JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2015
- Medium
- MCQ
Diameter of a steel ball is measured using a Vernier callipers which has divisions of \(0. 1\,cm\) on its main scale \((MS)\) and \(10\) divisions of its vernier scale \((VS)\) match \(9\) divisions on the main scale. Three such measurements for a ball are given as
| S.No. | \(MS\;(cm)\) | \(VS\) divisions |
| \((1)\) | \(0.5\) | \(8\) |
| \((2)\) | \(0.5\) | \(4\) |
| \((3)\) | \(0.5\) | \(6\) |
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(0.59\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Lets\,count\, = \frac{{0.1}}{{10}} = 0.01\,\,cm\\
{d_1} = 0.5 + 8 \times 0.01 + 0.03 = 0.61\,cm\\
{d_2} = 0.5 + 4 \times 0.01 + 0.03 = 0.57\,cm\\
{d_3} = 0.5 + 6 \times 0.01 + 0.03 = 0.59\,cm\\
Mean\,diameter\, = \frac{{0.61 + 0.57 + 0.59}}{3}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.59\,cm
\end{array}\)
Lets\,count\, = \frac{{0.1}}{{10}} = 0.01\,\,cm\\
{d_1} = 0.5 + 8 \times 0.01 + 0.03 = 0.61\,cm\\
{d_2} = 0.5 + 4 \times 0.01 + 0.03 = 0.57\,cm\\
{d_3} = 0.5 + 6 \times 0.01 + 0.03 = 0.59\,cm\\
Mean\,diameter\, = \frac{{0.61 + 0.57 + 0.59}}{3}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.59\,cm
\end{array}\)
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