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 121 questions; sign in to attempt the remaining 116 in the exam simulator.

1. Q1JEE Main 2026 (06 Apr, Shift 1)Errors of Measurement
   The density $\rho$ of a uniform cylinder is determined by measuring its mass $m$, length $l$ and diameter $d$. The measured values of $m$, $l$ and $d$ are $97.42 \pm 0.02$ g, $8.35 \pm 0.05$ mm and $20.20 \pm 0.02$ mm, respectively. Calculated percentage fractional error in $\rho$ is _______.
   1. A.$0.63\%$
   2. B.$0.82\%$
   3. C.$0.72\%$
   4. D.$0.25\%$
   Show answer & solution

   Answer: (B)

   The density of a uniform cylinder is given by $\rho = \dfrac{m}{V} = \dfrac{4m}{\pi d^2 l}$ The percentage fractional error in density is given by: $\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{\Delta m}{m} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta l}{l} \right) \times 100$ Substituting the given values: $\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{0.02}{97.42} + 2 \times \dfrac{0.02}{20.20} + \dfrac{0.05}{8.35} \right) \times 100$ $\dfrac{\Delta \rho}{\rho} \times 100 = \left( 0.000205 + 0.00198 + 0.005988 \right) \times 100$ $\dfrac{\Delta \rho}{\rho} \times 100 = 0.008173 \times 100 = 0.8173\%$ Rounding off to two decimal places, the percentage error is $0.82\%$. Answer: $0.82\%$
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 _____ .
   Show answer & solution

   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 (30 Jan, Shift 2)Fundamentals of Vectors
   A vector has magnitude same as that of $\vec{A}=3\hat{j}+4\hat{j}$ and is parallel to $\vec{B}=4\hat{i}+3\hat{j}$. The $x$ and $y$ components of this vector in first quadrant are $x$ and $3$ respectively where $x=$____.
   Show answer & solution

   Answer: 4

   Magnitude of vector $\left|\vec{A}\right|=\sqrt{{3}^{2}+{4}^{2}}=5$ Now, unit vector along $\vec{B}=\dfrac{(4\hat{i}+3\hat{j})}{\sqrt{{4}^{2}+{3}^{2}}}=\dfrac{4}{5}\hat{i}+\dfrac{3}{5}\hat{j}$ Now, $\vec{N}=\left|\vec{A}\right|\hat{B}=5\times \left(\dfrac{4}{5}\hat{i}+\dfrac{3}{5}\hat{j}\right)=4\hat{i}+3\hat{j}$ $∴x=4$
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
   Show answer & solution

   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 2026 (23 Jan, Shift 1)Errors of Measurement
   Four persons measure the length of a rod as $20.00 \mathrm{~cm}, 19.75 \mathrm{~cm}, 17.01 \mathrm{~cm}$ and 18.25 cm. The relative error in the measurement of average length of the rod is :
   1. A.0.08
   2. B.0.06
   3. C.0.18
   4. D.0.24
   Show answer & solution

   Answer: (B)

   Mean $= \frac{20.00+19.75+17.01+18.25}{4} = \frac{75.01}{4} = 18.7525$ cm. Absolute errors: $|20.00-18.7525|=1.2475$, $|19.75-18.7525|=0.9975$, $|17.01-18.7525|=1.7425$, $|18.25-18.7525|=0.5025$. Mean absolute error $= \frac{1.2475+0.9975+1.7425+0.5025}{4} = \frac{4.49}{4} = 1.1225$. Relative error $= \frac{1.1225}{18.7525} \approx 0.06$.