JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
Asseretion \(A:\) If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is \(5\, {mm}\) and there are \(50\) total divisions on circular scale, then least count is \(0.001\, {cm}\). Reason \(R:\) Least Count \(=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}\) In the light of the above statements, choose the most appropriate answer from the options given below:
- A
Both \(A\) and \(R\) are correct and \(R\) is the correct explanation of \(A\)
- B \(A\) is not correct but \(R\) is correct.
- C Both \(A\) and \(R\) are correct and \(R\) is NOT the correct explanation of \(A\).
- D \({A}\) is correct but \({R}\) is not correct.
Answer & Solution
Correct Answer
(B) \(A\) is not correct but \(R\) is correct.
Step-by-step Solution
Detailed explanation
Least count \(=\frac{\text { Pitch }}{\text { total division on circular scale }}\) In \(5\) revolution, distance travel, \(5 \,mm\) In \(1\) revolution, it will travel \(1 \,mm\). So least count \(=\frac{1}{50}=0.02\)
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