Experimental Physics Question 3
Question 3 - 30 January - Shift 1
In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by $0.5 mm$ on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while $46^{\text{th }}$ division the circular scale coincide with the reference line. The diameter of the wire is $\quad \times 10^{-2} mm$.
Show Answer
Answer: (22)
Solution:
Formula: Least count of screw gauge
Least count $=\frac{\text{ Pitch }}{\text{ No. of circular divisions }}$
$=\frac{0.5 mm}{100}$
Least count $=5 \times 10^{-3} mm$
Positive Error $=$ MSR + CSR $($ LC $)$
$=0 mm+6(5 \times 10^{-3} mm)$
Reading of Diameter $=$ MSR + CSR (LC)
Positive zero error
$=4 \times 0.5 mm+(46(5 \times 10^{-3}))-6(5 \times 10^{-3}) mm$
