Measurements and Units Practice Questions — JAMB UTME

Question 1Question →

A body of mass $50\text{ kg}$ is suspended from a spring balance attached to the ceiling of a lift. Calculate the reading registered by the spring balance, in Newtons, when the lift accelerates upward at $2.5\text{ m/s}^2$ . (Take acceleration due to gravity, $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 625

Answer

The reading of the spring balance is

625\text{ N}

.

When a lift accelerates upward, the effective acceleration experienced by an object relative to the lift is

g + a

. A spring balance measures the tension required to support and accelerate the object, which is

T = m(g + a)

. Substituting

m = 50\text{ kg}

,

g = 10\text{ m/s}^2

, and

a = 2.5\text{ m/s}^2

gives

T = 50 \times (10 + 2.5) = 625\text{ N}

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 50\text{ kg}

, acceleration due to gravity

g = 10\text{ m/s}^2

, and upward acceleration

a = 2.5\text{ m/s}^2

.

Spring balances measure weight (tension or normal force), which varies in an accelerating reference frame.

2

Apply Newton's second law to determine apparent weight in an upward accelerating lift

Apparent weight (scale reading)

T = m(g + a)

.

The upward force from the spring must overcome gravitational force

mg

and provide net upward accelerating force

ma

.

3

Compute the numerical value

T = 50 \times (10 + 2.5) = 50 \times 12.5 = 625\text{ N}

.

Multiplying mass by effective acceleration yields the correct balance reading.

Key Concept

Apparent weight measurement in an accelerating frame of reference using a spring balance

Question 2Question →

The derived SI unit of force is the newton ( $\text{N}$ ), which can be expressed in terms of fundamental SI base units as $\text{kg}\cdot\text{m}\cdot\text{s}^{x}$ . What is the numerical value of the exponent $x$ ?

Show answer & explanation

Answer: -2

Answer

The numerical value of the exponent

x

for time in the base SI expression of the newton is

-2

.

Force is defined by Newton's second law as mass times acceleration (

F = ma

). In SI fundamental units, mass is measured in kilograms (

\text{kg}

) and acceleration in meters per second squared (

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

). Combining these yields

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

. Therefore, the power of time in seconds (

x

) is

-2

.

Step-by-Step Solution

1

Identify the relationship between force and fundamental quantities

Force is defined as mass multiplied by acceleration (

F = ma

)

This relates force to fundamental quantities of mass, length, and time.

2

Substitute the SI base units of mass and acceleration

Mass unit is

\text{kg}

; acceleration unit is

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

Kilogram, meter, and second are standard fundamental SI units.

3

Determine the exponent of seconds

The combined base unit expression is

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

, giving

x = -2

Equating the powers of seconds gives

x = -2

.

Key Concept

Expressing derived units in terms of fundamental SI base units

Question 3Question →

The derived SI unit of electric potential, the volt ( $\text{V}$ ), can be expressed in terms of fundamental SI base units as $\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d$ . What is the value of the exponent $c$ ?

Show answer & explanation

Answer: -3

Answer

The exponent

c

corresponding to the base unit second (

\text{s}

) is

-3

.

Electric potential difference (

V

) is defined as work (

W

) per unit charge (

Q

). In terms of SI base units, work has units of

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

and charge has units of

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

. Dividing work by charge gives

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

. Equating this to

\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d

yields

c = -3

.

Step-by-Step Solution

1

Relate electric potential to work and charge

\text{1 V} = \frac{\text{1 J}}{\text{1 C}}

Electric potential difference is defined as work done per unit electric charge.

2

Break down Joules into base SI units

\text{1 J} = \text{1 N} \cdot \text{m} = (\text{1 kg} \cdot \text{m} \cdot \text{s}^{-2}) \cdot \text{m} = \text{1 kg} \cdot \text{m}^2 \cdot \text{s}^{-2}

Work is force multiplied by displacement, and force is mass multiplied by acceleration.

3

Break down Coulombs into base SI units

\text{1 C} = \text{1 A} \cdot \text{s}

Electric charge is electric current multiplied by time.

4

Substitute base unit expressions into the ratio for electric potential

\text{1 V} = \frac{\text{1 kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{\text{1 A} \cdot \text{s}} = \text{1 kg}^1 \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}

Combining powers of base units yields the complete fundamental representation.

5

Determine the exponent value

c

c = -3

Comparing

\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d

to

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

shows that the exponent of

\text{s}

is

-3

.

Key Concept

Expressing derived units in terms of fundamental SI base units

Question 4Question →

Which of the following physical quantities is classified as a fundamental quantity in the International System of Units (SI)?

Electric current

Force

Velocity

Density

Show answer & explanation

Answer: Electric current

Answer

Electric current

Electric current is one of the seven fundamental physical quantities defined by the International System of Units (SI). It serves as a base quantity from which other electrical quantities (such as charge and potential difference) are derived.

Step-by-Step Solution

1

Identify the definition of a fundamental quantity

Fundamental quantities are independent physical quantities that cannot be expressed in terms of other quantities.

SI defines seven basic quantities: length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.

2

Evaluate the options against the list of SI fundamental quantities

Electric current is a basic quantity (measured in amperes,

A

), whereas force, velocity, and density are derived from mass, length, and time.

Only electric current is an independent base quantity.

Key Concept

Fundamental quantities are basic physical quantities that are independent of other quantities.

Question 5Question →

A physical quantity $X$ is defined by the expression $X = \frac{P \cdot V}{m \cdot t}$ , where $P$ represents pressure, $V$ represents volume, $m$ represents mass, and $t$ represents time. Which of the following statements correctly classifies $X$ and expresses its unit strictly in terms of fundamental SI base units?

X

is a derived quantity, and its unit in SI base units is

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

.

X

is a fundamental quantity, and its unit in SI base units is

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

.

X

is a derived quantity, and its unit in fundamental units is

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

.

X

is a derived quantity, and its unit in SI base units is

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

.

Show answer & explanation

Answer: $X$ is a derived quantity, and its unit in SI base units is $\text{m}^2 \cdot \text{s}^{-3}$ .

Answer

The physical quantity

X

is a derived quantity, and its unit expressed strictly in fundamental SI base units is

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

.

The quantity

X

is defined via a mathematical formula involving pressure, volume, mass, and time, which classifies it as a derived quantity. Replacing each component with its fundamental SI base units gives pressure as

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

and volume as

\text{m}^3

, making the numerator

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

. Dividing by the denominator (

m \cdot t = \text{kg} \cdot \text{s}

) cancels out kilograms and leaves

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

, consisting purely of fundamental base units.

Step-by-Step Solution

1

Classify the physical quantity

X

X

is a derived physical quantity.

Fundamental physical quantities in the SI system are length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity. Because

X

is calculated from a combination of other quantities, it is derived.

2

Express pressure (

P

) and volume (

V

) in terms of fundamental SI base units

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

and

V = \text{m}^3

.

Pressure is defined as force per unit area

\left(\frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{\text{m}^2}\right)

, and volume has the base unit

\text{m}^3

.

3

Calculate the base unit expression for the numerator

P \cdot V

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

.

Combining the powers of length (metres) gives

-1 + 3 = 2

.

4

Divide by the denominator

m \cdot t

to obtain the base units of

X

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

.

The unit of mass (

\text{kg}

) cancels completely, and dividing by time (

\text{s}

) reduces the exponent of seconds from

-2

to

-3

.

Key Concept

Fundamental quantities are independent base quantities defined by the SI system, whereas derived quantities are defined algebraically from fundamental quantities. Reducing derived units to SI base units requires breaking down all non-base units into metres (m), kilograms (kg), seconds (s), amperes (A), kelvins (K), moles (mol), or candelas (cd).

Estimated Time:2m 0s

Question 6Question →

Match each derived SI unit listed on the left with its corresponding fundamental (base) SI unit representation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Joule (

\text{J}

) matches

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

; Pascal (

\text{Pa}

) matches

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

; Watt (

\text{W}

) matches

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

; Newton (

\text{N}

) matches

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

.

Each derived unit is systematically expressed in terms of the fundamental SI units of mass (kg), length (m), and time (s) by substituting definitions: Newton is

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

, Joule is

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

, Pascal is

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

, and Watt is

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

.

Step-by-Step Solution

1

Express Newton (N) in fundamental units

\text{Force} = \text{mass} \times \text{acceleration} \Rightarrow 1\text{ N} = 1\text{ kg}\cdot\text{m}\cdot\text{s}^{-2}

Force is defined as mass multiplied by acceleration.

2

Express Joule (J) in fundamental units

\text{Work} = \text{Force} \times \text{distance} \Rightarrow 1\text{ J} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2}

Work done is force multiplied by displacement in the direction of the force.

3

Express Pascal (Pa) in fundamental units

\text{Pressure} = \frac{\text{Force}}{\text{Area}} \Rightarrow 1\text{ Pa} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = 1\text{ kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

Pressure is defined as force applied perpendicular to a surface per unit area.

4

Express Watt (W) in fundamental units

\text{Power} = \frac{\text{Work}}{\text{time}} \Rightarrow 1\text{ W} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{s}} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-3}

Power is the rate at which work is done or energy is transferred.

Key Concept

Expressing derived SI units in terms of base SI fundamental units

Question 7Question →

Which of the following groups consists exclusively of fundamental physical quantities?

Mass, thermodynamic temperature, and luminous intensity

Mass, length, and electric charge

Electric current, force, and time

Amount of substance, weight, and temperature

Show answer & explanation

Answer: Mass, thermodynamic temperature, and luminous intensity

Answer

The group containing mass, thermodynamic temperature, and luminous intensity consists exclusively of fundamental physical quantities.

Mass, thermodynamic temperature, and luminous intensity are three of the seven internationally recognized SI base (fundamental) physical quantities.

Step-by-Step Solution

1

Identify the seven fundamental SI physical quantities

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

Fundamental quantities are independent quantities that cannot be defined in terms of other physical quantities.

2

Evaluate each provided option against the list of base quantities

Electric charge, force, and weight are derived quantities. Only mass, thermodynamic temperature, and luminous intensity are all fundamental.

Any quantity derived by multiplying or dividing base quantities is derived.

Key Concept

Fundamental quantities are basic physical quantities that are independent of one another and form the basis from which derived quantities are obtained.

Question 8Question →

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 9Question →

A physical quantity $Z$ is defined as the ratio of the product of impulse and linear velocity to the product of electric current and electric potential difference. When $Z$ is fully resolved into fundamental physical quantities, which fundamental quantity does $Z$ represent?

Time

Mass

Length

Electric current

Show answer & explanation

Answer: Time

Answer

Time

The correct answer is Time because reducing both the numerator (energy) and denominator (electrical power) to base fundamental dimensions yields

\frac{M L^2 T^{-2}}{M L^2 T^{-3}} = T

, which corresponds directly to the fundamental quantity of Time.

Step-by-Step Solution

1

Express impulse and velocity in terms of fundamental base quantities (Mass

M

, Length

L

, Time

T

)

\text{Impulse} = \text{Force} \times \text{Time} = M L T^{-1}

.

\text{Velocity} = L T^{-1}

. Product

= (M L T^{-1})(L T^{-1}) = M L^2 T^{-2}

.

To evaluate the numerator in fundamental base dimensions.

2

Express electric current and potential difference in terms of fundamental base quantities

\text{Current} = I

.

\text{Potential Difference} = \frac{\text{Work}}{\text{Charge}} = \frac{M L^2 T^{-2}}{I T} = M L^2 T^{-3} I^{-1}

. Product

= I \times (M L^2 T^{-3} I^{-1}) = M L^2 T^{-3}

.

To evaluate the denominator in fundamental base dimensions.

3

Divide the numerator by the denominator to simplify quantity

Z

Z = \frac{M L^2 T^{-2}}{M L^2 T^{-3}} = M^{1-1} L^{2-2} T^{-2 - (-3)} = T^1

.

Determining the net fundamental quantity after all derived units cancel out.

Key Concept

Fundamental and Derived Quantities

Question 10Question →

Match each physical quantity listed on the left with its correct SI classification or fundamental unit decomposition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electric current corresponds to the fundamental quantity measured in amperes (

\text{A}

); Pressure corresponds to the derived quantity with base SI units

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

; Thermodynamic temperature corresponds to the fundamental quantity measured in kelvin (

\text{K}

); Impulse corresponds to the derived quantity with base SI units

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

.

Electric current and thermodynamic temperature are fundamental SI quantities defined independently with units ampere (

\text{A}

) and kelvin (

\text{K}

). Pressure and impulse are derived quantities whose definitions rely on fundamental quantities, breaking down into

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

and

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

respectively.

Step-by-Step Solution

1

Identify fundamental physical quantities

Electric current and thermodynamic temperature are fundamental quantities with SI base units ampere (

\text{A}

) and kelvin (

\text{K}

) respectively.

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

2

Decompose pressure into base SI units

\text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

.

Derived quantities must be reduced to combinations of mass (

\text{kg}

), length (

\text{m}

), and time (

\text{s}

).

3

Decompose impulse into base SI units

\text{Impulse} = \text{Force} \times \text{Time} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{s} = \text{kg}\cdot\text{m}\cdot\text{s}^{-1}

.

Impulse is defined as change in momentum or force applied over a time interval.

Key Concept

Classification of fundamental quantities versus derived quantities and resolution of derived units into fundamental SI units.

Question 11Question →

Which of the following correctly expresses the derived SI unit of power, the watt ( $\text{W}$ ), in terms of fundamental SI base units?

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

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

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

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

Show answer & explanation

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

Answer

The watt (

\text{W}

) expressed in fundamental SI base units is

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

.

Power is defined as energy transferred or work done per unit time (

\text{P} = \frac{W}{t}

). Work is force times distance (

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

). Dividing work by time (

\text{s}

) yields

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

.

Step-by-Step Solution

1

Recall the defining formula for power

\text{Power} = \frac{\text{Work}}{\text{Time}} = \frac{\text{Force} \times \text{Displacement}}{\text{Time}}

Relating the derived quantity to simpler mechanical quantities.

2

Break down force into fundamental base units using Newton's second law (

F = ma

)

\text{Unit of Force (Newton)} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

Mass is measured in kilograms (

\text{kg}

), acceleration in meters per second squared (

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

).

3

Express work in fundamental base units

\text{Unit of Work (Joule)} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

Work is force multiplied by distance.

4

Divide the unit of work by the unit of time (seconds)

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

Dividing by

\text{s}

decreases the exponent of seconds by 1.

Key Concept

Expressing derived SI units in terms of fundamental SI base units

Estimated Time:1m 0s

Question 12Question →

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 13Question →

A student measures the length of a metal rod as $4.0\text{ cm}$ . If the actual length of the rod is $5.0\text{ cm}$ , what is the percentage error in the measurement?

Show answer & explanation

Answer: 20

Answer

The percentage error in the measurement is 20%.

The absolute error is the difference between the true length (

5.0\text{ cm}

) and the measured length (

4.0\text{ cm}

), which is

1.0\text{ cm}

. Dividing

1.0\text{ cm}

by the true length

5.0\text{ cm}

gives a fractional error of

0.20

. Expressed as a percentage,

0.20 \times 100\% = 20\%

.

Step-by-Step Solution

1

Find the error (difference between measured value and actual value)

Error =

|5.0\text{ cm} - 4.0\text{ cm}| = 1.0\text{ cm}

Absolute error is defined as the magnitude of the difference between the actual value and the measured value.

2

Calculate the percentage error

Percentage error =

\frac{1.0\text{ cm}}{5.0\text{ cm}} \times 100\% = 20\%

Percentage error is the ratio of absolute error to the actual value, expressed as a percentage.

Key Concept

Percentage Error Calculation

Question 14Question →

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 15Question →

In the dimensional equation for the period of oscillation of a simple pendulum, $T = k g^a l^b$ , where $T$ is the period, $g$ is the acceleration due to gravity, $l$ is the length of the pendulum, and $k$ 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 analysis to

T = k g^a l^b

, the dimension of the left-hand side is

\text{T}^1

. The right-hand side has dimensions

(\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}

. Equating the exponents of time

\text{T}

yields

1 = -2a

, which gives

a = -0.5

.

Step-by-Step Solution

1

Express the dimensions of all physical quantities involved in fundamental base dimensions (M, L, T).

The dimension of period

T

is

[\text{T}]

, length

l

is

[\text{L}]

, and gravitational acceleration

g

is

[\text{L T}^{-2}]

.

Dimensional analysis requires substituting each quantity with its fundamental dimensions.

2

Formulate the dimensional homogeneity equation.

\text{M}^0 \text{L}^0 \text{T}^1 = (\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}

.

The principle of dimensional homogeneity states that the exponents of base dimensions on both sides of a physically correct equation must be equal.

3

Equate exponents of time

\text{T}

and solve for

a

.

1 = -2a \implies a = -\frac{1}{2} = -0.5

.

Comparing the powers of

\text{T}

gives a linear equation in

a

.

Key Concept

Principle of Dimensional Homogeneity

Estimated Time:1m 15s

Question 16Question →

The gravitational force $F$ between two point masses $m_1$ and $m_2$ separated by a distance $r$ is expressed by the equation $F = \frac{G m_1 m_2}{r^2}$ , where $G$ represents the universal gravitational constant. What is the dimensional formula for $G$ ?

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

M L^3 T^{-2}

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

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

Show answer & explanation

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

Answer

The dimensional formula for the universal gravitational constant

G

is

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

.

Isolating

G

gives

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

. Substituting the fundamental dimensions for force (

M L T^{-2}

), distance (

L

), and mass (

M

) yields

\frac{(M L T^{-2})(L^2)}{M^2} = M^{-1} L^3 T^{-2}

.

Step-by-Step Solution

1

Make

G

the subject of the formula in Newton's law of gravitation.

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

To derive the dimensions of

G

, isolate it in terms of force, distance, and mass.

2

Substitute fundamental dimensions for force, distance, and mass.

[G] = \frac{[F][r]^2}{[m_1][m_2]} = \frac{(M L T^{-2})(L^2)}{M \cdot M}

Force has dimensions

M L T^{-2}

, distance has dimension

L

, and mass has dimension

M

.

3

Simplify the powers of fundamental dimensions

M

,

L

, and

T

.

[G] = M^{1-2} L^{1+2} T^{-2} = M^{-1} L^3 T^{-2}

Applying exponent rules simplifies the combined base dimensions.

Key Concept

Dimensions of Physical Constants

Estimated Time:1m 15s

Question 17Question →

In an experiment to determine the density of a solid sphere, the mass is measured as $(50.0 \pm 0.5)\text{ g}$ and the radius is measured as $(2.00 \pm 0.05)\text{ cm}$ . What is the percentage error in the calculated density of the sphere?

Show answer & explanation

Answer: 8.5

Answer

The percentage error in the calculated density of the sphere is

8.5\%

.

The density formula

\rho = \frac{3m}{4\pi r^3}

dictates that maximum relative error is given by

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\left(\frac{\Delta r}{r}\right)

. Evaluating the percentage errors gives

1.0\%

for mass and

2.5\%

for radius. Summing

1.0\% + 3(2.5\%)

yields

8.5\%

.

Step-by-Step Solution

1

Determine the functional dependence of density on measured quantities.

Density

\rho = \frac{m}{V} = \frac{3m}{4\pi r^3}

.

The volume of a sphere of radius

r

is

V = \frac{4}{3}\pi r^3

.

2

Formulate the maximum fractional error relationship.

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\left(\frac{\Delta r}{r}\right).

When combining uncertainties, fractional errors add, and exponents act as multiplying factors.

3

Calculate the percentage error in the mass measurement.

\frac{0.5}{50.0} \times 100\% = 1.0\%.

Percentage error in mass is the absolute uncertainty divided by the measured value multiplied by

100\%

.

4

Calculate the percentage error in the radius measurement.

\frac{0.05}{2.00} \times 100\% = 2.5\%.

Percentage error in radius is the absolute uncertainty divided by the measured value multiplied by

100\%

.

5

Calculate total percentage error in density.

Percentage error = 1.0\% + 3(2.5\%) = 8.5\%.

The radius contributes three times its relative error because volume depends on

r^3

.

Key Concept

Error propagation in derived quantities involving powers

Question 18Question →

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 19Question →

The electric current passing through a resistor is measured as $(5.0 \pm 0.1)\text{ A}$ , and the resistance is measured as $(10.0 \pm 0.3)\text{ }\Omega$ . What is the maximum percentage error in the calculated electrical power dissipated by the resistor?

7.0\%

5.0\%

4.0\%

12.0\%

Show answer & explanation

Answer: $7.0\%$

Answer

The maximum percentage error in the calculated power is

7.0\%

.

Electrical power is calculated using the formula

P = I^2 R

. In error analysis, the maximum fractional error for a quantity

y = a^n b^m

is

\frac{\Delta y}{y} = n\frac{\Delta a}{a} + m\frac{\Delta b}{b}

. Therefore, the maximum percentage error in

P

is equal to

2 \times (\text{percentage error in } I) + (\text{percentage error in } R) = 2(2.0\%) + 3.0\% = 7.0\%

.

Step-by-Step Solution

1

Calculate the percentage error in the measured current

I

.

\frac{\Delta I}{I} \times 100\% = \frac{0.1}{5.0} \times 100\% = 2.0\%

Percentage error is the ratio of absolute uncertainty to the measured value expressed as a percentage.

2

Calculate the percentage error in the measured resistance

R

.

\frac{\Delta R}{R} \times 100\% = \frac{0.3}{10.0} \times 100\% = 3.0\%

Applying the same percentage error formula to the resistance measurement.

3

Apply the error propagation formula for power

P = I^2 R

.

\frac{\Delta P}{P} \times 100\% = 2\left(\frac{\Delta I}{I} \times 100\%\right) + \left(\frac{\Delta R}{R} \times 100\%\right) = 2(2.0\%) + 3.0\% = 7.0\%

When physical quantities are raised to a power and multiplied, their fractional/percentage errors are multiplied by the exponent and summed to get the maximum error.

Key Concept

Propagation of errors in derived quantities involving powers and products

Alternative Method

Calculate absolute values of minimum and maximum possible power, then compute percentage variation relative to nominal power.

Estimated Time:1m 30s

Question 20Question →

The viscous force $F$ acting on a small sphere of radius $r$ moving with velocity $v$ through a liquid is given by Stokes' law, $F = 6\pi \eta r v$ , where $\eta$ is the coefficient of viscosity. Which of the following expressions represents the base dimensions of $\eta$ ?

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

M L T^{-1}

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

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

Show answer & explanation

Answer: $M L^{-1} T^{-1}$

Answer

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

Rearranging Stokes' law gives

\eta = \frac{F}{6\pi r v}

. Substituting base dimensions

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

,

[r] = L

, and

[v] = L T^{-1}

yields

[\eta] = \frac{M L T^{-2}}{L^2 T^{-1}} = M L^{-1} T^{-1}

.

Step-by-Step Solution

1

Express Stokes' law in terms of the coefficient of viscosity

\eta = \frac{F}{6\pi r v}

Isolating

\eta

allows substitution of fundamental dimensions.

2

Substitute fundamental dimensions for force, radius, and velocity

[\eta] = \frac{M L T^{-2}}{L \cdot L T^{-1}}

The constant

6\pi

is dimensionless, while

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

,

[r] = L

, and

[v] = L T^{-1}

.

3

Simplify the powers of base quantities

M

,

L

, and

T

[\eta] = M L^{1 - 2} T^{-2 - (-1)} = M L^{-1} T^{-1}

Applying algebraic rules of exponents simplifies the expression.

Key Concept

Dimensions of Physical Quantities and Dimensional Analysis

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