Measurements and Experimentation

Introduction

Measurement is the foundation of physics. Accurate measurements and proper experimentation techniques are essential skills for ICSE Class 9 students. This chapter covers units, measuring instruments, and the simple pendulum.

SI System of Units

The International System of Units (SI) is the standard system of measurement used worldwide.

Base Quantities and Units

QuantitySI UnitSymbol
Lengthmetrem
Masskilogramkg
Timeseconds
TemperaturekelvinK
Electric currentampereA
Amount of substancemolemol
Luminous intensitycandelacd

Prefixes for SI Units

PrefixSymbolFactor
kilok10³
hectoh10²
decada10¹
decid10⁻¹
centic10⁻²
millim10⁻³
microμ10⁻⁶
nanon10⁻⁹

Vernier Callipers

A Vernier calliper is used for measuring lengths accurately up to 0.01 cm (0.1 mm).

Parts

  • Main scale: Graduated in cm and mm
  • Vernier scale: Sliding scale with divisions slightly smaller than main scale

Least Count

Least Count (LC) = Value of 1 main scale division / Number of divisions on Vernier scale

For standard Vernier: 1 MSD = 1 mm, 10 VSD = 9 mm, so 1 VSD = 0.9 mm LC = 1 mm - 0.9 mm = 0.1 mm = 0.01 cm

Reading

Total reading = Main scale reading + (Vernier division coinciding × LC)

<ICSEExample title="Vernier Calliper Reading"> The main scale reading is 3.4 cm and the 6th division on the Vernier scale coincides with a main scale division. LC = 0.01 cm. Find the length. <Solution> Length = MSR + (VCD × LC) = 3.4 + (6 × 0.01) = 3.4 + 0.06 = 3.46 cm </Solution> </ICSEExample>

Screw Gauge

A screw gauge measures small lengths accurately up to 0.001 cm (0.01 mm).

Parts

  • Pitch: Distance moved by the spindle per one complete rotation
  • Circular scale: Divisions on the thimble

Least Count

LC = Pitch / Number of divisions on circular scale

Standard: pitch = 1 mm, circular scale = 100 divisions LC = 1/100 = 0.01 mm = 0.001 cm

Reading

Total reading = Main scale reading + (Circular scale reading × LC)

<ICSEExample title="Screw Gauge Reading"> The pitch of a screw gauge is 0.5 mm and it has 50 circular scale divisions. The main scale reading is 2.5 mm and the 25th division on the circular scale coincides. Find the thickness. <Solution> LC = 0.5/50 = 0.01 mm Reading = 2.5 + (25 × 0.01) = 2.5 + 0.25 = 2.75 mm </Solution> </ICSEExample>

Simple Pendulum

A simple pendulum consists of a point mass (bob) suspended by a light, inextensible string from a rigid support.

Time Period

T = 2π√(L/g)

Where T = time period, L = effective length, g = acceleration due to gravity.

Determination of g

g = 4π²L/T²

<ICSEExample title="Find g Using Pendulum"> A simple pendulum of length 100 cm has a time period of 2.01 s. Find g. <Solution> T = 2π√(L/g) g = 4π²L/T² g = 4 × (3.14)² × 1.0/(2.01)² g = 4 × 9.8596 × 1.0/4.0401 g = 9.76 m/s² </Solution> </ICSEExample>

Tabulation of Data and Graphs

Good experimental practice requires:

  • Recording observations in a neat table
  • Plotting graphs with appropriate scales
  • Identifying relationships between variables

Common Mistakes With Fixes

MistakeCorrection
Confusing MSR and VSR in VernierMSR is before zero of Vernier; VSR is the coinciding division
Forgetting zero error in screw gaugeAlways note and apply zero correction
Counting oscillations incorrectly for pendulumCount the number of complete oscillations, not half-swings
Plotting dependent variable on x-axisIndependent variable on x-axis, dependent on y-axis

ICSE Exam Focus

TopicMarks (approx.)Frequency
Vernier calliper reading3 marksVery common
Screw gauge reading3 marksVery common
Simple pendulum calculations4 marksCommon
SI units and conversions2-3 marksVery common

Self-Test

Q1: What is the least count of a Vernier calliper with 1 MSD = 1 mm and 10 VSD = 9 mm?

Q2: A screw gauge has a pitch of 1 mm and 100 divisions on the circular scale. What is its LC?

Q3: A pendulum of length 1.5 m has a time period of 2.45 s. Find g.

Q4: The main scale reading of a Vernier is 5.8 cm and the 8th Vernier division coincides. Find the length (LC = 0.01 cm).

Q5: Name the SI units of mass, time, and temperature.

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