A student measured the diameter of a small steel ball using a screw gauge of least count $0.001 ext{ cm}$. Th

Question

A student measured the diameter of a small steel ball using a screw gauge of least count $0.001	ext{ cm}$. The main scale reading is $5	ext{ mm}$ and zero of circular scale division coincides with $25$ divisions above the reference level. If the screw gauge has a zero error of $-0.004	ext{ cm}$, the correct diameter of the ball is:

Accepted Answer

1. **Identify the Given Data:**
Least Count (LC) = $0.001	ext{ cm}$
Main Scale Reading (MSR) = $5	ext{ mm} = 0.5	ext{ cm}$
Circular Scale Reading (CSR) = $25$ divisions
Zero Error = $-0.004	ext{ cm}$

2. **Calculate the Measured Value:**
The measured diameter is given by the formula:
$	ext{Measured Value} = 	ext{MSR} + (	ext{CSR} 	imes 	ext{LC})$
$	ext{Measured Value} = 0.5	ext{ cm} + (25 	imes 0.001	ext{ cm})$
$	ext{Measured Value} = 0.5	ext{ cm} + 0.025	ext{ cm} = 0.525	ext{ cm}$

3. **Apply Zero Correction to find Correct Diameter:**
The correct diameter is obtained by subtracting the zero error from the measured value:
$	ext{Correct Diameter} = 	ext{Measured Value} - 	ext{Zero Error}$
$	ext{Correct Diameter} = 0.525	ext{ cm} - (-0.004	ext{ cm})$
$	ext{Correct Diameter} = 0.525	ext{ cm} + 0.004	ext{ cm} = 0.529	ext{ cm}$

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