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 70 questions; sign in to attempt the remaining 65 in the exam simulator.

1. Q1JEE Main 2026 (04 Apr, Shift 1)Experimental Physics
   In a screw gauge when the circular scale is given five complete rotations it moves linearly by $2.5$ mm. If the circular scale has $100$ divisions, the least count of screw gauge is _____ mm.
   1. A.$1 \times 10^{-2}$
   2. B.$1 \times 10^{-3}$
   3. C.$5 \times 10^{-2}$
   4. D.$5 \times 10^{-3}$
   Show answer & solution

   Answer: (D)

   Pitch of the screw gauge is the linear distance moved in one complete rotation. $\text{Pitch} = \dfrac{2.5}{5} = 0.5 \text{ mm}$ Least count of the screw gauge is given by the ratio of the pitch to the total number of divisions on the circular scale. $\text{Least count} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}$ $\text{Least count} = \dfrac{0.5}{100} = 0.005 \text{ mm}$ $\text{Least count} = 5 \times 10^{-3} \text{ mm}$ Answer: $5 \times 10^{-3}$
2. Q2JEE Main 2026 (02 Apr, Shift 2)Experimental Physics
   In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions in circular scale and pitch of screw gauge is $0.1$ mm. When diameter of a sphere is measured, the reading of main scale is $5$ mm and $50^{\text{th}}$ division of circular scale coincides with the reference line of main scale. The diameter of sphere is _______ mm.
   1. A.$5.045$
   2. B.$5.055$
   3. C.$5.450$
   4. D.$5.550$
   Show answer & solution

   Answer: (A)

   Least count of the screw gauge is given by: $\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}$ $\text{LC} = \dfrac{0.1}{100} = 0.001 \text{ mm}$ When the two studs are in contact, the $5^{\text{th}}$ division of the circular scale coincides with the reference line. This indicates a positive zero error. $\text{Zero error} = +5 \times \text{LC} = +5 \times 0.001 = +0.005 \text{ mm}$ For the measurement of the sphere's diameter: $\text{Main scale reading (MSR)} = 5 \text{ mm}$ $\text{Circular scale reading (CSR)} = 50$ $\text{Measured diameter} = \text{MSR} + \text{CSR} \times \text{LC}$ $\text{Measured diameter} = 5 + 50 \times 0.001 = 5.050 \text{ mm}$ The true diameter is obtained by subtracting the zero error from the measured value: $\text{True diameter} = \text{Measured diameter} - \text{Zero error}$ $\text{True diameter} = 5.050 - 0.005 = 5.045 \text{ mm}$ Answer: $5.045$
3. Q3JEE Main 2026 (28 Jan, Shift 2)Experimental Physics
   In an experiment, a set of reading are obtained as follows - $1.24 \mathrm{~mm}, 1.25 \mathrm{~mm}, 1.23 \mathrm{~mm}$, 1.21 mm. The expected least count of the instrument used in recording these readings is $\_\_\_\_$ mm.
   1. A.0.001
   2. B.0.1
   3. C.0.01
   4. D.0.05
   Show answer & solution

   Answer: (C)

   Least count is the smallest division readable on an instrument scale. Given readings are 1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm, all expressed to two decimal places. The precision of measurement shows differences as small as 0.01 mm (e.g., 1.25 - 1.24 = 0.01 mm). This indicates all readings can be recorded to the nearest 0.01 mm, which represents the smallest measurable division. Therefore, the least count of the instrument is 0.01 mm.
4. Q4JEE Main 2026 (28 Jan, Shift 1)Experimental Physics
   When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, $4^{\text {th }}$ mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes 15 divisions on main scale and $5^{\text {th }}$ division of vernier scale coincides with a main scale division. Measured length of cylinder is $\_\_\_\_$ mm. (Least count of Vernier calliper $=0.1 \mathrm{~mm}$)
   1. A.15.5
   2. B.15.9
   3. C.15.1
   4. D.15.4
   Show answer & solution

   Answer: (C)

   Given Least Count ($LC$) = $0.1$ mm. When jaws touch, the zero of the vernier scale is to the right of the main scale zero, indicating a positive zero error. Zero error ($e$) = $+ (n \times LC)$, where $n$ is the coinciding vernier division. $e = +(4 \times 0.1) = +0.4$ mm. While measuring the cylinder: Main Scale Reading ($MSR$) = $15$ divisions. Assuming $1$ MSD = $1$ mm (standard for $LC = 0.1$ mm), $MSR = 15$ mm. Vernier Scale Reading ($VSR$) = $5 \times LC = 5 \times 0.1 = 0.5$ mm. Observed Reading = $MSR + VSR = 15 + 0.5 = 15.5$ mm. True Measured Length = Observed Reading $-$ Zero Error True Length = $15.5 - 0.4 = 15.1$ mm.
5. Q5JEE Main 2026 (24 Jan, Shift 2)Experimental Physics
   In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $=0.05 \mathrm{~mm}$, then the least count of the vernier callipers is $\_\_\_\_$ mm.
   1. A.0.005
   2. B.0.05
   3. C.0.002
   4. D.0.02
   Show answer & solution

   Answer: (C)

   Given: 50 vernier divisions = 48 main divisions and 1 main division = 0.05 mm 50 vernier divisions = 48 × 0.05 = 2.4 mm 1 vernier division = 2.4/50 = 0.048 mm Least count = 1 main division - 1 vernier division = 0.05 - 0.048 = 0.002 mm