Mathematics in Physics JEE Main previous year questions with solutions

5 solved JEE Main questions on Mathematics in Physics, free to read — no sign-in needed. The full chapter has 106 questions; sign in to attempt the remaining 101 in the exam simulator.

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1. Q1JEE Main 2026 (28 Jan, Shift 2)Angular momentum and Angular impulse
   When the position vector $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ changes sign as $-\vec{r}$, which one of the following vector will not flip under sign change ?
   1. A.Angular momentum
   2. B.Velocity
   3. C.Acceleration
   4. D.Linear momentum
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   Answer: (A)

   The transformation $\vec{r} \to -\vec{r}$ is known as a parity transformation (inversion through the origin). Vectors that change sign under parity are called polar vectors, while those that do not change sign are called axial vectors (or pseudovectors). Velocity is defined as $\vec{v} = \frac{d\vec{r}}{dt}$. Since $\vec{r} \to -\vec{r}$, we have $\vec{v} \to -\vec{v}$. Acceleration is defined as $\vec{a} = \frac{d\vec{v}}{dt}$. Since $\vec{v} \to -\vec{v}$, we have $\vec{a} \to -\vec{a}$. Linear momentum is defined as $\vec{p} = m\vec{v}$. Since $\vec{v} \to -\vec{v}$, we have $\vec{p} \to -\vec{p}$. Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$. Under the transformation $\vec{r} \to -\vec{r}$ and $\vec{p} \to -\vec{p}$, the cross product becomes: $\vec{L}' = (-\vec{r}) \times (-\vec{p}) = \vec{r} \times \vec{p} = \vec{L}$. Thus, angular momentum does not flip its sign under the transformation $\vec{r} \to -\vec{r}$.
2. Q2JEE Main 2025 (23 Jan, Shift 1)Addition and Subtraction of Vectors
   Two particles are located at equal distance from origin. The position vectors of those are represented by $\bar{A}=2 \hat{i}+3 n \hat{j}+2 \hat{k}$ and $\bar{B}=2 \hat{i}-2 \hat{j}+4 p \hat{k}$, respectively. If both the vectors are at right angle to each other, the value of $\mathrm{n}^{-1}$ is _____ .
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   Answer: 3

   $\begin{aligned} & \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0 \\ & 4-6 \mathrm{n}+8 \mathrm{p}=0 \\ & |\overrightarrow{\mathrm{~A}}|=|\overrightarrow{\mathrm{B}}| \\ & 4+9 \mathrm{n}^2+4=4+4+16 \mathrm{p}^2 \\ & 9 \mathrm{n}^2=16 \mathrm{p}^2 \\ & \mathrm{P}=+\frac{3}{4} \mathrm{n} \\ & 4-6 \mathrm{n} \pm 6 \mathrm{n}=0 \\ & 12 \mathrm{n}=4 \\ & \mathrm{n}=\frac{1}{3}\end{aligned}$
3. Q3JEE Main 2024 (08 Apr, Shift 1)Errors of Measurement
   Young's modulus is determined by the equation given by $\mathrm{Y}=49000 \frac{\mathrm{m}}{\mathrm{l}} \frac{\mathrm{dyn}}{\mathrm{cm}^2}$ where $M$ is the mass and $l$ is the extension of wire used in the experiment. Now error in Young modules $(Y)$ is estimated by taking data from $M-l$ plot in graph paper. The smallest scale divisions are $5 \mathrm{~g}$ and $0.02 \mathrm{~cm}$ along load axis and extension axis respectively. If the value of $M$ and $l$ are $500 \mathrm{~g}$ and $2 \mathrm{~cm}$ respectively then percentage error of $Y$ is :
   1. A.$0.5 \%$
   2. B.$2 \%$
   3. C.$0.02 \%$
   4. D.$0.2 \%$
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   Answer: (B)

   $\begin{aligned} \frac{\Delta \mathrm{Y}}{\mathrm{Y}} & =\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \ell}{\ell} \\ & =\frac{5}{500}+\frac{0.02}{2}=0.01+0.01 \\ \frac{\Delta \mathrm{Y}}{\mathrm{Y}} & =0.02 \Rightarrow \% \frac{\Delta \mathrm{Y}}{\mathrm{Y}}=2 \%\end{aligned}$
4. Q4JEE Main 2023 (25 Jan, Shift 1)Multiplication of Vectors
   If $\vec{P}=3\hat{i}+\sqrt{3}\hat{j}+2\hat{k}$ and $\vec{Q}=4\hat{i}+\sqrt{3}\hat{j}+2.5\hat{k}$ then, the unit vector in the direction of $\vec{P}\times \vec{Q}$ is $\dfrac{1}{x}(\sqrt{3}\hat{i}+\hat{j}-2\sqrt{3}\hat{k})$. The value of $x$ is
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   Answer: 4

   Unit vector in the direction of $\vec{P}\times \vec{Q}\text{is}\hat{n}=\dfrac{\vec{P}\times \vec{Q}}{|\vec{P}\times \vec{Q}|}$. Here, $\vec{P}\times \vec{Q}=\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5\end{matrix}\right|=\sqrt{3}\dfrac{\hat{i}}{2}+\dfrac{\hat{j}}{2}-\sqrt{3}\hat{k}$ And $\left|\vec{P}\times \vec{Q}\right|=\sqrt{{\left(\dfrac{\sqrt{3}}{2}\right)}^{2}+{\left(\dfrac{1}{2}\right)}^{2}+{\left(\sqrt{3}\right)}^{2}}=\sqrt{4}=2$ $\Rightarrow \dfrac{\vec{P}\times \vec{Q}}{|\vec{P}\times \vec{Q}|}=\dfrac{1}{2}\left(\sqrt{3}\dfrac{\hat{i}}{2}+\dfrac{\hat{j}}{2}-\sqrt{3}\hat{k}\right)$ $=\dfrac{1}{4}(\sqrt{3}\hat{i}+\hat{j}-2\sqrt{3}\hat{k}) \Rightarrow x=4$
5. Q5JEE Main 2021 (25 Jul, Shift 2)Fundamentals of Vectors
   Two vectors $\vec{X}$ and $\vec{Y}$ have equal magnitude. The magnitude of $(\vec{X}-\vec{Y})$ is $n$ times the magnitude of $(\vec{X}+\vec{Y})$. The angle between $\vec{X}$ and $\vec{Y}$ is :
   1. A.${\cos }^{-1}\left(\dfrac{-{n}^{2}-1}{{n}^{2}-1}\right)$
   2. B.${\cos }^{-1}\left(\dfrac{{n}^{2}-1}{-{n}^{2}-1}\right)$
   3. C.${\cos }^{-1}\left(\dfrac{{n}^{2}+1}{-{n}^{2}-1}\right)$
   4. D.${\cos }^{-1}\left(\dfrac{{n}^{2}+1}{{n}^{2}-1}\right)$
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   Answer: (B)

   Given $X=Y$ $\sqrt{{X}^{2}+{Y}^{2}-2\times Y\cos \theta }$ $=n\sqrt{{X}^{2}+{Y}^{2}+2\times Y\cos \theta }$ Square both sides $2{X}^{2}(1-\cos \theta )={n}^{2}\cdot 2{X}^{2}(1+\cos \theta )$ $1-\cos \theta ={n}^{2}+{n}^{2}\cos \theta$ $\cos \theta =\dfrac{1-{n}^{2}}{1+{n}^{2}}$ $\theta ={\cos }^{-1}\left[\dfrac{{n}^{2}-1}{-{n}^{2}-1}\right]$