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JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement

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
   \((1)\)      \(0.5\)       \(8\)
   \((2)\)     \(0.5\)       \(4\)
   \((3)\)     \(0.5\)       \(6\)
If the zero error is \(- 0.03\,cm,\) then mean corrected diameter is  ........... \(cm\)

  1. A \(0.52\)
  2. B \(0.59\)
  3. C \(0.56\)
  4. D \(0.53\)
Verified Solution

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}\\…

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