JEE Advanced · Physics · 29. Experimental Physics
The circular scale of a screw gauge has 50 divisions and pitch of \(0.5 \mathrm{~mm}\). Find the diameter of sphere. Main scale reading is 2 .
- A \(1.2\)
- B \(1.25\)
- C \(2.20\)
- D \(2.25\)
Answer & Solution
Correct Answer
(A) \(1.2\)
Step-by-step Solution
Detailed explanation
Least count \(\mathrm{LC}=\frac{\text { Pitch }}{\text { Number of divisions on circular scale }}\) \(=\frac{0.1}{50}=0.01 \mathrm{~mm}\)
Now, diameter of ball \(=(2 \times 0.5 \mathrm{~mm})~+\) \((25-5)(0.01)=1.2 \mathrm{~mm}\)
Now, diameter of ball \(=(2 \times 0.5 \mathrm{~mm})~+\) \((25-5)(0.01)=1.2 \mathrm{~mm}\)
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