mathrm{L}=300 \mathrm{~cm}, \mathrm{~d}=0.03 \mathrm{~mm}$

(4) L $=400 \mathrm{~cm}, \mathrm{~d}=0.01 \mathrm{~mm}$

Solution:

$1 \alpha \frac{L}{r^2}(Y$ and $F$ are constant)

The maximum expansion occurs in the wire for which the ratio $\frac{\mathrm{L}}{\mathrm{r}^2}$ will be maximum.

Hence, the correct answer is option (4).

Question: At what temperature does the average translational kinetic energy of a molecule in a gas equal to the kinetic energy of an electron accelerated from rest through a potential difference of 5 volt.

(1) $386.5 \times 10^3 \mathrm{~K}$

(2) $3.865 \times 10^3 \mathrm{~K}$

(3) $.38 \times 10^3 \mathrm{~K}$

(4) $38.65 \times 10^3 \mathrm{~K}$

Solution:

K.E. of the electron is

$\begin{aligned} & 5 \mathrm{eV}=5 \times 1.6 \times 10^{-19} \mathrm{~J} \\ & \text { But } \quad \mathrm{K} . \mathrm{E} .=3 / 2 \mathrm{KT} \\ & \therefore 5 \times 1.6 \times 10^{-19}=3 / 2\left(1.38 \times 10^{-23}\right) \times \mathrm{T} \\ & \Rightarrow \mathrm{T}=\frac{5 \times 1.6 \times 10^{-19} \times 2}{3 \times 1.38 \times 10^{-23}} \\ & \Rightarrow \mathrm{~T}=38.65 \times 10^3 \mathrm{~K}\end{aligned}$

Hence, the correct answer is option (4).

Question: An adiabatic change in represented by the equation

(1) $\mathrm{VP} \gamma=$ constant

(2) $\mathrm{PT} \gamma=$ constant

(3) $\mathrm{TV} \gamma=$ constant

(4) $\mathrm{PV} \gamma=$ constant

Solution:

This is the actual process equation

$\mathrm{PV} \gamma=\text { constant }$

Hence, the correct answer is option (4).

Question: The coefficient of static friction between a block of mass $m$ and an incline is $\mu_{\mathrm{s}}=0.3$. What can be the maximum angle $\theta$ of the incline with the horizontal so that the block does not slip on the plane?

Solution:

Angle of repose $\theta=\tan ^{-1}(\mu)$

$\therefore \theta=\tan ^{-1}(0.3)=16.7^{\circ}$

Hence, the answer is $16.7^{\circ}$

Question: A block rests on an inclined plane. The force of friction acting on the block is $(1 / n)$ times. The force required to move the block up the inclined plane with a uniform velocity $(n>1)$. If $\mu$ be the coefficient of friction, then the inclination of the plane with the horizontal is

(1) $\tan ^{-1}[(n-1) / \mu]$

(2) $\tan ^{-1}[\mu(n-1)]$

(3) $\tan ^{-1}[\mu /(n-1)]$

(4) $\tan ^{-1}(\mu)$

Solution: $\mathbf{C}$

$\begin{aligned} & m g \sin \theta=\frac{1}{n}(m g \sin \theta+\mu m g \cos \theta) \\ & \Rightarrow m g \sin \theta(n-1)=\mu m g \cos \theta \\ & \Rightarrow \tan \theta=\frac{\mu}{(n-1)} \\ & \theta=\tan ^{-1}\left[\frac{\mu}{(n-1)}\right]\end{aligned}$

Hence, the correct answer is option (3).

Question: If wattless current flows in the AC circuit, then the circuit is :

(1) Purely Resistive circuit

(2) Purely Inductive circuit

(3) LCR series circuit

(4) RC series circuit only

Solution:

If wattless current flows in the AC circuit, that means the phase difference between current voltage is 90^o
It is possible only for pure inductive or capacitive circuit
Hence, the correct answer is option (2)

Question: A body of mass 2 kg is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in 4s is $\frac{1}{3} \alpha^2 \sqrt{P} m$. The value of $\alpha$ will be_______.

Solution:

$\frac{1}{2} m v^2=p t$
$\mathrm{V}=\sqrt{\frac{2 \mathrm{pt}}{\mathrm{m}}}$
$\frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\frac{2 \mathrm{pt}}{\mathrm{m}}}$

$\int \mathrm{dx}=\sqrt{\frac{2 \mathrm{p}}{\mathrm{m}}} \int \sqrt{\mathrm{t}} \mathrm{dt}$

$\mathrm{x}=\sqrt{\frac{2 \mathrm{p}}{\mathrm{m}}}\left[\mathrm{t}^{3 / 2}\right]_0^4$
$\mathrm{x}=\frac{1}{3} \times 16 \sqrt{\mathrm{p}}$
$\propto=4$
Hence, the answer is 4

Question: A steel wire of length $3.2 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{s}}=2.0 \times 10^{11} \mathrm{Nm}^{-2}\right)$ and a copper wire of length $4.4 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \mathrm{Nm}^{-2}\right)$, both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm . The load applied, in Newton, will be: ( Given $\pi=\frac{22}{7}$ )

(1) 360

(2) 180

(3) 1080

(4) 154

Solution:

$\begin{aligned} & \ell_{\mathrm{s}}=3.2 \mathrm{~m} \\ & \mathrm{Y}_{\mathrm{s}}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \\ & \ell_{\mathrm{c}}=4.4 \mathrm{~m} \\ & \mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \\ & \mathrm{r}=1.4 \mathrm{~mm}=1.4 \times 10^{-3} \mathrm{~m} \\ & \Delta \ell=1.4 \mathrm{~mm}=1.4 \times 10^{-3} \mathrm{~m} \\ & \mathrm{~F}=\left(\frac{\mathrm{YA}}{\ell}\right) \Delta \ell\end{aligned}$

$
\begin{aligned}
& \mathrm{K}=\left(\frac{\mathrm{YA}}{\ell}\right) \Delta \ell \\
& \frac{1}{\mathrm{k}_{\mathrm{eq}}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2} \\
& \frac{1}{\mathrm{~K}_{\mathrm{eq}}}=\frac{\ell_{\mathrm{s}}}{\mathrm{Y}_{\mathrm{s}} \mathrm{~A}_{\mathrm{s}}}+\frac{\ell_{\mathrm{c}}}{\mathrm{Y}_{\mathrm{c}} \mathrm{~A}_{\mathrm{C}}} \\
& \frac{1}{\mathrm{~K}_{\mathrm{eq}}}=\frac{3.2}{\mathrm{~A} \times 2 \times 10^{11}}+\frac{4.4}{\mathrm{~A} \times 1.1 \times 10^{11}} \\
& \text { taking } \mathrm{A}_{\mathrm{s}}=\mathrm{A}_{\mathrm{c}}=\mathrm{A} \\
& \frac{1}{\mathrm{~K}_{\mathrm{eq}}}=\frac{1.6 \times 10^{-11}}{\mathrm{~A}}+\frac{4 \times 10^{-11}}{\mathrm{~A}} \\
& \mathrm{~K}_{\mathrm{eq}}=\frac{\mathrm{A}}{5.6 \times 10^{-11}} \\
& \mathrm{~K}_{\mathrm{eq}}=\frac{\mathrm{A} \times 10^{11}}{5.6} \\
& \mathrm{~F}=\mathrm{K}_{\mathrm{eq}}(\Delta \ell) \\
& =\frac{\mathrm{A} \times 10^{11}}{5.6} \times 1.4 \times 10^{-3} \\
& =\frac{3.14 \times\left(1.4 \times 10^{-3}\right)^3 \times 10^{11}}{5.6} \\
& \mathrm{~F}=\frac{22 \times(1.4)^2 \times 10^2}{7 \times 4}
\end{aligned}
$
,F = 154N

Hence, the correct answer is option (4).

Question: The elongation of a wire on the surface of the earth is $10^{-4} \mathrm{~m}$. The same wire of the same dimensions is elongated $6 \times 10^{-5} \mathrm{~m}$ on another planet. The acceleration due to gravity on the planet will be___________$\mathrm{ms}^{-2}$. (Take acceleration due to gravity on the surface of the earth = 10 $\mathrm{ms}^{-2}$ )

(1) 6

(2) 7

(3) 5

(4) 4

Solution:

For a given material, Y there will be constant
$\left.\begin{gathered}\Delta l_E=10^{-4} m \\ \Delta l_p=6 \times 10^{-5} m\end{gathered} \right\rvert\, g_P \rightarrow$ acceleration due to gravity on planet
$\mathrm{Y}=\frac{\mathrm{mg}}{\mathrm{A}} \times \frac{\mathrm{l}}{\Delta \mathrm{l}}$
$\mathrm{g} \propto \Delta \mathrm{l}$
$\frac{g_{\mathrm{E}}}{g_{\mathrm{p}}}=\frac{\Delta \mathrm{l}_{\mathrm{E}}}{\Delta \mathrm{l}_{\mathrm{P}}}$$\frac{10}{g_p}=\frac{10^{-4}}{6 \times 10^{-5}} \Rightarrow g_p=6 \frac{m}{s^2}$

Hence, the answer is option (1).

For more practice solve JEE Main 10 Mock test according to latest pattern.

Top 10 Most Important Chapters For JEE Mains Physics 2026 as Per Previous Years

Acing the exam is all about studying smart these days due to the huge competition. Smart study requires strategy and in depth research. The good thing is we have already done the research for you. Again like the topics, we have curated a list of top 10 most important chapters for JEE mains physics 2026. This data is also researched and curated by experts from 2016 to 2025 all JEE Mains physics question papers.

This table has the name of the most important chapters for JEE Mains physics 2026 ( as “Chapter Name”) and to its right we have associated the total number of questions asked from that particular chapter over the 10 years.