Measurement of Length and Measuring Instruments Practice Questions — JAMB UTME

Question 1Question →

A micrometer screw gauge with a positive zero error of $+0.04\text{ mm}$ gives an observed reading of $2.46\text{ mm}$ when measuring the diameter of a thin wire. What is the actual diameter of the wire?

2.42\text{ mm}

2.46\text{ mm}

2.50\text{ mm}

2.38\text{ mm}

Show answer & explanation

Answer: $2.42\text{ mm}$

Answer

The actual diameter of the wire is

2.42\text{ mm}

.

To calculate the actual reading on a measuring instrument, use the relationship

\text{Actual Reading} = \text{Observed Reading} - \text{Zero Error}

. Subtracting

+0.04\text{ mm}

from

2.46\text{ mm}

yields

2.42\text{ mm}

.

Step-by-Step Solution

1

Identify the given readings

Observed reading =

2.46\text{ mm}

, Zero error =

+0.04\text{ mm}

Extract values needed for zero error correction.

2

Apply zero error correction formula

\text{Actual Reading} = \text{Observed Reading} - \text{Zero Error} = 2.46\text{ mm} - (+0.04\text{ mm}) = 2.42\text{ mm}

A positive zero error means the instrument reads higher than true zero, so the zero error must be subtracted from the observed reading.

Key Concept

Instrument Zero Error Correction

Estimated Time:45s

Question 2Question →

Match each length measurement instrument setup on the left with its corresponding true (corrected) measurement value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match each instrument's corrected reading: micrometer with positive zero error to

2.74\text{ mm}

, Vernier caliper with negative zero error to

3.48\text{ cm}

, metre rule measurement to

12.5\text{ cm}

, and micrometer with negative zero error to

1.98\text{ mm}

.

Each instrument pairing correctly applies the instrument's least count formula and the standard zero error relationship:

\text{Corrected Reading} = \text{Observed Reading} - \text{Zero Error}

.

Step-by-Step Solution

1

Calculate the corrected reading for the first micrometer screw gauge.

Least count =

0.01\text{ mm}

. Observed =

2.5\text{ mm} + 0.28\text{ mm} = 2.78\text{ mm}

. Corrected reading =

2.78\text{ mm} - 0.04\text{ mm} = 2.74\text{ mm}

.

True Reading = Observed Reading - (Zero Error).

2

Calculate the corrected reading for the Vernier caliper.

Observed =

3.4\text{ cm} + 0.06\text{ cm} = 3.46\text{ cm}

. Corrected reading =

3.46\text{ cm} - (-0.02\text{ cm}) = 3.48\text{ cm}

.

Subtracting a negative zero error is equivalent to adding the absolute error magnitude.

3

Calculate the length of the wooden rod measured with the metre rule.

Length =

14.8\text{ cm} - 2.3\text{ cm} = 12.5\text{ cm}

.

Subtracting the initial scale alignment point from the final scale alignment point eliminates end-wear errors.

4

Calculate the corrected reading for the second micrometer screw gauge.

Observed =

1.5\text{ mm} + 0.45\text{ mm} = 1.95\text{ mm}

. Corrected reading =

1.95\text{ mm} - (-0.03\text{ mm}) = 1.98\text{ mm}

.

Subtracting the negative zero error compensates for the instrument reading below zero prior to measurement.

Key Concept

Instrument Least Count and Zero Error Corrections

Question 3Question →

A micrometer screw gauge has a pitch of $0.5\text{ mm}$ and $50$ divisions on its circular thimble scale. When the anvil and spindle are brought into contact without an object between them, the $47\text{th}$ division on the thimble scale aligns exactly with the datum line of the main scale. When used to measure the diameter of a uniform metal rod, the main scale reads $1.5\text{ mm}$ and the $18\text{th}$ division on the thimble scale aligns with the datum line. What is the true diameter of the metal rod?

1.65\text{ mm}

1.68\text{ mm}

1.71\text{ mm}

2.15\text{ mm}

Show answer & explanation

Answer: $1.71\text{ mm}$

Answer

The true diameter of the metal rod is

1.71\text{ mm}

.

The least count of the micrometer is

0.01\text{ mm}

. Because the zero position when closed shows the

47\text{th}

division aligned, it has a negative zero error of

-0.03\text{ mm}

. Subtracting this negative error from the observed reading of

1.68\text{ mm}

gives

1.68 - (-0.03) = 1.71\text{ mm}

.

Step-by-Step Solution

1

Calculate the least count (precision) of the micrometer screw gauge.

\text{Least count} = \frac{\text{Pitch}}{\text{Number of thimble divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}

.

The least count determines the value of each thimble division.

2

Determine the zero error of the instrument.

Since the

47\text{th}

division coincides with the datum line when closed, the zero mark lies above the reference line.

\text{Zero error} = -(50 - 47) \times 0.01\text{ mm} = -0.03\text{ mm}

.

When the thimble zero mark has passed the datum line in the reverse direction, the instrument exhibits a negative zero error.

3

Calculate the observed reading from the main scale and thimble scale.

\text{Observed reading} = 1.5\text{ mm} + (18 \times 0.01\text{ mm}) = 1.5\text{ mm} + 0.18\text{ mm} = 1.68\text{ mm}

.

The observed reading is the sum of the main scale reading and the thimble scale reading.

4

Compute the true reading by correcting for zero error.

\text{True reading} = \text{Observed reading} - \text{Zero error} = 1.68\text{ mm} - (-0.03\text{ mm}) = 1.71\text{ mm}

.

True value is found by subtracting the zero error (including its sign) from the observed value.

Key Concept

Zero Error Correction in Micrometer Screw Gauge

Question 4Question →

A Vernier caliper with a least count of $0.01\text{ cm}$ is used to measure the internal diameter of a beaker. Before making the measurement, the jaws are brought together tightly; the zero mark of the Vernier scale lies to the left of the zero mark of the main scale, and the $6\text{th}$ division on the Vernier scale coincides with a main scale division. If the observed reading for the internal diameter of the beaker is $3.43\text{ cm}$ , what is the actual internal diameter of the beaker?

3.39\text{ cm}

3.43\text{ cm}

3.47\text{ cm}

3.49\text{ cm}

Show answer & explanation

Answer: $3.47\text{ cm}$

Answer

The actual internal diameter of the beaker is

3.47\text{ cm}

.

The instrument possesses a negative zero error because the Vernier zero mark lies to the left of the main scale zero mark when closed. The magnitude of this negative error is calculated as

-(10 - 6) \times 0.01\text{ cm} = -0.04\text{ cm}

. Applying the standard correction formula

\text{Actual Value} = \text{Observed Value} - \text{Zero Error}

gives

3.43\text{ cm} - (-0.04\text{ cm}) = 3.47\text{ cm}

.

Step-by-Step Solution

1

Determine the zero error of the Vernier caliper

Zero error =

- (10 - 6) \times 0.01\text{ cm} = -0.04\text{ cm}

When the zero mark of the Vernier scale lies to the left of the main scale zero mark, the zero error is negative. For a 10-division Vernier scale, the magnitude is given by

(10 - N) \times \text{least count}

, where

N=6

is the coinciding division.

2

Apply the zero error correction formula

Actual Reading = Observed Reading - Zero Error

The true reading is obtained by subtracting the zero error (with its sign) from the observed value.

3

Calculate the actual internal diameter

Actual Diameter =

3.43\text{ cm} - (-0.04\text{ cm}) = 3.47\text{ cm}

Subtracting a negative zero error is mathematically equivalent to adding its magnitude to the observed reading.

Key Concept

Vernier Caliper Zero Error Correction

Estimated Time:1m 30s

Question 5Question →

A micrometer screw gauge with a pitch of $0.5\text{ mm}$ and $50$ divisions on its circular thimble scale has a negative zero error such that when the jaws are fully closed, the $46\text{th}$ division on the thimble scale aligns with the main scale datum line. When this instrument is used to measure the total thickness of a stack of $40$ identical sheets of paper, the main scale reads $10.0\text{ mm}$ and the thimble scale reads $16$ divisions. What is the actual thickness of a single sheet of paper?

0.253\text{ mm}

0.255\text{ mm}

0.254\text{ mm}

0.266\text{ mm}

Show answer & explanation

Answer: $0.255\text{ mm}$

Answer

The actual thickness of a single sheet of paper is

0.255\text{ mm}

.

The least count of the micrometer is

\frac{0.5\text{ mm}}{50} = 0.01\text{ mm}

. Because the 46th division coincides with the datum line upon closure, the zero error is negative:

-(50 - 46) \times 0.01\text{ mm} = -0.04\text{ mm}

. The observed reading is

10.0\text{ mm} + 0.16\text{ mm} = 10.16\text{ mm}

. Correcting for zero error gives a true total thickness of

10.16\text{ mm} - (-0.04\text{ mm}) = 10.20\text{ mm}

. Dividing this total thickness by 40 sheets gives an individual sheet thickness of

0.255\text{ mm}

.

Step-by-Step Solution

1

Determine the least count (LC) of the micrometer screw gauge.

\text{LC} = \frac{\text{Pitch}}{\text{Number of thimble divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}

.

The least count is the minimum distance measurable by one circular scale division.

2

Calculate the zero error of the instrument.

\text{Zero error} = -(50 - 46) \times 0.01\text{ mm} = -0.04\text{ mm}

.

Since the 46th division aligns with the datum line when jaws are closed, the zero mark lies 4 divisions below the line, indicating a negative zero error.

3

Find the observed reading for the stack of 40 sheets.

\text{Observed reading} = \text{Main scale} + (\text{Thimble reading} \times \text{LC}) = 10.0\text{ mm} + (16 \times 0.01\text{ mm}) = 10.16\text{ mm}

.

The total observed value combines the main scale reading and the fractional thimble scale reading.

4

Calculate the true (corrected) total thickness of the stack.

\text{True reading} = \text{Observed reading} - \text{Zero error} = 10.16\text{ mm} - (-0.04\text{ mm}) = 10.20\text{ mm}

.

Zero error correction requires subtracting the zero error (with sign) from the observed reading.

5

Calculate the thickness of a single sheet of paper.

\text{Thickness of one sheet} = \frac{10.20\text{ mm}}{40} = 0.255\text{ mm}

.

Dividing the total corrected thickness by the total number of sheets gives the thickness per sheet.

Key Concept

Measurement of length using micrometer screw gauge with negative zero error correction

Estimated Time:2m 0s

Question 6Question →

A Vernier caliper with a least count of $0.01\text{ cm}$ is used to determine the outer diameter of a cylindrical tube. When the measuring jaws are brought together without any object between them, the zero mark of the vernier scale lies to the right of the main scale zero mark, with the $3\text{rd}$ vernier division coinciding with a main scale line. When measuring the tube, the main scale reads $4.20\text{ cm}$ and the $6\text{th}$ vernier division coincides with a main scale line. What is the actual outer diameter of the tube?

4.23\text{ cm}

4.29\text{ cm}

4.26\text{ cm}

4.17\text{ cm}

Show answer & explanation

Answer: $4.23\text{ cm}$

Answer

The actual outer diameter of the tube is

4.23\text{ cm}

.

The observed reading of the Vernier caliper is calculated as Main Scale Reading + (Coinciding Division × Least Count) =

4.20\text{ cm} + (6 \times 0.01\text{ cm}) = 4.26\text{ cm}

. Because the vernier zero lies to the right when closed, it has a positive zero error of

+0.03\text{ cm}

. Subtracting this zero error from the observed reading gives the true measurement of

4.23\text{ cm}

.

Step-by-Step Solution

1

Determine the zero error of the Vernier caliper

Zero Error

= + (3 \times 0.01\text{ cm}) = +0.03\text{ cm}

Since the zero of the vernier scale lies to the right of the main scale zero, the instrument has a positive zero error.

2

Calculate the observed reading

Observed Reading

= 4.20\text{ cm} + (6 \times 0.01\text{ cm}) = 4.26\text{ cm}

The observed measurement is the main scale reading plus the product of the coinciding vernier division and the least count.

3

Calculate the actual (corrected) reading

Actual Reading

= \text{Observed Reading} - \text{Zero Error} = 4.26\text{ cm} - (+0.03\text{ cm}) = 4.23\text{ cm}

Zero error must always be subtracted algebraically from the observed reading to obtain the accurate measurement.

Key Concept

Zero Error Correction in Vernier Calipers

Estimated Time:1m 30s

Question 7Question →

Match each length measuring instrument listed on the left with its standard precision or suitable measurement application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Micrometer screw gauge matches with precision

0.01\text{ mm}

for fine wire diameter; Vernier caliper matches with precision

0.01\text{ cm}

for tube diameters; Metre rule matches with precision

0.1\text{ cm}

for laboratory lengths; Tape measure matches with flexible long-distance measuring.

Each instrument is correctly matched to its standard least count and characteristic physical measurement task: Micrometer screw gauge (

0.01\text{ mm}

), Vernier caliper (

0.01\text{ cm}

), Metre rule (

0.1\text{ cm}

), and Tape measure for long flexible lengths.

Step-by-Step Solution

1

Identify the least count and primary application for each measuring tool.

Micrometer screw gauge:

0.01\text{ mm}

(wires); Vernier caliper:

0.01\text{ cm}

(internal/external tube diameters); Metre rule:

0.1\text{ cm}

(standard lab objects); Tape measure: large flexible distances.

Matching instruments requires recalling their resolution (least count) and physical design features.

Key Concept

Instrument least counts and appropriate selection based on object dimensions.

Question 8Question →

A micrometer screw gauge has a pitch of $0.5\text{ mm}$ and $100$ equal divisions on its circular thimble scale. When the anvil and spindle are fully closed without any object, the zero mark on the circular scale lies $4$ divisions below the datum line. The instrument is then used to measure the total thickness of a tightly bound stack of $10$ identical metal sheets. The main scale reading is $2.5\text{ mm}$ and the $38\text{th}$ division on the circular scale aligns with the datum line. What is the actual mean thickness of a single metal sheet?

0.269\text{ mm}

0.271\text{ mm}

0.267\text{ mm}

0.284\text{ mm}

Show answer & explanation

Answer: $0.267\text{ mm}$

Answer

The actual mean thickness of a single metal sheet is

0.267\text{ mm}

.

The least count of the micrometer is

\frac{0.5\text{ mm}}{100} = 0.005\text{ mm}

. With the zero mark 4 divisions below the datum line, there is a positive zero error of

+0.020\text{ mm}

. The observed total reading for 10 sheets is

2.5\text{ mm} + (38 \times 0.005\text{ mm}) = 2.690\text{ mm}

. Subtracting the positive zero error yields an actual thickness of

2.690\text{ mm} - 0.020\text{ mm} = 2.670\text{ mm}

for 10 sheets, which gives

0.267\text{ mm}

per sheet.

Step-by-Step Solution

1

Calculate the least count of the micrometer screw gauge

\text{Least count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.5\text{ mm}}{100} = 0.005\text{ mm}

The least count determines the value of each division on the circular scale.

2

Determine the zero error

\text{Zero error} = +4 \times 0.005\text{ mm} = +0.020\text{ mm}

Since the zero mark on the thimble is below the main scale datum line when closed, the error is positive.

3

Calculate the observed reading for 10 sheets

\text{Observed reading} = \text{Main scale reading} + (\text{Circular scale division} \times \text{Least count}) = 2.5\text{ mm} + (38 \times 0.005\text{ mm}) = 2.5\text{ mm} + 0.190\text{ mm} = 2.690\text{ mm}

Combines the main scale and circular scale readings to find the raw measured value.

4

Calculate the corrected reading for 10 sheets

\text{Corrected reading} = \text{Observed reading} - \text{Zero error} = 2.690\text{ mm} - 0.020\text{ mm} = 2.670\text{ mm}

Subtracting a positive zero error gives the true thickness of the 10 sheets.

5

Find the thickness of a single sheet

\text{Thickness per sheet} = \frac{2.670\text{ mm}}{10} = 0.267\text{ mm}

Dividing the total corrected thickness by the total number of sheets gives the average thickness per sheet.

Key Concept

Micrometer Screw Gauge Least Count and Zero Error Correction

Question 9Question →

A Vernier caliper with a least count of $0.01\text{ cm}$ is used to measure the thickness of a uniform metal plate. When the jaws are closed together without the plate inserted, the zero line of the Vernier scale lies to the right of the main scale zero, with the $3\text{rd}$ Vernier division coinciding with a main scale mark. When clamped around the metal plate, the main scale reading is $1.8\text{ cm}$ and the $4\text{th}$ Vernier division aligns perfectly with a main scale mark. What is the correct thickness of the metal plate?

1.81\text{ cm}

1.84\text{ cm}

1.87\text{ cm}

1.77\text{ cm}

Show answer & explanation

Answer: $1.81\text{ cm}$

Answer

The correct thickness of the metal plate is

1.81\text{ cm}

.

When the Vernier zero is positioned to the right of the main scale zero when jaws are closed, it indicates a positive zero error equal to

+ (3 \times 0.01\text{ cm}) = +0.03\text{ cm}

. The observed measurement with the metal plate is

1.8\text{ cm} + (4 \times 0.01\text{ cm}) = 1.84\text{ cm}

. Applying the standard correction formula

\text{Correct Reading} = \text{Observed Reading} - \text{Zero Error}

gives

1.84\text{ cm} - 0.03\text{ cm} = 1.81\text{ cm}

.

Step-by-Step Solution

1

Determine the zero error of the Vernier caliper

Zero error

= +3 \times 0.01\text{ cm} = +0.03\text{ cm}

Because the Vernier zero lies to the right of the main scale zero, the instrument has a positive zero error.

2

Calculate the observed reading

Observed reading

= 1.8\text{ cm} + (4 \times 0.01\text{ cm}) = 1.84\text{ cm}

The total observed value is the sum of the main scale reading and the Vernier scale reading.

3

Apply the zero error correction to obtain the actual reading

Correct reading

= 1.84\text{ cm} - (+0.03\text{ cm}) = 1.81\text{ cm}

True Reading = Observed Reading - Zero Error (with proper sign).

Key Concept

Vernier Caliper Zero Error Correction

Estimated Time:1m 15s

Question 10Question →

Match each length measuring instrument listed on the left with its standard precision and appropriate physical measurement application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Metre rule matches precision

0.1\text{ cm}

for pendulum rod; Vernier caliper matches precision

0.01\text{ cm}

for tube diameters; Micrometer screw gauge matches precision

0.01\text{ mm}

for wire diameter; Flexible measuring tape matches precision

0.1\text{ cm}

for flexible/large distances.

Each instrument is matched according to its fundamental physical least count and specialized geometry: Metre rule measures rigid lengths to

0.1\text{ cm}

, Vernier caliper measures internal/external diameters to

0.01\text{ cm}

, Micrometer screw gauge measures fine dimensions to

0.01\text{ mm}

, and flexible measuring tape measures long or curved distances to

0.1\text{ cm}

.

Step-by-Step Solution

1

Identify the least count (precision) of each measuring instrument.

Metre rule:

0.1\text{ cm}

, Vernier caliper:

0.01\text{ cm}

, Micrometer screw gauge:

0.01\text{ mm}

, Measuring tape:

0.1\text{ cm}

.

Least count defines the minimum measurable dimension and precision limit for each tool.

2

Match each instrument to its specific physical design feature and suitable application.

Metre rule

\rightarrow

rigid straight measurements (pendulum rod); Vernier caliper

\rightarrow

internal/external diameters (jaws); Micrometer screw gauge

\rightarrow

very thin objects (wire diameter); Tape measure

\rightarrow

curved/large dimensions.

The mechanical structure of the instrument dictates its intended physical measurement domain.

Key Concept

Instrument least count and application domain in length measurement

Question 11Question →

A Vernier caliper has $10$ divisions on its Vernier scale that coincide with $9$ main scale divisions of $1\text{ mm}$ each. When the measuring jaws are fully closed without any object between them, the zero mark of the Vernier scale lies to the left of the main scale zero mark, and the $7\text{th}$ Vernier division coincides precisely with a main scale mark. When used to measure the internal diameter of a hollow brass ring, the main scale reads $3.5\text{ cm}$ and the $4\text{th}$ Vernier division coincides with a main scale mark. What is the actual corrected internal diameter of the ring in cm?

Show answer & explanation

Answer: 3.57

Answer

The corrected internal diameter of the hollow brass ring is

3.57\text{ cm}

.

The least count of the Vernier caliper is

0.01\text{ cm}

. The negative zero error is

-(10 - 7) \times 0.01\text{ cm} = -0.03\text{ cm}

. The observed reading is

3.5\text{ cm} + (4 \times 0.01\text{ cm}) = 3.54\text{ cm}

. Correcting for zero error gives

3.54\text{ cm} - (-0.03\text{ cm}) = 3.57\text{ cm}

.

Step-by-Step Solution

1

Calculate the least count of the Vernier caliper.

Least count =

0.01\text{ cm}

(

0.1\text{ mm}

).

One main scale division is

1\text{ mm} = 0.1\text{ cm}

. Ten Vernier divisions equal nine main scale divisions (

0.9\text{ mm}

), so one Vernier division =

0.09\text{ cm}

. Least count =

0.1\text{ cm} - 0.09\text{ cm} = 0.01\text{ cm}

.

2

Determine the zero error of the instrument.

Zero Error =

-0.03\text{ cm}

.

For a negative zero error where the Vernier zero lies to the left of the main scale zero, Zero Error =

-(N - n) \times \text{least count}

, where

N = 10

and

n = 7

. Thus, Zero Error =

-(10 - 7) \times 0.01\text{ cm} = -0.03\text{ cm}

.

3

Calculate the uncorrected observed reading.

Observed Reading =

3.54\text{ cm}

.

Observed Reading = Main scale reading + (Vernier coincided division \times Least count) =

3.5\text{ cm} + (4 \times 0.01\text{ cm}) = 3.54\text{ cm}

.

4

Apply zero error correction to obtain the actual reading.

Corrected Reading =

3.57\text{ cm}

.

Corrected Reading = Observed Reading - Zero Error =

3.54\text{ cm} - (-0.03\text{ cm}) = 3.57\text{ cm}

.

Key Concept

Measurement of length using a Vernier caliper with negative zero error correction

Question 12Question →

A micrometer screw gauge with a pitch of $0.5\text{ mm}$ and $50$ circular scale divisions is used to measure the diameter of a uniform brass sphere. When the anvil and spindle are brought into contact without the sphere, the $45\text{th}$ division on the thimble scale aligns with the main scale index line. When the sphere is clamped between the anvil and spindle, the main scale reads $3.5\text{ mm}$ and the $28\text{th}$ division on the thimble scale coincides with the index line. What is the actual diameter of the brass sphere?

3.83\text{ mm}

3.78\text{ mm}

3.73\text{ mm}

3.33\text{ mm}

Show answer & explanation

Answer: $3.83\text{ mm}$

Answer

The actual diameter of the brass sphere is

3.83\text{ mm}

.

The correct answer of

3.83\text{ mm}

is derived by first establishing the least count (

0.01\text{ mm}

). When the jaws are closed, the

45\text{th}

mark lies below the reference line, giving a negative zero error of

-0.05\text{ mm}

. Adding the main scale (

3.5\text{ mm}

) to the thimble reading (

0.28\text{ mm}

) gives an observed value of

3.78\text{ mm}

. Subtracting the negative zero error yields

3.78\text{ mm} - (-0.05\text{ mm}) = 3.83\text{ mm}

.

Step-by-Step Solution

1

Determine the least count (precision) of the micrometer screw gauge.

\text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}

.

Least count defines the minimum measurement value represented by one thimble scale division.

2

Calculate the zero error of the instrument.

Since the

45\text{th}

division is aligned when closed, it is

5

divisions below the zero line (

45 - 50 = -5

). Thus,

\text{Zero Error (ZE)} = -5 \times 0.01\text{ mm} = -0.05\text{ mm}

.

When the zero mark on the thimble lies below the index line, the instrument has a negative zero error.

3

Calculate the observed reading of the sphere.

\text{Observed Reading (OR)} = 3.5\text{ mm} + (28 \times 0.01\text{ mm}) = 3.5\text{ mm} + 0.28\text{ mm} = 3.78\text{ mm}.

The total observed reading combines the main scale reading and the circular thimble reading.

4

Apply the zero error correction to find the true diameter.

\text{True Diameter} = \text{Observed Reading} - \text{Zero Error} = 3.78\text{ mm} - (-0.05\text{ mm}) = 3.78\text{ mm} + 0.05\text{ mm} = 3.83\text{ mm}.

True measurement is always obtained by subtracting the zero error (including its sign) from the observed reading.

Key Concept

Negative zero error correction in micrometer screw gauge measurements

Estimated Time:2m 0s

Question 13Question →

A micrometer screw gauge has a pitch of $0.5\text{ mm}$ and $50$ divisions on its circular thimble scale. When its anvil and spindle are fully brought together without any object, the zero mark of the thimble scale lies $4$ divisions past the main scale index line in the tightening direction (a positive zero error). When the wall thickness of a hollow metal pipe is measured with this micrometer, the main scale reading is $3.5\text{ mm}$ and the $32\text{nd}$ thimble division aligns with the index line. If the outer diameter of the pipe is separately measured to be $25.40\text{ mm}$ , what is the corrected inner diameter of the metal pipe in millimeters?

Show answer & explanation

Answer: 17.84

Answer

17.84 mm

The least count of the micrometer screw gauge is

\frac{0.5\text{ mm}}{50} = 0.01\text{ mm}

. The observed wall thickness is

3.5\text{ mm} + (32 \times 0.01\text{ mm}) = 3.82\text{ mm}

. Accounting for the positive zero error of

+0.04\text{ mm}

gives a true wall thickness of

3.82\text{ mm} - 0.04\text{ mm} = 3.78\text{ mm}

. Subtracting twice the wall thickness from the outer diameter yields the inner diameter:

25.40\text{ mm} - 2(3.78\text{ mm}) = 17.84\text{ mm}

.

Step-by-Step Solution

1

Calculate the least count (precision) of the micrometer screw gauge

Least count = 0.01 mm

Least count is defined as pitch divided by total circular scale divisions: 0.5 mm / 50 = 0.01 mm.

2

Determine the observed wall thickness from the instrument readings

Observed thickness = 3.82 mm

Observed reading = Main scale reading + (Thimble division * Least count) = 3.5 mm + (32 * 0.01 mm) = 3.82 mm.

3

Correct the observed thickness for positive zero error

Corrected wall thickness = 3.78 mm

Positive zero error (+0.04 mm) must be subtracted from the observed reading: 3.82 mm - 0.04 mm = 3.78 mm.

4

Calculate the corrected inner diameter of the hollow pipe

Inner diameter = 17.84 mm

The inner diameter equals the outer diameter minus twice the wall thickness: 25.40 mm - 2(3.78 mm) = 25.40 mm - 7.56 mm = 17.84 mm.

Key Concept

Measurement of length, micrometer zero error correction, and geometry of hollow cylinders

Estimated Time:2m 30s

Question 14Question →

A Vernier caliper with a least count of $0.01\text{ cm}$ is used to measure the internal diameter of a cylindrical pipe. When the jaws are fully closed without any object between them, the zero mark of the Vernier scale lies to the right of the zero mark on the main scale, and the $3\text{rd}$ Vernier division coincides with a main scale division. During measurement, the main scale reads $2.40\text{ cm}$ and the $6\text{th}$ Vernier division coincides with a main scale mark. What is the corrected internal diameter of the pipe?

2.43\text{ cm}

2.46\text{ cm}

2.49\text{ cm}

2.37\text{ cm}

Show answer & explanation

Answer: $2.43\text{ cm}$

Answer

The corrected internal diameter of the pipe is

2.43\text{ cm}

.

The Vernier caliper has a positive zero error because the zero mark on the Vernier scale lies to the right of the zero mark on the main scale when closed. The magnitude of this error is

3 \times 0.01\text{ cm} = +0.03\text{ cm}

. The observed reading is

2.40\text{ cm} + (6 \times 0.01\text{ cm}) = 2.46\text{ cm}

. Subtracting the zero error from the observed reading yields

\text{Corrected reading} = 2.46\text{ cm} - 0.03\text{ cm} = 2.43\text{ cm}

.

Step-by-Step Solution

1

Determine the zero error of the Vernier caliper.

Zero error

= + (3 \times 0.01\text{ cm}) = +0.03\text{ cm}

.

Since the zero of the Vernier scale lies to the right of the main scale zero mark, the instrument has a positive zero error.

2

Calculate the uncorrected (observed) reading.

\text{Observed reading} = 2.40\text{ cm} + (6 \times 0.01\text{ cm}) = 2.46\text{ cm}$.

The observed reading is the sum of the main scale reading and the Vernier scale coincidence value.

3

Apply the zero error correction formula.

\text{Corrected reading} = 2.46\text{ cm} - (+0.03\text{ cm}) = 2.43\text{ cm}$.

Corrected reading is given by subtracting the zero error (including its sign) from the observed reading.

Key Concept

Zero Error Correction in Length Measurement Instruments

Question 15Question →

A Vernier caliper has $20$ divisions on its Vernier scale that coincide with $19$ main scale divisions of $1\text{ mm}$ each. When the jaws of the instrument are brought together without any object, the zero of the Vernier scale lies to the right of the main scale zero mark, and the $3\text{rd}$ Vernier division coincides with a main scale mark. When used to measure the thickness of a wooden block, the main scale reads $3.5\text{ cm}$ and the $12\text{th}$ Vernier division coincides with a main scale line. What is the actual thickness of the wooden block?

3.560\text{ cm}

3.575\text{ cm}

3.545\text{ cm}

3.530\text{ cm}

Show answer & explanation

Answer: $3.545\text{ cm}$

Answer

The actual thickness of the wooden block is

3.545\text{ cm}

.

The least count of a 20-division Vernier caliper with

1\text{ mm}

main scale divisions is

\frac{0.1\text{ cm}}{20} = 0.005\text{ cm}

. Since the Vernier zero lies to the right of the main scale zero when closed, it has a positive zero error of

+ (3 \times 0.005\text{ cm}) = +0.015\text{ cm}

. The observed measurement is

3.5\text{ cm} + (12 \times 0.005\text{ cm}) = 3.560\text{ cm}

. Subtracting the positive zero error gives the actual thickness:

3.560\text{ cm} - 0.015\text{ cm} = 3.545\text{ cm}

.

Step-by-Step Solution

1

Determine the least count (precision) of the Vernier caliper.

\text{Least Count (LC)} = \frac{1\text{ Main Scale Division (MSD)}}{\text{Number of Vernier Divisions}} = \frac{1\text{ mm}}{20} = 0.05\text{ mm} = 0.005\text{ cm}

.

The least count is the smallest value that can be measured directly by the instrument.

2

Calculate the zero error of the instrument.

\text{Zero Error} = +(3 \times 0.005\text{ cm}) = +0.015\text{ cm}

.

Because the Vernier zero lies to the right of the main scale zero mark when closed, the instrument has a positive zero error.

3

Calculate the total observed reading from the scales.

\text{Observed Reading} = 3.5\text{ cm} + (12 \times 0.005\text{ cm}) = 3.5\text{ cm} + 0.060\text{ cm} = 3.560\text{ cm}

.

The observed reading combines the main scale reading and the coincidental Vernier scale division multiplied by the least count.

4

Apply zero error correction to obtain the actual reading.

\text{Actual Reading} = \text{Observed Reading} - \text{Zero Error} = 3.560\text{ cm} - (+0.015\text{ cm}) = 3.545\text{ cm}

.

Zero error correction always requires subtracting the zero error (with its sign) from the observed measurement.

Key Concept

Measurement of length using Vernier calipers and zero error correction

Estimated Time:1m 30s

Question 16Question →

Match each length measuring instrument on the left with its appropriate application and precision on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Metre rule pairs with measuring laboratory desk height (precision

1\text{ mm}

); Vernier caliper pairs with measuring internal tube diameter (precision

0.1\text{ mm}

); Micrometer screw gauge pairs with measuring glass cover slip thickness (precision

0.01\text{ mm}

); Measuring tape pairs with measuring distances over several metres.

Each instrument is correctly matched to its standard precision and intended application: the metre rule measures moderate lengths to within

1\text{ mm}

; the Vernier caliper measures internal and external dimensions to within

0.1\text{ mm}

; the micrometer screw gauge measures small thicknesses to within

0.01\text{ mm}

; and the measuring tape measures long or curved distances.

Step-by-Step Solution

1

Determine the least count (precision) and functional design of each measuring instrument.

Metre rule has a least count of

1\text{ mm}

; Vernier caliper has a least count of

0.1\text{ mm}

(

0.01\text{ cm}

) and possesses internal jaws; Micrometer screw gauge has a least count of

0.01\text{ mm}

; Measuring tape is flexible and suitable for long distances.

Each instrument is engineered for a specific range of dimensions and required degree of precision.

2

Match each instrument to the scenario requiring its specific precision and physical capability.

Metre rule matches desk height; Vernier caliper matches internal tube diameter; Micrometer screw gauge matches thin glass sheet thickness; Measuring tape matches large room dimensions.

Selecting the proper instrument minimizes measurement uncertainty and matches physical constraints.

Key Concept

Instrument selection based on least count, precision requirements, and physical geometry

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