JEE Main 2026 — Experimental Physics Question with Solution

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.