Physics · 2015
JEE Main · 10 April 2015, Shift 1 · Q30
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…
Diameter of a steel ball is measured using a Vernier callipers which has divisions of $\displaystyle 0.1$ cm on its main scale (MS) and $\displaystyle 10$ divisions of its vernier scale (VS) match $\displaystyle 9$ divisions on the main scale. Three such measurements for a ball are given as :
If the zero error is -$\displaystyle 0.03$ cm, then mean corrected diameter is :
| S.No. | MS (cm) | VS divisions |
| 1. | $\displaystyle 0.5$ | $\displaystyle 8$ |
| 2. | $\displaystyle 0.5$ | $\displaystyle 4$ |
| 3. | $\displaystyle 0.5$ | $\displaystyle 6$ |
Official answer
(2)
0.$\displaystyle 59$ cm
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JEE Main 2015 Physics question, with the answer from NTA’s final answer key. Where our answers come from.