Measurements and Units Practice Questions — JAMB UTME — Page 3

Question 41Question →

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 42Question →

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 43Question →

The speed of sound $v$ in a gas depends on the gas pressure $P$ and density $\rho$ according to the relation $v = k P^a \rho^b$ , where $k$ is a dimensionless constant. What is the numerical value of the exponent $a$ ?

Show answer & explanation

Answer: 0.5

Answer

The numerical value of the exponent

a

is

0.5

.

Applying dimensional homogeneity to the relation

v = k P^a \rho^b

yields

[M^0 L^1 T^{-1}] = [M L^{-1} T^{-2}]^a [M L^{-3}]^b

. Equating the powers of time

T

gives

-2a = -1

, which simplifies to

a = 0.5

.

Step-by-Step Solution

1

Determine the base dimensions of all physical quantities in the relationship

[v] = M^0 L T^{-1}

,

[P] = M L^{-1} T^{-2}

, and

[\rho] = M L^{-3}

Physical quantities must be expressed in fundamental dimensions (

M, L, T

) to apply dimensional analysis.

2

Formulate the dimensional balance equation

M^0 L^1 T^{-1} = M^{a+b} L^{-a-3b} T^{-2a}

By the principle of dimensional homogeneity, the total dimensions on the left side must equal those on the right side.

3

Equate exponents of

T

to solve for

a

-2a = -1 \implies a = 0.5

Comparing powers of time

T

directly isolates the variable

a

.

Key Concept

Dimensional Homogeneity and Derivation of Exponents

Question 44Question →

During an investigation on electrical circuits, a student measures electric current, potential difference, time, and thermodynamic temperature. Which of the measured physical quantities is a derived quantity?

Electric current

Potential difference

Time

Thermodynamic temperature

Show answer & explanation

Answer: Potential difference

Answer

Potential difference

Potential difference is a derived physical quantity because it is derived from work (energy) and electric charge (

V = \frac{W}{Q} = \frac{W}{I \cdot t}

), unlike electric current, time, and thermodynamic temperature which are fundamental SI quantities.

Step-by-Step Solution

1

Identify fundamental (base) physical quantities

The seven SI fundamental quantities are mass, length, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.

Fundamental quantities are basic quantities that do not depend on any other physical quantity.

2

Classify the given quantities

Electric current, time, and thermodynamic temperature are base quantities. Potential difference (

V = \frac{W}{Q}

) is derived from work and electric charge.

Derived quantities are physical quantities defined by combining fundamental quantities.

Key Concept

Fundamental and Derived Quantities

Estimated Time:1m 0s

Question 45Question →

According to Newton's law of universal gravitation, the gravitational force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2}$ . Which of the following correctly expresses the derived SI unit of the universal gravitational constant, $G$ , in terms of fundamental (base) units?

\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}

\text{kg}\cdot\text{m}^3\cdot\text{s}^{-2}

\text{kg}^{-1}\cdot\text{m}^2\cdot\text{s}^{-2}

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

Show answer & explanation

Answer: $\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}$

Answer

The SI unit of the universal gravitational constant

G

expressed in fundamental base units is

\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}

.

The expression

\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}

is correct because rearranging

F = \frac{G m_1 m_2}{r^2}

yields

G = \frac{F r^2}{m_1 m_2}

. Replacing

F

with its base equivalent

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

,

r

with

\text{m}

, and

m_1, m_2

with

\text{kg}

gives

\frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}^2}{\text{kg}^2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}

.

Step-by-Step Solution

1

Rearrange the gravitational formula to solve for

G

G = \frac{F \cdot r^2}{m_1 \cdot m_2}

To express the unit of

G

, we need it in terms of quantities with known units.

2

Substitute the SI unit for force in fundamental base units

\text{Unit of } F = \text{N} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

Force is mass times acceleration (

F = ma

), so its base unit is

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

.

3

Substitute all base units into the formula for

G

\text{Unit of } G = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \cdot \text{m}^2}{\text{kg} \cdot \text{kg}} = \frac{\text{kg}\cdot\text{m}^3\cdot\text{s}^{-2}}{\text{kg}^2}

Combining length terms (

\text{m} \cdot \text{m}^2 = \text{m}^3

) and mass terms in the denominator.

4

Simplify the mass exponent using laws of indices

\text{Unit of } G = \text{kg}^{1 - 2}\cdot\text{m}^3\cdot\text{s}^{-2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}

Dividing by

\text{kg}^2

subtracts 2 from the mass exponent.

Key Concept

Derivation of Derived Units in Base SI Units

Estimated Time:1m 15s

Question 46Question →

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 47Question →

In electrostatics, Coulomb's law states that the force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a vacuum is given by $F = \frac{q_1 q_2}{4\pi \varepsilon_0 r^2}$ , where $\varepsilon_0$ is the permittivity of free space. What are the fundamental dimensions of $\varepsilon_0$ expressed in terms of mass ( $M$ ), length ( $L$ ), time ( $T$ ), and electric current ( $I$ )?

M^{-1} L^{-3} T^4 I^2

M^{-1} L^{-3} T^2 I^2

M L^3 T^{-4} I^{-2}

M^{-1} L^{-2} T^4 I^2

Show answer & explanation

Answer: $M^{-1} L^{-3} T^4 I^2$

Answer

The dimensions of permittivity of free space

\varepsilon_0

are

M^{-1} L^{-3} T^4 I^2

.

The correct answer is derived by isolating

\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}

. Substituting

[q] = I T

,

[F] = M L T^{-2}

, and

[r] = L

gives

[\varepsilon_0] = \frac{I^2 T^2}{M L^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2

.

Step-by-Step Solution

1

Rearrange Coulomb's Law to isolate permittivity of free space

\varepsilon_0

\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}

Isolating

\varepsilon_0

allows us to substitute the dimensions of each constituent physical quantity.

2

Determine the dimensions of charge

q

, force

F

, distance

r

, and the constant

4\pi

[q] = I T

,

[F] = M L T^{-2}

,

[r^2] = L^2

, and

[4\pi] = 1

(dimensionless)

Electric current is a base unit (

I

), so electric charge is current multiplied by time (

I T

). Force is mass times acceleration (

M L T^{-2}

).

3

Substitute the fundamental dimensions into the rearranged equation and simplify exponent powers

[\varepsilon_0] = \frac{(I T)(I T)}{(M L T^{-2})(L^2)} = \frac{I^2 T^2}{M L^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2

Applying exponent laws:

T^2 / T^{-2} = T^{2 - (-2)} = T^4

,

1 / M = M^{-1}

, and

1 / L^3 = L^{-3}

.

Key Concept

Dimensional analysis of physical constants in electromagnetism

Question 48Question →

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 49Question →

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 50Question →

An object has a mass of $12\text{ kg}$ on Earth, where the acceleration due to gravity is $10\text{ m s}^{-2}$ . What are the mass and weight of the object on the Moon, where the acceleration due to gravity is $1.6\text{ m s}^{-2}$ ?

12\text{ kg}

and

19.2\text{ N}

1.92\text{ kg}

and

19.2\text{ N}

12\text{ kg}

and

120\text{ N}

1.92\text{ kg}

and

120\text{ N}

Show answer & explanation

Answer: $12\text{ kg}$ and $19.2\text{ N}$

Answer

The mass on the Moon is

12\text{ kg}

and the weight on the Moon is

19.2\text{ N}

.

Mass is constant everywhere in the universe, so the object maintains a mass of

12\text{ kg}

on the Moon. Weight is the gravitational force exerted on the object, given by

W = mg

. On the Moon,

W = 12 \cdot 1.6 = 19.2\text{ N}

. Thus, the option specifying

12\text{ kg}

and

19.2\text{ N}

is correct.

Step-by-Step Solution

1

Determine the mass of the object on the Moon.

Mass

= 12\text{ kg}

.

Mass is a scalar physical quantity representing the quantity of matter in a body; it is constant and independent of location or gravitational field.

2

Calculate the weight of the object on the Moon using

W = m \cdot g_{\text{moon}}

.

W = 12\text{ kg} \times 1.6\text{ m s}^{-2} = 19.2\text{ N}

.

Weight is the force of gravitational attraction acting on a mass and varies directly with local gravitational field strength.

Key Concept

Mass is an intrinsic property that remains constant across different locations, whereas weight is a force that depends on local gravitational acceleration (

W = mg

).

Estimated Time:45s

Question 51Question →

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 52Question →

A spring balance calibrated in newtons has a zero error of $+2.0\text{ N}$ (it displays $+2.0\text{ N}$ before any load is attached). When an object is suspended from the balance in a location where the acceleration due to gravity is $10\text{ m s}^{-2}$ , the scale displays a reading of $42.0\text{ N}$ . What is the true mass of the object?

4.2\text{ kg}

4.0\text{ kg}

4.4\text{ kg}

40.0\text{ kg}

Show answer & explanation

Answer: $4.0\text{ kg}$

Answer

4.0\text{ kg}

To obtain the true weight from a spring balance with a positive zero error, the zero offset must be subtracted from the indicated reading:

\text{True Weight} = 42.0\text{ N} - 2.0\text{ N} = 40.0\text{ N}

. Dividing this true weight by the gravitational acceleration (

10\text{ m s}^{-2}

) gives the true mass of

4.0\text{ kg}

.

Step-by-Step Solution

1

Correct the scale reading for instrument zero error to obtain true weight

True Weight

W = 42.0\text{ N} - 2.0\text{ N} = 40.0\text{ N}

A positive zero error means the balance overcounts force, so the initial reading must be subtracted from the scale display.

2

Calculate the true mass using the formula

W = mg

Mass

m = \frac{W}{g} = \frac{40.0\text{ N}}{10\text{ m s}^{-2}} = 4.0\text{ kg}

Mass is found by dividing the true force of gravity (weight) by the acceleration due to gravity.

Key Concept

Instrument zero error correction and mass-weight relationship

Estimated Time:1m 0s

Question 53Question →

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 54Question →

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 55Question →

Which of the following combinations consists entirely of derived physical quantities?

Length, acceleration, and thermodynamic temperature

Volume, momentum, and electrical potential difference

Mass, velocity, and luminous intensity

Time, force, and electric current

Show answer & explanation

Answer: Volume, momentum, and electrical potential difference

Answer

The combination comprising volume, momentum, and electrical potential difference consists entirely of derived physical quantities.

The combination containing volume, momentum, and electrical potential difference consists entirely of derived physical quantities because volume depends on length (

L^3

), momentum depends on mass, length, and time (

M\cdot L\cdot T^{-1}

), and electrical potential difference depends on mass, length, time, and electric current (

M\cdot L^2\cdot T^{-3}\cdot I^{-1}

). None of these three are base/fundamental quantities.

Step-by-Step Solution

1

Identify the seven fundamental physical quantities in SI units.

The seven fundamental quantities are mass, length, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.

Fundamental quantities serve as the basic foundation from which all other physical quantities are defined.

2

Classify each physical quantity present in the given combinations.

Volume (

m^3

), momentum (

kg\cdot m/s

), and potential difference (

V

or

kg\cdot m^2/(A\cdot s^3)

) are all derived from fundamental units. Other combinations contain fundamental quantities such as length, mass, time, temperature, or electric current.

Derived quantities are physical quantities defined by mathematical combinations of fundamental quantities.

3

Select the option where every listed quantity is a derived quantity.

The group containing volume, momentum, and electrical potential difference is the only set where all items are derived quantities.

It fulfills the requirement of containing exclusively derived physical quantities.

Key Concept

Fundamental quantities are independent quantities defined by international standard protocols, whereas derived quantities are formed by mathematical combinations of fundamental quantities.

Question 56Question →

The volume flow rate $Q$ of a viscous liquid through a pipe depends on the radius $r$ of the pipe, the coefficient of viscosity $\eta$ , and the pressure gradient $\frac{\Delta P}{L}$ according to the dimensional equation $Q = k r^x \eta^y \left(\frac{\Delta P}{L}\right)^z$ , where $k$ is a dimensionless constant. What is the value of the exponent $x$ ?

Show answer & explanation

Answer: 4

Answer

The value of the exponent

x

is 4.

Applying the principle of dimensional homogeneity, the dimensions of volume flow rate

[Q] = L^3 T^{-1}

are equated to

[r]^x [\eta]^y \left[\frac{\Delta P}{L}\right]^z = L^x (M L^{-1} T^{-1})^y (M L^{-2} T^{-2})^z

. Equating powers yields

y + z = 0

for mass,

-y - 2z = -1

for time, and

x - y - 2z = 3

for length. Solving these simultaneous equations gives

z = 1

,

y = -1

, and

x = 4

.

Step-by-Step Solution

1

Identify the base dimensions of each physical quantity in the given equation.

Flow rate

[Q] = M^0 L^3 T^{-1}

, radius

[r] = L

, viscosity

[\eta] = M L^{-1} T^{-1}

, and pressure gradient

\left[\frac{\Delta P}{L}\right] = M L^{-2} T^{-2}

.

Expressing each quantity in terms of fundamental dimensions (

M

,

L

,

T

) is necessary for dimensional analysis.

2

Substitute dimensions into the power-law equation and collect powers of base dimensions.

M^0 L^3 T^{-1} = M^{y+z} L^{x-y-2z} T^{-y-2z}

.

The principle of dimensional homogeneity requires both sides of a physically valid equation to have identical dimensions.

3

Set up and solve linear equations for the exponents

x

,

y

, and

z

.

Solving

y + z = 0

,

-y - 2z = -1

, and

x - y - 2z = 3

yields

z = 1

,

y = -1

, and

x = 4

.

Equating powers of

M

,

T

, and

L

allows step-by-step determination of each unknown exponent.

Key Concept

Dimensions of Physical Quantities and Dimensional Analysis

Question 57Question →

A ticker-tape timer connected to a $50\text{ Hz}$ alternating current mains supply is used to measure the time interval of a laboratory cart moving down an inclined plane. On the printed tape, $10$ distinct tick spaces are recorded between the initial and final position. A electronic stopwatch running concurrently has a positive zero error of $+0.04\text{ s}$ . What is the true elapsed time interval represented by the ticker tape after properly correcting for the stopwatch zero error?

0.16\text{ s}

0.24\text{ s}

0.20\text{ s}

0.22\text{ s}

Show answer & explanation

Answer: $0.16\text{ s}$

Answer

The true elapsed time interval is

0.16\text{ s}

.

The ticker-tape frequency of

50\text{ Hz}

gives a time interval of

0.02\text{ s}

per space. Ten tick spaces correspond to an uncorrected time of

0.20\text{ s}

. To correct for a positive zero error of

+0.04\text{ s}

, the zero error must be subtracted from the uncorrected reading, yielding a true elapsed time of

0.16\text{ s}

.

Step-by-Step Solution

1

Calculate the period of a single tick interval

T = \frac{1}{f} = \frac{1}{50\text{ Hz}} = 0.02\text{ s}

Frequency

f

is the inverse of period

T

for a ticker-tape timer.

2

Determine the uncorrected total elapsed time from tick spaces

t_{\text{uncorrected}} = 10 \times 0.02\text{ s} = 0.20\text{ s}

Elapsed time equals the number of tick spaces multiplied by the period of one space.

3

Apply the zero error correction

t_{\text{true}} = t_{\text{uncorrected}} - \text{zero error} = 0.20\text{ s} - 0.04\text{ s} = 0.16\text{ s}

True Reading = Observed Reading - (Zero Error).

Key Concept

Measurement of time using ticker-tape timers and instrument zero error correction

Estimated Time:1m 30s

Question 58Question →

A student records the period of oscillation of a simple pendulum by measuring the time taken for 20 complete swings as $40.0\text{ s}$ . If the absolute error in this time measurement is $\pm 0.8\text{ s}$ , what is the percentage error in the measured time?

0.2\%

2.0\%

5.0\%

8.0\%

Show answer & explanation

Answer: $2.0\%$

Answer

2.0\%

The correct answer is

2.0\%

. Percentage error is defined as the absolute error divided by the measured value, expressed as a percentage:

\frac{0.8\text{ s}}{40.0\text{ s}} \times 100\% = 2.0\%

.

Step-by-Step Solution

1

Identify the given measurement and absolute error.

Measured time

t = 40.0\text{ s}

and absolute uncertainty

\Delta t = 0.8\text{ s}

.

These quantities are needed to determine the relative error of the time measurement.

2

Apply the percentage error formula:

\text{Percentage Error} = \left(\frac{\Delta t}{t}\right) \times 100\%

.

\text{Percentage Error} = \left(\frac{0.8\text{ s}}{40.0\text{ s}}\right) \times 100\% = 0.02 \times 100\% = 2.0\%

.

Multiplying the fractional error by 100 converts it into a percentage representation.

Key Concept

Percentage Error in Physical Measurements

Estimated Time:45s

Question 59Question →

Match each of the physical quantities given on the left with its corresponding fundamental SI status or base unit resolution on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Luminous intensity matches with Fundamental quantity measured in candela (cd); Linear momentum matches with Derived quantity expressed in kg·m·s⁻¹; Thermodynamic temperature matches with Fundamental quantity measured in kelvin (K); Specific heat capacity matches with Derived quantity expressed in m²·s⁻²·K⁻¹.

Luminous intensity and thermodynamic temperature are fundamental physical quantities with SI units candela (

\text{cd}

) and kelvin (

\text{K}

) respectively. Linear momentum (

p = mv

) and specific heat capacity (

c = \frac{Q}{m \Delta T}

) are derived physical quantities whose fundamental SI base unit resolutions are

\text{kg}\cdot\text{m}\cdot\text{s}^{-1}

and

\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}

respectively.

Step-by-Step Solution

1

Identify fundamental physical quantities and their base SI units

Luminous intensity (measured in

\text{cd}

) and thermodynamic temperature (measured in

\text{K}

) are base physical quantities that cannot be expressed in terms of other quantities.

The standard SI system establishes 7 base independent quantities.

2

Resolve derived quantities into fundamental SI base units

Linear momentum (

p = mv

) has units

\text{kg}\cdot\text{m}\cdot\text{s}^{-1}

. Specific heat capacity (

c = \frac{Q}{m\Delta T}

) has units

\frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}\cdot\text{K}} = \text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}

.

Derived physical quantities are formed by combining fundamental quantities algebraically according to physical laws.

3

Match each physical quantity to its correct description

Luminous intensity

\rightarrow

candela (

\text{cd}

); Linear momentum

\rightarrow

\text{kg}\cdot\text{m}\cdot\text{s}^{-1}

; Thermodynamic temperature

\rightarrow

kelvin (

\text{K}

); Specific heat capacity

\rightarrow

\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}

.

Each pair directly aligns the physical quantity with its base classification and unit derivation.

Key Concept

Classification of fundamental and derived physical quantities and their resolution into SI base units

Question 60Question →

Pair each physical quantity in List I with its equivalent SI unit expressed purely in terms of fundamental base units in List II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electric Charge matches

\text{A}\cdot\text{s}

; Linear Momentum matches

\text{kg}\cdot\text{m}\cdot\text{s}^{-1}

; Young's Modulus matches

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

; Magnetic Flux Density matches

\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}

.

Each physical quantity is resolved to fundamental SI base units using defining equations: Electric charge (

Q=It

) yields

\text{A}\cdot\text{s}

, linear momentum (

p=mv

) yields

\text{kg}\cdot\text{m}\cdot\text{s}^{-1}

, Young's modulus (stress/strain) yields

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

, and magnetic flux density (

B=\frac{F}{IL}

) yields

\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}

.

Step-by-Step Solution

1

Derive base units for Electric Charge

\text{A}\cdot\text{s}

From

Q = I t

, electric current has the fundamental unit ampere (A) and time has the fundamental unit second (s).

2

Derive base units for Linear Momentum

\text{kg}\cdot\text{m}\cdot\text{s}^{-1}

From

p = m v

, mass is in kilograms (kg) and velocity is in meters per second (m/s).

3

Derive base units for Young's Modulus

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

Tensile strain is dimensionless. Tensile stress is force per area:

\frac{\text{N}}{\text{m}^2} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

.

4

Derive base units for Magnetic Flux Density

\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}

From magnetic force

F = I L B

, solving for

B = \frac{F}{I L}

gives

\frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{A}\cdot\text{m}} = \text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}

.

Key Concept

Expressing derived physical quantities in terms of fundamental SI base units

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