FORMULA CHAPTER
Chapter 01 – Measurements and Experimentation
Physical Quantity
Worked example
Given: Length = 5 metres Calculation: Numerical value = 5 and unit = m Answer: Length = 5 m
Vernier Calipers Least Count
Worked example
Given: 1 MSD = 1 mm and 10 Vernier divisions Calculation: LC = 1/10 mm Answer: LC = 0.1 mm
Vernier Scale Reading
Worked example
Given: p = 6 and LC = 0.01 cm Calculation: Vernier reading = 6 × 0.01 Answer: Vernier reading = 0.06 cm
Vernier Calipers Total Reading
Worked example
Given: Main-scale reading = 2.3 cm; Vernier reading = 0.06 cm Calculation: Total reading = 2.3 + 0.06 Answer: Total reading = 2.36 cm
Corrected Vernier Reading
Worked example
Given: Observed reading = 2.36 cm; zero error = +0.02 cm Calculation: Correct reading = 2.36 − (+0.02) Answer: Correct reading = 2.34 cm
Screw Gauge Least Count
Worked example
Given: Pitch = 1 mm; circular-scale divisions = 100 Calculation: LC = 1/100 mm Answer: LC = 0.01 mm
Corrected Screw Gauge Reading
Worked example
Given: Observed reading = 4.28 mm; zero error = −0.03 mm Calculation: Correct reading = 4.28 − (−0.03) Answer: Correct reading = 4.31 mm
Time Period from Frequency
Worked example
Given: f = 5 Hz Calculation: T = 1/5 Answer: T = 0.2 s
Frequency from Time Period
Worked example
Given: T = 0.25 s Calculation: f = 1/0.25 Answer: f = 4 Hz
Time Period of a Simple Pendulum
Worked example
Given: l = 1 m; g = 9.8 m/s² Calculation: T = 2π√(1/9.8) Answer: T ≈ 2.01 s