Experimental Physics JEE Main previous year questions with solutions

5 solved JEE Main questions on Experimental Physics, free to read — no sign-in needed. The full chapter has 78 questions; sign in to attempt the remaining 73 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (05 Apr, Shift 2)Electric Current and Drift of Electrons
   In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as $10\text{ V}$ and $5\text{ A}$, respectively. The least counts of voltmeter and ammeter are $500\text{ mV}$ and $200\text{ mA}$, respectively. The estimated error in the resistance measurement is _______ $\Omega$.
   1. A.$0.25$
   2. B.$2$
   3. C.$2.5$
   4. D.$0.18$
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   Answer: (D)

   Given $V = 10\text{ V}$ and $I = 5\text{ A}$ Least count of voltmeter, $\Delta V = 500\text{ mV} = 0.5\text{ V}$ Least count of ammeter, $\Delta I = 200\text{ mA} = 0.2\text{ A}$ Resistance is given by Ohm's law: $R = \dfrac{V}{I} = \dfrac{10}{5} = 2\text{ }\Omega$ The maximum permissible relative error in resistance is: $\dfrac{\Delta R}{R} = \dfrac{\Delta V}{V} + \dfrac{\Delta I}{I}$ $\dfrac{\Delta R}{2} = \dfrac{0.5}{10} + \dfrac{0.2}{5}$ $\dfrac{\Delta R}{2} = 0.05 + 0.04 = 0.09$ $\Delta R = 2 \times 0.09 = 0.18\text{ }\Omega$ Answer: $0.18$
2. Q2JEE Main 2025 (24 Jan, Shift 1)Experimental Physics
   The least count of a screw guage is 0.01 mm . If the pitch is increased by $75 \%$ and number of divisions on the circular scale is reduced by $50 \%$, the new least count will be _____ $\times 10^{-3} \mathrm{~mm}$
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   Answer: 35

   $\begin{aligned} & \text { Given least count of Screw Gauge }=0.01 \mathrm{~mm} \\ & \text { L.C }=\frac{(\text { pitch })}{\text { No. of circular turn }}=\frac{\mathrm{P}}{\mathrm{N}}=0.01 \mathrm{~mm} \\ & \text { New pitch }=\frac{\mathrm{P}(1+0.75)}{\mathrm{N}(1-0.5)}=\frac{\mathrm{P}}{\mathrm{N}}\left[\frac{1.75}{0.5}\right] \\ & =(0.01) 3.5 \\ & =0.035 \mathrm{~mm} \\ & =35 \times 10^{-3} \mathrm{~mm}\end{aligned}$
3. Q3JEE Main 2023 (29 Jan, Shift 2)Errors of Measurement
   In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as $5.25mm$ and apparent thickness of the glass slab at $5.00mm$. Travelling microscope has $20$ divisions in one $cm$ on main scale and $50$ divisions on Vernier scale is equal to $49$ divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is $\dfrac{x}{10}\times {10}^{-3}$, where $x$ is ______
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   Answer: 41

   Given: $50VSD=49MSD \Rightarrow 1VSD=\dfrac{49}{50}MSD$ So least count will be $=1MSD-1VSD =\dfrac{1}{20}cm-\dfrac{49}{50}\cdot \dfrac{1}{20}cm$ $=\dfrac{1}{50}\times \dfrac{1}{20}cm =0.01mm$ As we know, $\mu =\dfrac{h}{h'}=\dfrac{5.25}{5.00}$ Applying log on both sides, we get $ln\mu =lnh–lnh’$ Differentiating both sides, we get $\dfrac{d\mu }{\mu }=\dfrac{dh}{h}-\dfrac{dh'}{h'}$ For calculation, all errors will be added, therefore $d\mu =\left[\dfrac{0.01}{5.25}+\dfrac{0.01}{5.00}\right]\mu$ $=\dfrac{41}{10}\times {10}^{-3}$ Hence, $x=41$
4. Q4JEE Main 2022 (28 Jun, Shift 1)Diodes
   For using a multimeter to identify diode from electrical components, choose the correct statement out of the following about the diode
   1. A.It is two terminal device which conducts current in both directions.
   2. B.It is two terminal device which conducts current in one direction only
   3. C.It does not conduct current gives an initial deflection which decays to zero.
   4. D.It is three terminal device which conducts current in one direction only between central terminal and either of the remaining two terminals
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   Answer: (B)

   A diode is a two-terminal device that conducts current in forward bias and does not conduct current in reverse bias i.e. it conducts current in one direction only.
5. Q5JEE Main 2021 (01 Sep, Shift 2)Resistance and Resistivity
   Two resistors ${R}_{1}=(4\pm 0.8)\Omega$ and ${R}_{2}=(4\pm 0.4)\Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be :
   1. A.$(4\pm 0.4)\Omega$
   2. B.$(2\pm 0.4)\Omega$
   3. C.$(4\pm 0.3)\Omega$
   4. D.$(2\pm 0.3)\Omega$
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   Answer: (D)

   $\dfrac{\Delta R}{{R}^{2}}=\dfrac{\Delta {R}_{1}}{{R}_{1}^{2}}+\dfrac{\Delta {R}_{2}}{{R}_{2}^{2}}$ $\Delta R={\left(R\right)}^{2}\left[\dfrac{\Delta {R}_{1}}{{R}_{1}^{2}}+\dfrac{\Delta {R}_{2}}{{R}_{2}^{2}}\right]$ $={\left(2\right)}^{2}\left[\dfrac{0.8}{{4}^{2}}+\dfrac{0.4}{{4}^{2}}\right]$ $=[0.2+0.1]=0.3$ ${R}_{\text{net}}=R+\Delta R=2\pm 0.3$