Experimental Skills Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Experimental Skills — screw gauge, Vernier calipers, and error analysis — with step-by-step solutions.
Solved JEE Main 2026 previous year questions from Experimental Skills, covering least count, zero error, and error propagation with concise worked solutions.
Solutions are AI-generated and pending review.
Questions
Solution
Step 1 — Find the pitch. The pitch is the linear distance moved in one full rotation:
$$\text{pitch} = \frac{2.5\ \text{mm}}{5\ \text{rotations}} = 0.5\ \text{mm}$$Step 2 — Least count. With $100$ divisions on the circular scale:
$$\text{LC} = \frac{\text{pitch}}{\text{no. of circular divisions}} = \frac{0.5}{100} = 5 \times 10^{-3}\ \text{mm}$$Answer: D
Solution
Step 1 — Young’s modulus expression. With area $A = \frac{\pi}{4}d^2$,
$$Y = \frac{F\,L}{A\,\Delta L} = \frac{4 F L}{\pi\, d^2\, \Delta L}$$Using SI units ($F = 100\ \text{N}$, $L = 1.5\ \text{m}$, $d = 8\times10^{-4}\ \text{m}$, $\Delta L = 5\times10^{-3}\ \text{m}$):
$$Y = \frac{4(100)(1.5)}{\pi (8\times10^{-4})^2 (5\times10^{-3})} \approx 5.97 \times 10^{10}\ \text{N/m}^2$$Step 2 — Relative error (load ignored). Since $d$ appears squared:
$$\frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta L}{L} + \frac{\Delta(\Delta L)}{\Delta L}$$$$= 2\left(\frac{0.001}{0.08}\right) + \frac{0.1}{150} + \frac{0.001}{0.5} = 0.025 + 0.00067 + 0.002 = 0.02767$$Step 3 — Absolute error.
$$\Delta Y = 0.02767 \times 5.97 \times 10^{10} \approx 1.65 \times 10^{9}\ \text{N/m}^2$$So $\alpha \approx 1.65$.
Answer: B
Solution
Step 1 — Least count.
$$\text{LC} = \frac{\text{pitch}}{\text{divisions}} = \frac{0.1}{100} = 0.001\ \text{mm}$$Step 2 — Zero error. With the studs in contact the $5^{\text{th}}$ circular division sits on the reference line, so the instrument reads a value above zero. This is a positive zero error:
$$\text{zero error} = +5 \times \text{LC} = +0.005\ \text{mm}$$Step 3 — Observed reading.
$$\text{observed} = 5 + 50 \times 0.001 = 5.050\ \text{mm}$$Step 4 — Corrected diameter (subtract positive zero error):
$$d = 5.050 - 0.005 = 5.045\ \text{mm}$$Answer: A
Solution
Step 1 — Least count.
$$\text{LC} = \frac{1\ \text{MSD}}{\text{no. of VSD}} = \frac{1\ \text{mm}}{10} = 0.1\ \text{mm} = 0.01\ \text{cm}$$Step 2 — Sign of zero error. With the jaws closed the Vernier zero lies to the right of the main-scale zero, so the reading is greater than zero when it should be zero. This is a positive zero error.
Step 3 — Magnitude. The $7^{\text{th}}$ Vernier division coincides:
$$\text{zero error} = +7 \times \text{LC} = 7 \times 0.01 = 0.07\ \text{cm (positive)}$$Answer: C
Solution
Step 1 — Nominal resistance.
$$R = \frac{V}{I} = \frac{10}{5} = 2\ \Omega$$Step 2 — Relative error. For $R = V/I$:
$$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} = \frac{0.5}{10} + \frac{0.2}{5} = 0.05 + 0.04 = 0.09$$Step 3 — Absolute error.
$$\Delta R = 0.09 \times 2 = 0.18\ \Omega$$Answer: D