Measurements and Units Practice Questions — JAMB UTME — Page 4

Question 61Question →

A body is suspended from a spring balance inside an elevator that is accelerating upwards at $2.0\text{ m s}^{-2}$ on Earth, giving a scale reading of $60\text{ N}$ . The body is then taken to a planet where the acceleration due to gravity is $6.0\text{ m s}^{-2}$ and placed on an ideal beam balance against standard masses calibrated on Earth. Taking $g = 10.0\text{ m s}^{-2}$ on Earth, what is the reading obtained on the beam balance?

5.0\text{ kg}

6.0\text{ kg}

7.5\text{ kg}

3.0\text{ kg}

Show answer & explanation

Answer: $5.0\text{ kg}$

Answer

The reading obtained on the beam balance is

5.0\text{ kg}

.

In an upward accelerating elevator, the spring balance registers an apparent weight of

W_{app} = m(g + a)

. Substituting the given values gives

60\text{ N} = m(10\text{ m s}^{-2} + 2\text{ m s}^{-2})

, which yields a true mass

m = 5.0\text{ kg}

. When measured on another planet using an ideal beam balance, the gravitational force on the unknown mass balances against standard masses (

m \cdot g_p = m_{standard} \cdot g_p

). Since local gravity

g_p

cancels from both sides, the beam balance measures the invariant mass of

5.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the true mass of the body using the scale reading in the accelerating elevator.

m = 5.0\text{ kg}

Inside an upward accelerating elevator, the apparent weight registered by the spring balance is

W_{app} = m(g + a)

. Substituting

W_{app} = 60\text{ N}

,

g = 10.0\text{ m s}^{-2}

, and

a = 2.0\text{ m s}^{-2}

yields

60 = m(10 + 2)

, which gives

m = 5.0\text{ kg}

.

2

Determine the mass reading on a beam balance on the new planet.

The beam balance reads

5.0\text{ kg}

.

A beam balance compares the gravitational force on the object with standard masses:

m \cdot g_p = m_{standard} \cdot g_p \implies m_{standard} = m

. Because local acceleration due to gravity

g_p

cancels out on both sides, a beam balance measures true invariant mass regardless of location or gravity.

Key Concept

Mass vs Weight Measurement: Accelerating Frames and Beam Balance Invariance

Estimated Time:1m 0s

Question 62Question →

A student investigating the physical properties of a uniform metallic rod records its mass, length, thermodynamic temperature, and mass density. Which of the recorded physical quantities is classified as a derived quantity?

Mass density

Mass

Length

Thermodynamic temperature

Show answer & explanation

Answer: Mass density

Answer

Mass density is the derived physical quantity.

Mass density is defined as mass per unit volume (

\rho = \frac{m}{V}

). Because it is obtained by combining the fundamental quantities of mass and length, it is a derived physical quantity.

Step-by-Step Solution

1

Identify the seven fundamental SI physical quantities

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

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

2

Classify each physical quantity given in the scenario

Mass, length, and thermodynamic temperature are fundamental quantities. Mass density is defined as mass divided by volume (

\rho = \frac{m}{V}

).

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

Key Concept

Classification of Fundamental and Derived Quantities

Question 63Question →

A simple pendulum completes $20$ full oscillations in a time interval of $40\text{ s}$ . What is the period of oscillation of the pendulum in seconds?

Show answer & explanation

Answer: 2

Answer

The period of oscillation of the pendulum is

2.0\text{ s}

.

The period

T

of a repeating motion is the time taken to complete one full cycle. Dividing the total measured time (

40\text{ s}

) by the number of oscillations (

20

) gives

2.0\text{ s}

.

Step-by-Step Solution

1

Identify the given total time and total number of oscillations

Total time

t = 40\text{ s}

, number of oscillations

N = 20

The period is defined as the time taken for one single complete oscillation.

2

Apply the period formula

T = \frac{t}{N}

T = \frac{40\text{ s}}{20} = 2.0\text{ s}

Dividing the total measured time by the total count of oscillations yields the time per oscillation.

Key Concept

Period of a Simple Pendulum

Question 64Question →

Match each physical quantity listed on the left with its corresponding classification and physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electric potential difference matches derived quantity defined as work done per unit electric charge; Thermodynamic temperature matches fundamental quantity representing thermal state, measured in kelvins; Impulse matches derived quantity defined as force times time interval; Luminous intensity matches fundamental quantity measuring perceived light power per unit solid angle.

Thermodynamic temperature and luminous intensity are two of the seven fundamental SI quantities. Electric potential difference and impulse are derived quantities because they are expressed through equations involving base physical quantities.

Step-by-Step Solution

1

Identify fundamental physical quantities

Thermodynamic temperature and luminous intensity are basic independent physical quantities defined by SI standards.

Fundamental quantities cannot be defined in terms of other physical quantities.

2

Identify derived physical quantities and their defining expressions

Electric potential difference (

V = \frac{W}{Q}

) and impulse (

I = F \Delta t

) are derived from basic quantities.

Derived quantities are defined by mathematical combinations of fundamental quantities.

Key Concept

Fundamental and Derived Quantities

Question 65Question →

A voltmeter records a potential difference of $12.6\text{ V}$ across a resistor. If the percentage error in this measurement is $5.0\%$ , what is the absolute error in the reading?

0.63\text{ V}

0.05\text{ V}

2.52\text{ V}

6.30\text{ V}

Show answer & explanation

Answer: $0.63\text{ V}$

Answer

The absolute error in the reading is

0.63\text{ V}

.

The correct response is obtained by taking

5.0\%

of the measured potential difference

12.6\text{ V}

. Evaluating

\frac{5.0}{100} \times 12.6\text{ V}

yields an absolute error of

0.63\text{ V}

.

Step-by-Step Solution

1

Identify the given parameters

Measured potential difference

V = 12.6\text{ V}

, Percentage error

= 5.0\%

.

These are the measured values provided in the problem statement.

2

Recall the percentage error formula

\text{Percentage Error} = \left(\frac{\text{Absolute Error}}{\text{Measured Reading}}\right) \times 100\%

.

This formula defines the relationship between absolute error, measured quantity, and percentage error.

3

Calculate the absolute error

\text{Absolute Error} = \frac{5.0 \times 12.6\text{ V}}{100} = 0.63\text{ V}

.

Multiplying the percentage error fraction by the measured potential difference yields the absolute uncertainty in volts.

Key Concept

Absolute Error and Percentage Error in Measurements

Question 66Question →

The rate of heat transfer through a uniform metallic rod of cross-sectional area $A$ and length $d$ is given by $\frac{Q}{t} = \frac{k A \Delta T}{d}$ , where $Q$ is heat energy, $t$ is time, $\Delta T$ is temperature difference, and $k$ is thermal conductivity. What is the dimensional formula for $k$ in terms of mass ( $M$ ), length ( $L$ ), time ( $T$ ), and temperature ( $\Theta$ )?

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

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

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

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

Show answer & explanation

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

Answer

The dimensional formula for thermal conductivity is

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

.

Expressing thermal conductivity explicitly gives

k = \frac{Q d}{A \Delta T t}

. Substituting fundamental dimensions yields

[k] = \frac{(M L^2 T^{-2})(L)}{(L^2)(\Theta)(T)} = M L T^{-3} \Theta^{-1}

. Thus, the dimensional formula

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

is correct.

Step-by-Step Solution

1

Make thermal conductivity

k

the subject of the formula.

k = \frac{Q d}{A \Delta T t}

Isolating

k

enables direct substitution of base dimensional quantities.

2

Substitute fundamental dimensions for each physical quantity.

[k] = \frac{[M L^2 T^{-2}][L]}{[L^2][\Theta][T]}

Heat energy

Q

has dimensions

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

, distance

d

is

[L]

, area

A

is

[L^2]

, temperature change

\Delta T

is

[\Theta]

, and time

t

is

[T]

.

3

Combine exponents for each fundamental dimension.

[k] = M^{1} L^{2+1-2} T^{-2-1} \Theta^{-1} = M L T^{-3} \Theta^{-1}

Applying exponent laws simplifies the expression to base units.

Key Concept

Dimensional Analysis of Physical Constants

Question 67Question →

During a laboratory experiment, a student measures the thickness of a glass slide as $2.5\text{ mm}$ . If the actual thickness of the slide is $2.0\text{ mm}$ , what is the percentage error in the measurement?

20.0\%

25.0\%

0.25\%

5.0\%

Show answer & explanation

Answer: $25.0\%$

Answer

The percentage error in the measurement is

25.0\%

.

Percentage error is determined by dividing the absolute error (

|2.5\text{ mm} - 2.0\text{ mm}| = 0.5\text{ mm}

) by the actual value (

2.0\text{ mm}

) and multiplying by

100\%

. This gives

\frac{0.5}{2.0} \times 100\% = 25.0\%

.

Step-by-Step Solution

1

Calculate the absolute error in the measurement

\text{Absolute error} = |\text{Measured value} - \text{Actual value}| = |2.5\text{ mm} - 2.0\text{ mm}| = 0.5\text{ mm}

Absolute error measures the magnitude of discrepancy between measured and true values.

2

Calculate the percentage error relative to the actual value

\text{Percentage error} = \frac{\text{Absolute error}}{\text{Actual value}} \times 100\% = \frac{0.5\text{ mm}}{2.0\text{ mm}} \times 100\% = 25.0\%

Percentage error is defined as the absolute error divided by the accepted actual value, expressed as a percentage.

Key Concept

Percentage Error Calculation

Estimated Time:45s

Question 68Question →

In a laboratory experiment, the radius of a circular wire is measured as $(2.00 \pm 0.05)\text{ mm}$ . What is the percentage error in the calculated cross-sectional area of the wire?

2.5\%

5.0\%

6.25\%

1.25\%

Show answer & explanation

Answer: $5.0\%$

Answer

The percentage error in the calculated area is

5.0\%

.

The cross-sectional area of a wire is given by

A = \pi r^2

. The percentage error in a calculated quantity

y = x^n

is given by

n \times (\text{percentage error in } x)

. Here, the percentage error in the radius

r

is

(0.05 / 2.00) \times 100\% = 2.5\%

. Multiplying by the power

n = 2

gives

2 \times 2.5\% = 5.0\%

.

Step-by-Step Solution

1

Calculate the percentage error in the measured radius

r

.

\text{Percentage error in } r = \left(\frac{0.05}{2.00}\right) \times 100\% = 2.5\%

Relative error is the absolute error divided by the measured value.

2

Apply the rule of error propagation for quantities raised to a power (

A = \pi r^2

).

\text{Percentage error in } A = 2 \times (\text{Percentage error in } r) = 2 \times 2.5\% = 5.0\%

When a physical quantity is raised to the power

n

, its fractional or percentage error is multiplied by

n

.

Key Concept

Error Propagation in Derived Quantities

Estimated Time:1m 30s

Question 69Question →

An unknown object is evaluated inside a space probe descending towards the surface of a distant planet. A beam balance calibrated on Earth registers a reading of $15.0\text{ kg}$ for the object. The probe has a downward acceleration of $2.0\text{ m s}^{-2}$ in a region where the local gravitational acceleration of the planet is $6.0\text{ m s}^{-2}$ . What reading will a spring balance display when the same object is suspended from it inside the probe?

90.0\text{ N}

120.0\text{ N}

60.0\text{ N}

150.0\text{ N}

Show answer & explanation

Answer: $60.0\text{ N}$

Answer

60.0\text{ N}

A beam balance compares gravitational forces on equal balance arms, meaning local gravity and frame acceleration affect both sides equally. Therefore, the beam balance measures the true invariant mass of

15.0\text{ kg}

. A spring balance measures the tension or apparent weight

W_{\text{app}} = m(g - a)

. Substituting

m = 15.0\text{ kg}

, local gravity

g = 6.0\text{ m s}^{-2}

, and downward acceleration

a = 2.0\text{ m s}^{-2}

gives

W_{\text{app}} = 15.0 \times (6.0 - 2.0) = 60.0\text{ N}

.

Step-by-Step Solution

1

Determine the true mass of the object using the beam balance measurement

True mass

m = 15.0\text{ kg}

A beam balance compares unknown mass with standard masses under the same local gravitational field. Because local gravity acts equally on both pans, beam balance readings yield the true invariant mass regardless of local gravitational acceleration or uniform reference frame acceleration (provided net effective gravity is greater than zero).

2

Calculate the effective acceleration experienced inside the accelerating reference frame

g_{\text{eff}} = g_{\text{planet}} - a = 6.0\text{ m s}^{-2} - 2.0\text{ m s}^{-2} = 4.0\text{ m s}^{-2}

When a reference frame accelerates downward at rate

a

, the effective apparent gravitational acceleration felt by objects suspended inside is reduced by

a

.

3

Calculate the apparent weight registered by the spring balance

W_{\text{app}} = m \cdot g_{\text{eff}} = 15.0\text{ kg} \times 4.0\text{ m s}^{-2} = 60.0\text{ N}

A spring balance measures tension force (apparent weight) exerted on its spring, which equals

m(g - a)

in a downward accelerating system.

Key Concept

Mass measurement via beam balance vs. apparent weight measurement via spring balance in accelerating frames

Question 70Question →

A spring balance and an equal-arm beam balance are used to measure an object at a location where the local acceleration due to gravity is $8.0\text{ m s}^{-2}$ . If the spring balance registers a weight of $40\text{ N}$ , what mass will the equal-arm beam balance register at this location?

5.0\text{ kg}

4.0\text{ kg}

320\text{ kg}

50\text{ kg}

Show answer & explanation

Answer: $5.0\text{ kg}$

Answer

The equal-arm beam balance will register a mass of

5.0\text{ kg}

.

Weight is given by

W = mg

. Given

W = 40\text{ N}

and

g = 8.0\text{ m s}^{-2}

, the mass of the object is

m = \frac{40}{8.0} = 5.0\text{ kg}

. An equal-arm beam balance balances the unknown mass against standard masses under the exact same local gravitational field, so local gravity cancels out and it accurately measures the object's mass as

5.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the true mass of the object from the spring balance reading

m = \frac{W}{g} = \frac{40\text{ N}}{8.0\text{ m s}^{-2}} = 5.0\text{ kg}

Weight is the gravitational force acting on a mass (

W = mg

), so mass equals weight divided by local acceleration due to gravity.

2

Determine the reading on the equal-arm beam balance

Mass registered =

5.0\text{ kg}

An equal-arm beam balance compares the gravitational force on the unknown mass against standard masses. Because local gravity affects both sides equally, it measures true mass independent of local gravitational acceleration.

Key Concept

Measurement of Mass and Weight

Estimated Time:1m 0s

Question 71Question →

The length and width of a rectangular metal sheet are measured as $(8.0 \pm 0.2)\text{ cm}$ and $(5.0 \pm 0.1)\text{ cm}$ , respectively. What is the percentage error in the calculated area of the metal sheet?

Show answer & explanation

Answer: 4.5

Answer

The percentage error in the calculated area of the metal sheet is

4.5\%

.

For a calculated quantity involving multiplication (

A = L \times W

), the total relative error is the sum of the relative errors of the individual measurements. The fractional error in length is

\frac{0.2}{8.0} = 0.025

(

2.5\%

) and in width is

\frac{0.1}{5.0} = 0.020

(

2.0\%

). Summing these gives

0.045

, which is equivalent to

4.5\%

.

Step-by-Step Solution

1

Determine the fractional error in the length measurement

\frac{0.2\text{ cm}}{8.0\text{ cm}} = 0.025

Fractional error is given by the ratio of absolute error to the measured value.

2

Determine the fractional error in the width measurement

\frac{0.1\text{ cm}}{5.0\text{ cm}} = 0.020

Fractional error is calculated as the absolute uncertainty divided by the measured dimension.

3

Combine the fractional errors for the calculated area

\frac{\Delta A}{A} = 0.025 + 0.020 = 0.045

For quantities multiplied together (

A = L \times W

), individual fractional errors sum to give the total fractional error.

4

Convert the fractional error to percentage error

0.045 \times 100\% = 4.5\%

Multiplying the relative error by 100 yields the percentage error.

Key Concept

Error Propagation in Products

Question 72Question →

A ticker-tape timer connected to an alternating current supply operates at a frequency of $50\text{ Hz}$ . If a tape pulled through the timer displays $6$ consecutive dots, what is the total time interval represented by this section of tape?

0.10\text{ s}

0.12\text{ s}

0.02\text{ s}

0.25\text{ s}

Show answer & explanation

Answer: $0.10\text{ s}$

Answer

The total time interval represented by the section of tape is

0.10\text{ s}

.

The frequency of

50\text{ Hz}

means each tick interval takes

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

. For

6

consecutive dots, there are

5

intervals between them. The total time elapsed is

5 \times 0.02\text{ s} = 0.10\text{ s}

.

Step-by-Step Solution

1

Calculate the period

T

of a single tick interval

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

The period of oscillation of the timer represents the time elapsed between two successive dots.

2

Determine the number of time intervals (spaces) between 6 consecutive dots

Number of intervals =

6 - 1 = 5\text{ intervals}

Time measurement on a ticker tape is based on the number of spaces between dots, not the count of dots themselves.

3

Calculate the total time interval

t

t = 5 \times 0.02\text{ s} = 0.10\text{ s}

Multiply the number of spaces by the time period per space.

Key Concept

Measurement of time using a ticker-tape timer

Question 73Question →

A stopwatch with a zero error of $-0.25\text{ s}$ is used to record the time taken for an object to travel down an inclined path. If the stopwatch displays a time reading of $14.65\text{ s}$ at the end of the motion, what is the true time elapsed in seconds?

Show answer & explanation

Answer: 14.9

Answer

The true time elapsed is 14.90 seconds.

The correct answer is obtained by subtracting the zero error from the observed reading. Since the stopwatch has a negative zero error of

-0.25\text{ s}

, the true time is

14.65\text{ s} - (-0.25\text{ s}) = 14.90\text{ s}

.

Step-by-Step Solution

1

Apply the zero error correction formula for measurement instruments

Actual Time = Displayed Time - Zero Error

Zero error represents a constant bias on the instrument that must be subtracted from the uncorrected reading.

2

Substitute the given values into the formula

Actual Time = 14.65 s - (-0.25 s)

The instrument starts behind zero by 0.25 s, so the zero error value is negative.

3

Perform the calculation

Actual Time = 14.90 s

Subtracting a negative quantity is mathematically equivalent to adding its positive magnitude.

Key Concept

Zero Error Correction in Stopwatch Time Measurement

Question 74Question →

An object of unknown mass is suspended from a spring balance possessing a zero error of $+1.5\text{ N}$ inside a lift on an unexplored planet. When the lift accelerates vertically upwards at $2.0\text{ m s}^{-2}$ , the spring balance displays a reading of $33.5\text{ N}$ . Simultaneously, an equal-arm beam balance calibrated with standard masses measures the mass of the object inside the accelerating lift to be $4.0\text{ kg}$ . What is the local acceleration due to gravity on this planet and the true weight of the object when at rest on its surface?

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

and

24.0\text{ N}

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

and

25.5\text{ N}

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

and

40.0\text{ N}

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

and

32.0\text{ N}

Show answer & explanation

Answer: $6.0\text{ m s}^{-2}$ and $24.0\text{ N}$

Answer

The local acceleration due to gravity on the planet is

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

and the true weight of the object at rest is

24.0\text{ N}

.

The correct response identifies that an equal-arm beam balance measures invariant mass (

4.0\text{ kg}

) because acceleration affects both balance pans equally. Subtracting the

+1.5\text{ N}

zero error from the scale reading yields a true apparent weight of

32.0\text{ N}

. Applying Newton's second law in an upward accelerating lift gives

W_{\text{app}} = m(g + a)

, which yields

32.0 = 4.0(g + 2.0)

, resulting in

g = 6.0\text{ m s}^{-2}

. The true weight at rest is therefore

W = mg = 4.0 \times 6.0 = 24.0\text{ N}

.

Step-by-Step Solution

1

Determine the true mass of the object from the beam balance measurement.

m = 4.0\text{ kg}

An equal-arm beam balance operates by comparing gravitational moments on standard masses and the test object. Because the effective acceleration

(g + a)

acts equally on both pans, it cancels out, making the beam balance measure the true, invariant mass regardless of frame acceleration or location.

2

Correct the spring balance scale reading for zero error to find the true apparent weight.

W_{\text{app}} = 33.5\text{ N} - 1.5\text{ N} = 32.0\text{ N}

A positive zero error means the balance reads

+1.5\text{ N}

when unloaded, so the true force exerted on the spring is the scale reading minus the zero error.

3

Relate apparent weight to local gravity

g

in an upward accelerating lift.

g = 6.0\text{ m s}^{-2}

In an upward accelerating frame with acceleration

a = 2.0\text{ m s}^{-2}

, the normal force/apparent weight is

W_{\text{app}} = m(g + a)

. Substituting values gives

32.0 = 4.0(g + 2.0) \implies 8.0 = g + 2.0 \implies g = 6.0\text{ m s}^{-2}

.

4

Calculate the true weight of the object when at rest on the planet.

W_{\text{true}} = 24.0\text{ N}

True weight is the force of gravity acting on the mass at rest:

W_{\text{true}} = m \cdot g = 4.0\text{ kg} \times 6.0\text{ m s}^{-2} = 24.0\text{ N}

.

Key Concept

Distinction between mass (measured by beam balance, frame-invariant) and weight (measured by spring balance, dependent on frame acceleration and zero error).

Estimated Time:2m 0s

Question 75Question →

The mass $m$ of an object is measured as $(2.0 \pm 0.1)\text{ kg}$ and its speed $v$ is measured as $(5.0 \pm 0.2)\text{ m s}^{-1}$ . What is the maximum percentage error in the calculated kinetic energy of the object?

13\%

9\%

8\%

21\%

Show answer & explanation

Answer: $13\%$

Answer

The maximum percentage error in the calculated kinetic energy is

13\%

.

For a calculated quantity

E = \frac{1}{2}mv^2

, the constant factor

\frac{1}{2}

has no error. The fractional error propagation formula gives

\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}

. Substituting

\frac{\Delta m}{m} = 0.05

(

5\%

) and

\frac{\Delta v}{v} = 0.04

(

4\%

) yields

5\% + 2(4\%) = 13\%

. Thus, the option stating

13\%

is correct.

Step-by-Step Solution

1

Calculate the fractional error and percentage error in mass

m

\frac{\Delta m}{m} = \frac{0.1}{2.0} = 0.05 = 5\%

Percentage error in mass is the absolute error divided by the measured value multiplied by 100.

2

Calculate the fractional error and percentage error in speed

v

\frac{\Delta v}{v} = \frac{0.2}{5.0} = 0.04 = 4\%

Percentage error in speed is the absolute error divided by the measured value multiplied by 100.

3

Apply the error propagation formula for kinetic energy

E = \frac{1}{2}m v^2

\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2\left(\frac{\Delta v}{v}\right)

When physical quantities are raised to a power and multiplied, their fractional errors are multiplied by the respective power index and added.

4

Compute the maximum percentage error in kinetic energy

\%\text{ error in } E = 5\% + 2(4\%) = 5\% + 8\% = 13\%

Combining the weighted percentage errors yields the total maximum percentage error.

Key Concept

Error Propagation in Power and Product Relationships

Estimated Time:2m 0s

Question 76Question →

The radius $r$ of a solid cylinder is measured as $(2.0 \pm 0.1)\text{ cm}$ and its height $h$ is measured as $(5.0 \pm 0.1)\text{ cm}$ . What is the maximum percentage error in the calculated volume of the cylinder?

7.0\%

8.0\%

10.0\%

12.0\%

Show answer & explanation

Answer: $12.0\%$

Answer

12.0\%

The formula for the volume of a cylinder is

V = \pi r^2 h

. According to the principles of error propagation, the fractional error in

V

is

\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta h}{h}

. Substituting the values:

\frac{\Delta r}{r} = \frac{0.1}{2.0} = 0.05

(

5.0\%

) and

\frac{\Delta h}{h} = \frac{0.1}{5.0} = 0.02

(

2.0\%

). Thus, the maximum percentage error is

2(5.0\%) + 2.0\% = 12.0\%

.

Step-by-Step Solution

1

Calculate the percentage error in the radius measurement

\text{Percentage error in } r = \left(\frac{0.1}{2.0}\right) \times 100\% = 5.0\%

Percentage error is given by the ratio of absolute uncertainty to the measured value multiplied by 100.

2

Calculate the percentage error in the height measurement

\text{Percentage error in } h = \left(\frac{0.1}{5.0}\right) \times 100\% = 2.0\%

Percentage error of a single linear measurement.

3

Apply the error propagation formula for the volume of a cylinder

\frac{\Delta V}{V} \times 100\% = 2\left(\frac{\Delta r}{r} \times 100\%\right) + \left(\frac{\Delta h}{h} \times 100\%\right)

Since

V = \pi r^2 h

, fractional errors add up, with the power of any variable acting as a multiplier for its fractional error.

4

Compute the total maximum percentage error in volume

\text{Percentage error in } V = 2(5.0\%) + 2.0\% = 10.0\% + 2.0\% = 12.0\%

Adding the individual fractional error contributions gives the maximum percentage error.

Key Concept

Error propagation in derived quantities with exponent powers

Question 77Question →

A spring balance is attached to the ceiling of an elevator accelerating downwards at $2.0\text{ m s}^{-2}$ . Suspended from the hook of the spring balance is a light, frictionless pulley carrying two masses of $3.0\text{ kg}$ and $1.0\text{ kg}$ connected by a light inextensible string. Taking the acceleration due to gravity $g = 10.0\text{ m s}^{-2}$ , what is the reading registered by the spring balance in newtons?

Show answer & explanation

Answer: 24

Answer

The reading registered by the spring balance is 24 N.

In a frame accelerating downwards at

a = 2.0\text{ m s}^{-2}

, the effective acceleration due to gravity is reduced to

g' = g - a = 8.0\text{ m s}^{-2}

. Within this frame, the tension in the Atwood machine string is

T = \frac{2(3.0)(1.0)}{3.0 + 1.0} \times 8.0 = 12.0\text{ N}

. Since two string segments act downward on the light pulley suspended from the spring balance, the total tension force registered by the balance scale is

2T = 24.0\text{ N}

.

Step-by-Step Solution

1

Determine the effective local acceleration due to gravity inside the accelerating elevator

g' = 8.0\text{ m s}^{-2}

Because the elevator accelerates downward at

a = 2.0\text{ m s}^{-2}

, objects inside experience an apparent gravitational acceleration of

g' = g - a

.

2

Compute the tension in the string supporting the two masses in the modified gravitational field

T = 12.0\text{ N}

For an Atwood machine system in effective gravity

g'

, string tension is

T = \frac{2 m_1 m_2}{m_1 + m_2} g' = \frac{2(3.0)(1.0)}{4.0} \times 8.0 = 12.0\text{ N}

.

3

Calculate the downward pull on the spring balance

F = 24.0\text{ N}

The spring balance supports the frictionless pulley, which experiences a downward force from two upward string segments, making the total measured weight force equal to

2T = 24.0\text{ N}

.

Key Concept

Apparent weight measurement and tension forces in accelerating frames

Question 78Question →

A student uses a stopwatch with a positive zero error of $+0.30\text{ s}$ to measure the time taken for a trolley to travel down an inclined plane. If the stopwatch display reads $15.70\text{ s}$ at the end of the trial, what is the actual time taken by the trolley?

15.40\text{ s}

16.00\text{ s}

15.70\text{ s}

15.4\text{ s}

Show answer & explanation

Answer: $15.40\text{ s}$

Answer

The actual time taken by the trolley is

15.40\text{ s}

.

The correct answer is

15.40\text{ s}

. When a measuring instrument has a positive zero error, it means the scale reads a value greater than zero before measurement begins. Therefore, the true value is found by subtracting the zero error from the observed reading:

15.70\text{ s} - 0.30\text{ s} = 15.40\text{ s}

.

Step-by-Step Solution

1

Identify the observed reading and the zero error of the instrument.

Observed reading =

15.70\text{ s}

, Zero error =

+0.30\text{ s}

.

Instrument readings must be corrected for systematic errors before recording actual values.

2

Apply the zero error correction formula:

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

.

\text{Actual Time} = 15.70\text{ s} - (+0.30\text{ s}) = 15.40\text{ s}

.

A positive zero error indicates the timer started above zero, so the initial offset must be subtracted.

Key Concept

Zero Error Correction in Time Measurement Instruments

Question 79Question →

A simple pendulum suspended in a physics laboratory has a length of $0.64\text{ m}$ . Given that acceleration due to gravity $g = 10\text{ m/s}^2$ and taking $\pi^2 = 10$ , what is the period of oscillation of the pendulum in seconds?

Show answer & explanation

Answer: 1.6

Answer

The period of oscillation of the simple pendulum is

1.6\text{ s}

.

Using the pendulum period relationship

T = 2\pi \sqrt{\frac{L}{g}}

, substituting

L = 0.64\text{ m}

,

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

, and

\pi = \sqrt{10}

yields

T = 2\sqrt{10}\sqrt{\frac{0.64}{10}} = 2\sqrt{0.64} = 1.6\text{ s}

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum.

The period formula is

T = 2\pi \sqrt{\frac{L}{g}}

.

The period depends on the length of the pendulum

L

and acceleration due to gravity

g

.

2

Substitute the given values into the formula.

T = 2\pi \sqrt{\frac{0.64\text{ m}}{10\text{ m/s}^2}}

.

Given parameters are

L = 0.64\text{ m}

and

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

.

3

Simplify the equation using

\pi = \sqrt{10}

.

T = 2\sqrt{10} \times \sqrt{0.064} = 2 \sqrt{10 \times 0.064} = 2 \sqrt{0.64} = 2 \times 0.8 = 1.6\text{ s}

.

Using

\pi^2 = 10

simplifies the calculation cleanly without requiring a calculator.

Key Concept

Simple Pendulum Period of Oscillation

Question 80Question →

A particle moves along a circular track of radius $r$ with a constant speed $v$ . If the radius of the track is measured with a percentage error of $1.2\%$ , and the speed is measured as $(10.0 \pm 0.3)\text{ m s}^{-1}$ , what is the percentage error in the calculated centripetal acceleration of the particle?

Show answer & explanation

Answer: 7.2

Answer

The percentage error in the calculated centripetal acceleration is 7.2%.

The centripetal acceleration is

a_c = \frac{v^2}{r}

. The percentage error in

v

is

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

. According to error propagation rules, the percentage error in

a_c

is

2 \times (\text{percentage error in } v) + (\text{percentage error in } r) = 2(3.0\%) + 1.2\% = 7.2\%

.

Step-by-Step Solution

1

Calculate the percentage error in the speed measurement

v

.

\text{Percentage error in } v = \frac{0.3}{10.0} \times 100\% = 3.0\%

Percentage error is given by dividing the absolute error by the measured value and multiplying by 100%.

2

Write the relationship for centripetal acceleration

a_c

in terms of

v

and

r

.

a_c = \frac{v^2}{r}

Centripetal acceleration is directly proportional to the square of speed and inversely proportional to radius.

3

Apply the fractional error propagation law for a quantity involving powers and quotients.

\frac{\Delta a_c}{a_c} = 2\left(\frac{\Delta v}{v}\right) + \frac{\Delta r}{r}

When a measured quantity is raised to a power

n

, its fractional error contribution is multiplied by

n

. Errors accumulate additively.

4

Substitute the individual percentage errors to find the total percentage error in

a_c

.

\text{Percentage error in } a_c = 2(3.0\%) + 1.2\% = 7.2\%

Multiplying the speed's percentage error by 2 and adding the radius percentage error yields the total relative error.

Key Concept

Propagation of percentage errors in physical formulas containing powers and ratios.

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