Back to Directory
NEET PHYSICSMedium

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm0.001\text{ cm}. The main scale reading is 5 mm5\text{ mm} and zero of circular scale division coincides with 2525 divisions above the reference level. If the screw gauge has a zero error of −0.004 cm-0.004\text{ cm}, the correct diameter of the ball is:

A

0.521 cm0.521\text{ cm}

B

0.525 cm0.525\text{ cm}

C

0.053 cm0.053\text{ cm}

D

0.529 cm0.529\text{ cm}

Step-by-Step Solution

  1. Identify the Given Data: Least Count (LC) = 0.001 cm0.001\text{ cm} Main Scale Reading (MSR) = 5 mm=0.5 cm5\text{ mm} = 0.5\text{ cm} Circular Scale Reading (CSR) = 2525 divisions Zero Error = −0.004 cm-0.004\text{ cm}

  2. Calculate the Measured Value: The measured diameter is given by the formula: Measured Value=MSR+(CSR×LC)\text{Measured Value} = \text{MSR} + (\text{CSR} \times \text{LC}) Measured Value=0.5 cm+(25×0.001 cm)\text{Measured Value} = 0.5\text{ cm} + (25 \times 0.001\text{ cm}) Measured Value=0.5 cm+0.025 cm=0.525 cm\text{Measured Value} = 0.5\text{ cm} + 0.025\text{ cm} = 0.525\text{ cm}

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

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started