PhysicsICSE Class 9 · Chapter 1

Measurements and Experimentation ICSE Class 9: Vernier Callipers, Screw Gauge and Pendulum

Updated 15 October 2026 · Fact-checked

This chapter is about measuring accurately. ICSE Class 9 covers SI units (with cgs and fps), least count, reading Vernier callipers and a screw gauge of pitch 1 mm and least count 0.01 mm, correcting zero error, and the simple pendulum: time period, frequency, the l–T² graph and T = 2π√(l/g).

What Measurements and Experimentation is about

Physics depends on measurement, and every measurement has a unit and a limit to its accuracy. The CISCE 2028 Class IX syllabus covers:

the International System of Units (SI), with the cgs and fps systems for comparison;
the least count (LC) of an instrument and why a smaller least count gives a more accurate reading;
measuring length with Vernier callipers and a micrometre screw gauge (focus on pitch 1 mm and LC 0.01 mm), with zero error as a basic idea; only pictorial numericals (reading a drawn scale) are asked;
the simple pendulum: time period, frequency, the graph of **length $l$ against $T^{2}$ ** and its slope, and $T = 2 π \frac{l}{g}$ (no derivation); simple numericals only.

The Class IX paper (80 marks, 2 hours) is set by your school. Expect MCQs on units and least count, 2 to 3-mark scale readings with zero error, and 3 to 4-mark pendulum calculations or graph questions.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Physics. Class IX is examined by your school: one 80-mark paper of 2 hours plus 20 marks of internal assessment. Measurements and Experimentation is in the 1. Measurements and Experimentation (i)-(ii) part of the 2028 syllabus.Source: CISCE ICSE 2028 Physics syllabus, Class IX

Key concepts

SI, cgs and fps units

A physical quantity is written as a number times a unit, e.g. 2.5 m.

Fundamental quantities and their SI units: length (metre, m), mass (kilogram, kg), time (second, s), temperature (kelvin, K), electric current (ampere, A), amount of substance (mole, mol), luminous intensity (candela, cd). Other units, such as the newton or joule, are derived from these.

Older systems: cgs (centimetre, gram, second) and fps (foot, pound, second). Useful links: 1 m = 100 cm, 1 kg = 1000 g.

Writing rules: unit symbols are not pluralised (5 kg, not 5 kgs), have no full stop, and a unit named after a person has a lower-case name but a capital symbol (newton, N).

Worked example

Convert a speed of 72 km h⁻¹ into m s⁻¹ and into cm s⁻¹.

Try it first, then show the solutionSolution

$72 \times \frac{1000}{3600} = 20$ m s⁻¹.
$20 \times 100 = 2000$ cm s⁻¹.

Answer: 20 m s⁻¹ = 2000 cm s⁻¹

Least count and accuracy

The least count (LC) of an instrument is the smallest measurement it can read correctly. A metre rule has LC 1 mm; Vernier callipers commonly 0.1 mm; a screw gauge 0.01 mm.

A smaller least count gives a more accurate measurement, because the reading can be closer to the true value. That is why a wire's diameter is measured with a screw gauge, not a ruler.

A reading should be recorded to the least count of the instrument used: a screw gauge reading is written to two decimal places of a millimetre, e.g. 2.47 mm.

Worked example

Which instrument would you choose to measure (i) the length of a room (ii) the internal diameter of a test tube (iii) the thickness of a sheet of paper?

Try it first, then show the solutionSolution

(i) Metre rule or measuring tape (LC 1 mm is enough).
(ii) Vernier callipers (they have internal jaws; LC 0.1 mm).
(iii) Screw gauge (LC 0.01 mm).

Answer: Tape/rule; Vernier callipers; screw gauge

Vernier callipers

A Vernier calliper has a main scale (usually in mm) and a sliding Vernier scale. Commonly 10 Vernier divisions equal 9 main-scale divisions.

Vernier constant (least count): $L C = 1 MSD - 1 VSD = \frac{value of 1 MSD}{number of Vernier divisions}$ . With 1 MSD = 1 mm and 10 divisions: $L C = 0.1$ mm $= 0.01$ cm.

Reading $=$ main-scale reading (just before the Vernier zero) $+$ (number of the Vernier division that coincides with a main-scale line) $\times$ LC.

Parts: outer jaws (external diameter), inner jaws (internal diameter), strip at the end (depth).

Worked example

A Vernier calliper has LC 0.01 cm. The Vernier zero lies just past 3.4 cm on the main scale and the 6th Vernier division coincides with a main-scale line. Find the reading.

Try it first, then show the solutionSolution

Main-scale reading $= 3.4$ cm.
Vernier reading $= 6 \times 0.01 = 0.06$ cm.
Total $= 3.46$ cm.

Answer: 3.46 cm

Screw gauge (pitch 1 mm, LC 0.01 mm)

A screw gauge has a linear (pitch) scale on the sleeve and a circular (head) scale on the thimble.

Pitch: distance the spindle moves in one full turn of the thimble (here 1 mm).
Least count $= \frac{pitch}{number of circular divisions} = \frac{1 mm}{100} = 0.01$ mm.

Reading $=$ linear-scale reading (last visible mark) $+$ (circular-scale division on the reference line) $\times$ LC.

Use the ratchet to close the spindle on the object gently, so it is not squeezed and the reading is repeatable.

Worked example

For a screw gauge of pitch 1 mm with 100 circular divisions, the linear scale shows 4 mm and the 37th circular division is on the reference line. Find the diameter of the wire.

Try it first, then show the solutionSolution

LC $= \frac{1}{100} = 0.01$ mm.
Reading $= 4 + 37 \times 0.01 = 4.37$ mm.

Answer: 4.37 mm

Zero error (basic idea)

When the jaws (or the stud and spindle) are closed, the reading should be zero. If it is not, the instrument has a zero error.

Vernier callipers: if the Vernier zero is to the right of the main-scale zero, the error is positive (readings are too large); to the left, negative.
Screw gauge: if the circular-scale zero is below the reference line (a reading like 3 divisions appears), the error is positive ( $+ 3 \times L C$ ); if it is above (e.g. the 97th division is on the line), the error is negative: $- (100 - 97) \times L C$ .

Corrected reading $=$ observed reading $-$ zero error (with its sign). Subtracting a negative error adds to the reading.

Worked example

A screw gauge (LC 0.01 mm) shows the 97th circular division on the reference line when closed. A wire then reads 2.45 mm. Find the zero error and the correct diameter.

Try it first, then show the solutionSolution

Zero is above the line: error $= - (100 - 97) \times 0.01 = - 0.03$ mm.
Correct reading $= 2.45 - (- 0.03) = 2.48$ mm.

Answer: −0.03 mm; 2.48 mm

The simple pendulum

A simple pendulum is a small heavy bob hung by a light, inextensible string from a rigid support.

Effective length $l$ : from the point of suspension to the centre of the bob (string length + radius of bob).
Time period $T$ : time for one complete oscillation; find it by timing 20 oscillations and dividing.
Frequency $f = \frac{1}{T}$ (unit hertz, Hz).

For small amplitudes: $T = 2 π \frac{l}{g}$ (no derivation). So $T$ depends on $l$ and $g$ but not on the mass of the bob or the (small) amplitude; $T \propto l$ .

Graph: $l$ against $T^{2}$ is a straight line through the origin. Its slope $= \frac{l}{T ^{2}} = \frac{g}{4 π ^{2}}$ , so $g = 4 π^{2} \times slope$ .

A seconds pendulum has $T = 2$ s; its length is about 1 m on the Earth.

Worked example

A pendulum makes 20 oscillations in 40 s. Find its time period, its frequency and its effective length ( $g = 9.8$ m s⁻², $π^{2} \approx 9.87$ ).

Try it first, then show the solutionSolution

$T = \frac{40}{20} = 2$ s; $f = \frac{1}{2} = 0.5$ Hz.
$l = \frac{g T ^{2}}{4 π ^{2}} = \frac{9.8 \times 4}{4 \times 9.87}$ .
$l \approx 0.993$ m.

Answer: 2 s; 0.5 Hz; about 0.99 m

Physics in real life: Measurements and Experimentation

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Choosing the right instrument

Suppose you are asked to find the thickness of one page of your textbook, which has 400 pages (200 sheets) and is 2.2 cm thick without its covers.

Find the thickness of one sheet, and say which instrument could check a single sheet directly.

Try it first, then show the solutionSolution

Thickness of one sheet $= \frac{2.2 cm}{200} = 0.011$ cm $= 0.11$ mm.
A screw gauge (LC 0.01 mm) can measure a single sheet directly.

Answer: 0.11 mm; a screw gauge.

Memory hook: Too thin to measure? Measure many and divide, or use a finer instrument.

Zero error correction

Imagine a friend's Vernier calliper shows 0.04 cm when its jaws are fully closed, with the Vernier zero to the right of the main zero. She measures a coin as 2.47 cm.

What is the coin's real diameter?

Try it first, then show the solutionSolution

Zero error $= + 0.04$ cm (positive).
True diameter $= 2.47 - 0.04 = 2.43$ cm.

Answer: 2.43 cm.

Memory hook: A positive error makes every reading too big.

Pendulum clock

Suppose a pendulum clock runs slow because its pendulum's period has become 2.02 s instead of 2.00 s after the rod expanded slightly in summer.

Should the bob be raised or lowered to correct it, and why?

Try it first, then show the solutionSolution

$T \propto l$ ; a longer period means the effective length has increased.
Raising the bob shortens $l$ , which reduces $T$ back to 2 s.

Answer: Raise the bob (shorten the effective length).

Memory hook: Slow clock: shorten the pendulum.

Topics in this chapter

Measurements and Experimentation formula sheet

Vernier least count: $L C = 1 MSD - 1 VSD = \frac{1 MSD}{n}$; n Vernier divisions.
Vernier reading: $MSR + (coinciding VSD) \times L C$; Then subtract zero error.
Screw gauge least count: $L C = \frac{pitch}{circular divisions}$; 1 mm/100 = 0.01 mm.
Screw gauge reading: $PSR + (CSR) \times L C$; Then subtract zero error.
Corrected reading: $true = observed - zero error$; Keep the sign of the error.
Pendulum period: $T = 2 π \frac{l}{g}$; No derivation; small amplitude.
Frequency: $f = \frac{1}{T}$; Unit Hz.
Slope of l–T² graph: $\frac{l}{T ^{2}} = \frac{g}{4 π ^{2}}$; g = 4π² × slope.

Measurements and Experimentation: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Vernier callipers

Draw and label the parts, or read a drawn scale; 2 to 3 marks.

Draw

A long main scale (marked in cm and mm) with a fixed jaw at its left end.
A sliding Vernier scale with a movable jaw, below the main scale.
Outer (large) jaws at the bottom and inner (small) jaws at the top.
A thin strip coming out of the right end, and a locking screw on the slider.

Label

Main scale
Vernier scale
Outer jaws
Inner jaws
Strip (depth)
Locking screw

Watch out: Say what each jaw measures: outer for external diameter, inner for internal diameter, strip for depth.

Micrometre screw gauge

Draw and label, or read a drawn gauge; 2 to 3 marks.

Draw

A U-shaped frame with a fixed stud on the left arm.
A spindle passing through the right arm, facing the stud.
A sleeve with a linear (pitch) scale and a reference line.
A thimble with a circular scale of 100 divisions, ending in a ratchet.

Label

Frame
Stud (anvil)
Spindle
Sleeve with linear scale
Reference (index) line
Thimble with circular scale
Ratchet

Watch out: State the use of the ratchet: it slips when the grip is firm, preventing overtightening.

Graph of l against T²

Plot a pendulum's data and find g from the slope; 3 to 4 marks.

Draw

Make a table of l and T, and add a column for T².
Plot T² (s²) on the x-axis and l (m or cm) on the y-axis with a stated scale.
Draw the best straight line through the origin.
Choose two far-apart points on the line to find the slope.

Label

x-axis: T² (s²)
y-axis: l (m)
scale
slope triangle with values

Watch out: g = 4π² × slope; use points on the line, not raw data points, for the slope.

How the school exam asks Measurements and Experimentation

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Units	Name SI units of fundamental quantities; convert between SI and cgs; spot wrongly written units.	Recall the seven base units; use 1 m = 100 cm, 1 kg = 1000 g.
Least count	Find the LC of a Vernier or screw gauge from its scale data; say why a smaller LC is better.	LC = 1 MSD ÷ number of Vernier divisions, or pitch ÷ circular divisions.
Pictorial reading	A drawing of a Vernier or screw gauge scale is given; read it and correct for zero error.	Main or linear reading + coinciding division × LC; subtract the zero error with its sign.
Pendulum calculations	Find T, f, l or the effect of changing l; find the length of a seconds pendulum.	T = time ÷ oscillations; T ∝ √l; use T = 2π√(l/g).
l–T² graph	State the shape of the graph, use its slope to find g, or read a value.	Straight line through the origin; slope = g/4π².

Tricks to remember Measurements and Experimentation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Smaller LC, sharper eyes
The smaller the least count, the finer the detail an instrument sees.
Use it when: Explaining accuracy.
Error out
True = Observed − Error. A negative error therefore gets added.
Use it when: Zero-error correction.
Below is plus
Screw gauge: circular zero below the line → positive error; above the line → negative.
Use it when: Sign of screw-gauge zero error.
Four times longer, twice as slow
T ∝ √l: quadruple the length to double the period.
Use it when: Pendulum ratio questions.

Measurements and Experimentation: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Least count of a screw gauge · 1 mark · MCQExam level

A screw gauge has a pitch of 1 mm and 100 divisions on its circular scale. Its least count is:
(a) 0.1 mm
(b) 0.01 mm
(c) 0.001 mm
(d) 1 mm

Try it first, then show the solutionSolution

$L C = \frac{1}{100} = 0.01$ mm.

Answer: (b)

Example 2: Vernier with 20 divisions · 2 marks · Short answerExam level

On a Vernier calliper, 20 Vernier divisions coincide with 19 main-scale divisions of 1 mm each. Find its least count. If the main-scale reading is 2.3 cm and the 12th Vernier division coincides, find the length.

Try it first, then show the solutionSolution

$1 VSD = \frac{19}{20}$ mm $= 0.95$ mm; $L C = 1 - 0.95 = 0.05$ mm $= 0.005$ cm.
Length $= 2.3 + 12 \times 0.005 = 2.36$ cm.

Answer: 0.05 mm; 2.36 cm

Example 3: Pendulum ratio · 2 marks · Short answerExam level

A simple pendulum has a time period of 2 s. What will its time period be if its effective length is made one-fourth?

Try it first, then show the solutionSolution

$T \propto l$ , so $\frac{T _{2}}{T _{1}} = \frac{1}{4} = \frac{1}{2}$ .
$T_{2} = 1$ s.

Answer: 1 s

Example 4: Vernier reading with zero error · 3 marks · Short answerHarder exam level

A Vernier calliper has LC 0.01 cm. With the jaws closed, the Vernier zero is to the left of the main-scale zero and the 7th Vernier division coincides. While measuring a cylinder, the main-scale reading is 2.8 cm and the 4th Vernier division coincides. Find the zero error and the correct diameter.

Try it first, then show the solutionSolution

Vernier zero to the left: negative error. With 10 divisions, error $= - (10 - 7) \times 0.01 = - 0.03$ cm.
Observed reading $= 2.8 + 4 \times 0.01 = 2.84$ cm.
Correct $= 2.84 - (- 0.03) = 2.87$ cm.

Answer: −0.03 cm; 2.87 cm

Example 5: Effective length · 3 marks · Short answerHarder exam level

A bob of diameter 2.0 cm hangs on a string 98.0 cm long (measured from the support to the top of the bob). Find the effective length and the time period ( $g = 9.8$ m s⁻², $π = 3.14$ ).

Try it first, then show the solutionSolution

$l = 98.0 + 1.0 = 99.0$ cm $= 0.99$ m.
$T = 2 \times 3.14 \times \frac{0.99}{9.8} = 6.28 \times 0.3178$ .
$T \approx 2.0$ s.

Answer: 0.99 m; about 2.0 s

Example 6: g from the graph · 3 marks · Short answerHarder exam level

In an experiment, the graph of $l$ (in m) against $T^{2}$ (in s²) is a straight line through the origin passing through $(4.0, 0.99)$ . Find the slope and the value of $g$ ( $π^{2} = 9.87$ ).

Try it first, then show the solutionSolution

Slope $= \frac{0.99}{4.0} = 0.2475$ m s⁻².
$g = 4 π^{2} \times slope = 4 \times 9.87 \times 0.2475$ .
$g \approx 9.77$ m s⁻².

Answer: 0.2475 m s⁻²; about 9.8 m s⁻²

Example 7: Two pendulums in step · 4 marks · Long answerHOTS / competency

Pendulum A has effective length 1.00 m and pendulum B has effective length 0.64 m. They are released together from the same side. Taking $T_{A} = 2.0$ s, find $T_{B}$ , and find when they will next be back at the starting side together, and how many oscillations each will have made.

Try it first, then show the solutionSolution

$T_{B} = T_{A} \frac{0.64}{1.00} = 2.0 \times 0.8 = 1.6$ s.
Both are back at the start together when the time is a whole number of periods of each: a common multiple of 2.0 s and 1.6 s.
The least common multiple is 8 s: A makes $\frac{8}{2} = 4$ oscillations and B makes $\frac{8}{1.6} = 5$ .

Answer: T_B = 1.6 s; after 8 s (A: 4, B: 5 oscillations)

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 9-mark question split into 2 + 3 + 4 marks.

A school lab session

In a lab, Riya measures a steel ball with a screw gauge (pitch 1 mm, 100 circular divisions). With the gauge closed, the 2nd circular division lies on the reference line (zero below the line). With the ball in place, the linear scale shows 12 mm and the 46th division is on the line. She then hangs the ball on a string to make a pendulum whose effective length is 64.0 cm and times 20 oscillations.

(i) · 2 marks
Find the least count and the zero error of the screw gauge.
Answer
Answer: 0.01 mm; +0.02 mm
LC $= \frac{1}{100} = 0.01$ mm. Zero below the line: error $= + 2 \times 0.01 = + 0.02$ mm.
(ii) · 3 marks
Find the correct diameter of the ball and its radius in cm.
Answer
Answer: 12.44 mm; 0.622 cm
Observed $= 12 + 46 \times 0.01 = 12.46$ mm. Correct $= 12.46 - 0.02 = 12.44$ mm. Radius $= 6.22$ mm $= 0.622$ cm.
(iii) · 4 marks
Predict the time for 20 oscillations ( $g = 9.8$ m s⁻², $π = 3.14$ ). If she wanted each oscillation to take exactly twice as long, what effective length would she need?
Answer
Answer: About 32 s; 2.56 m
$T = 6.28 \frac{0.64}{9.8} = 6.28 \times 0.2556 \approx 1.605$ s; 20 oscillations $\approx 32.1$ s. Doubling $T$ needs $4 l = 2.56$ m.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

State the least count of Vernier callipers in which 10 Vernier divisions equal 9 main-scale divisions of 1 mm, and explain why a smaller least count is preferred. (2 marks)

Step	Marks
$L C = 1 - 0.9 = 0.1$ mm $= 0.01$ cm	1
A smaller least count lets the instrument measure more accurately (closer to the true value)	1

A screw gauge (LC 0.01 mm) has a zero error of −0.04 mm. When measuring a wire, the linear scale reads 1 mm and the 28th circular division is on the reference line. Find the correct diameter. (3 marks)

Step	Marks
Observed $= 1 + 28 \times 0.01 = 1.28$ mm	1
Correct $=$ observed $-$ zero error $= 1.28 - (- 0.04)$	1
$= 1.32$ mm	1

A pendulum makes 30 oscillations in 60 s at a place where $g = 9.8$ m s⁻². Find (a) its time period (b) its frequency (c) its effective length (take $π^{2} = 9.8$ ) (d) the period if the length is made four times. (4 marks)

Step	Marks
(a) $T = \frac{60}{30} = 2$ s	1
(b) $f = 0.5$ Hz	1
(c) $l = \frac{g T ^{2}}{4 π ^{2}} = \frac{9.8 \times 4}{4 \times 9.8} = 1$ m	1
(d) $T \propto l$ : $T = 4$ s	1

Where students lose marks in Measurements and Experimentation

The slips examiners see most often, and what to write instead.

Mistake: Adding the zero error instead of subtracting it
Corrected reading = observed − zero error; students add a positive error and make the reading larger.
Wrong working
$2.47 + 0.04 = 2.51$ cm
Right working
$2.47 - 0.04 = 2.43$ cm
Do this: Always write 'true = observed − error' with the error's sign in brackets.
Usually costs: 1 mark.
Mistake: Wrong sign of zero error
Students guess the sign. For a screw gauge, zero below the reference line means positive; above means negative.
Do this: Ask: does the closed instrument read more than zero (positive) or less (negative)?
Usually costs: 1 mark.
Mistake: Using string length instead of effective length
The effective length is measured to the centre of the bob.
Do this: Add the radius of the bob to the string length.
Usually costs: 1 mark.
Mistake: Mixing units in T = 2π√(l/g)
Using $l$ in cm with $g$ in m s⁻² gives a wrong period.
Do this: Convert $l$ to metres before substituting.
Usually costs: The answer.
Mistake: Timing a single oscillation
Reaction time makes one swing inaccurate.
Do this: Time 20 oscillations and divide; this reduces the error of reaction time.
Usually costs: Practical marks.

Practice questions on Measurements and Experimentation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
Which of these is a fundamental SI unit?
(a) newton
(b) joule
(c) ampere
(d) watt
Hint
Base quantities.
Answer
(c)
Q2 · 1 mark · MCQExam level
The time period of a simple pendulum does not depend on:
(a) its length
(b) the acceleration due to gravity
(c) the mass of the bob
(d) the effective length
Hint
T = 2π√(l/g).
Answer
(c)
Q3 · 2 marks · Short answerExam level
A Vernier calliper reads 4.2 cm on the main scale with the 5th Vernier division coinciding (LC 0.01 cm). It has a zero error of +0.02 cm. Find the correct reading.
Hint
Observed − error.
Answer
$4.25 - 0.02 = 4.23$ cm.
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): A screw gauge measures the diameter of a thin wire more accurately than Vernier callipers.
Reason (R): The least count of a screw gauge (0.01 mm) is smaller than that of common Vernier callipers (0.1 mm).
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Smaller LC, higher accuracy.
Answer
(c)
Q5 · 3 marks · Short answerHarder exam level
A screw gauge (pitch 1 mm, 100 divisions) reads 3 mm on the linear scale and 72 on the circular scale for a ball bearing. When closed, the 96th division was on the reference line. Find the correct diameter.
Hint
Zero above the line: negative error.
Answer
Error $= - (100 - 96) \times 0.01 = - 0.04$ mm. Observed $= 3.72$ mm. Correct $= 3.76$ mm.
Q6 · 2 marks · Short answerHarder exam level
The length of a seconds pendulum at a place is 99.4 cm. Find $g$ there ( $π^{2} = 9.87$ ).
Hint
T = 2 s; g = 4π²l/T².
Answer
$g = \frac{4 \times 9.87 \times 0.994}{4} = 9.81$ m s⁻².
Q7 · 2 marks · Short answerHarder exam level
Two pendulums have lengths in the ratio $9 : 16$ . Find the ratio of their time periods and of their frequencies.
Hint
T ∝ √l; f = 1/T.
Answer
$T_{1} : T_{2} = 3 : 4$ ; $f_{1} : f_{2} = 4 : 3$ .
Q8 · 3 marks · Short answerHarder exam level
A pendulum takes 32 s for 20 oscillations. By what fraction of its length must it be changed so that it makes 20 oscillations in 40 s?
Hint
T goes from 1.6 s to 2 s.
Answer
$\frac{l _{2}}{l _{1}} = (\frac{2}{1.6})^{2} = 1.5625$ : increase the length by $56.25%$ (multiply by $\frac{25}{16}$ ).
Q9 · 3 marks · Short answerHarder exam level
In a Vernier calliper, 1 MSD = 0.5 mm and 25 Vernier divisions equal 24 MSD. Find the least count. If the main-scale reading is 31.5 mm and the 14th Vernier division coincides, find the length in cm.
Hint
LC = 1 MSD ÷ 25.
Answer
$L C = \frac{0.5}{25} = 0.02$ mm. Length $= 31.5 + 14 \times 0.02 = 31.78$ mm $= 3.178$ cm.
Q10 · 4 marks · Long answerHarder exam level
Values from a pendulum experiment: $l$ = 0.40, 0.60, 0.80, 1.00 m; $T$ = 1.27, 1.55, 1.79, 2.01 s. Find $T^{2}$ for each, state the shape of the $l$ – $T^{2}$ graph, and estimate $g$ from the first and last readings.
Hint
Slope = Δl/ΔT².
Answer
$T^{2} \approx 1.61, 2.40, 3.20, 4.04$ s². A straight line through the origin. Slope $\approx \frac{1.00 - 0.40}{4.04 - 1.61} = \frac{0.60}{2.43} \approx 0.247$ ; $g \approx 4 \times 9.87 \times 0.247 \approx 9.75$ m s⁻².
Q11 · 4 marks · Long answerHOTS / competency
A student uses a screw gauge with a negative zero error but forgets to correct for it, and also uses the string length (not the effective length) for a pendulum. State whether each measured value (diameter; length used in $T = 2 π l / g$ ) is too big or too small, and the effect on the calculated $g$ from $g = \frac{4 π ^{2} l}{T ^{2}}$ .
Hint
Negative error: readings are too small.
Answer
Negative zero error means readings are too small, so the diameter is too small. Using the string length leaves out the bob's radius, so $l$ is too small. A smaller $l$ with the correct measured $T$ makes the calculated $g$ too small.
Q12 · 2 marks · Short answerHOTS / competency
Why is a simple pendulum's amplitude kept small (a few degrees), and why should the string be light and inextensible?
Hint
Conditions for the formula.
Answer
$T = 2 π l / g$ holds only for small amplitudes. A light, inextensible string keeps the effective length fixed and the mass concentrated in the bob.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.SI unit of mass?Show answer
Kilogram (kg).
2.What is the least count of an instrument?Show answer
The smallest measurement it can read accurately.
3.LC of a screw gauge with pitch 1 mm and 100 divisions?Show answer
0.01 mm.
4.Formula for the corrected reading?Show answer
Observed reading − zero error.
5.What is the effective length of a pendulum?Show answer
Distance from the point of suspension to the centre of the bob.
6.Does the period of a simple pendulum depend on the mass of the bob?Show answer
No.
7.Shape of the l–T² graph?Show answer
A straight line through the origin.
8.Time period of a seconds pendulum?Show answer
2 s.

Best order to study it

Learn SI base units and the cgs and fps systems.
Understand least count and practise reading Vernier scales.
Practise screw-gauge readings with pitch 1 mm and LC 0.01 mm.
Learn zero error signs and corrections for both instruments.
Finish with the pendulum: T, f, effective length, T ∝ √l and the l–T² graph.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you list the SI base units and symbols?
Can you find the least count from scale data?
Can you read a drawn Vernier or screw-gauge scale?
Do you know the sign rules for zero error?
Do you subtract the zero error with its sign?
Can you find T, f and l for a pendulum?
Can you find g from the slope of an l–T² graph?

Measurements and Experimentation: frequently asked questions

Do I need to derive T = 2π√(l/g)?

No. The syllabus says no derivation; learn to use the formula in simple numericals.

Are numericals on Vernier callipers and the screw gauge in the syllabus?

Yes, but only pictorial numericals: you read a drawn scale and correct for zero error.

Why is the period found by timing many oscillations?

Timing 20 oscillations spreads the reaction-time error over many swings, so T is more accurate.

Why is the l–T² graph a straight line?

From T = 2π√(l/g), T² = (4π²/g)l, so l is proportional to T².

What is the ratchet on a screw gauge for?

It slips once the spindle grips the object, so the object is not squeezed and readings are consistent.

Based on: CISCE ICSE 2028 Science (52) Physics syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028