# Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0.1 \mathrm{~cm}$ on its main scale $(\mathrm{MS})$ and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as: \begin{tabular}{|c|c|c|} \hline S. No. & MS(cm) & VS divisions \\ \hline $1 .$ & $0.5$ & 8 \\ \hline $2 .$ & $0.5$ & 4 \\ \hline $3 .$ & $0.5$ & 6 \\ \hline \end{tabular} If the zero error is $-0.03 \mathrm{~cm}$, then mean corrected diameter is:

Question:

Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0.1 \mathrm{~cm}$ on its main scale $(\mathrm{MS})$ and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as:

\begin{tabular}{|c|c|c|}

\hline S. No. & MS(cm) & VS divisions \\

\hline $1 .$ & $0.5$ & 8 \\

\hline $2 .$ & $0.5$ & 4 \\

\hline $3 .$ & $0.5$ & 6 \\

\hline

\end{tabular}

If the zero error is $-0.03 \mathrm{~cm}$, then mean corrected diameter is:

1. $0.52 \mathrm{~cm}$

2. $0.59 \mathrm{~cm}$

3. $0.56 \mathrm{~cm}$

4. $0.53 \mathrm{~cm}$

JEE Main Previous Year Single Correct Question of JEE Main from Physics Physical World, Units and Measurements chapter.

JEE Main Previous Year April 10, 2015

Correct Option: 2

Solution:

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