Measurement of Time Practice Questions — JAMB UTME

Question 1Question →

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 2Question →

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 3Question →

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 4Question →

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 5Question →

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 6Question →

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 7Question →

A simple pendulum on Earth ( $g = 10\text{ m/s}^2$ ) completes $50$ full oscillations in $40\text{ s}$ . The pendulum is then transferred to a lunar station where the acceleration due to gravity is $1.6\text{ m/s}^2$ , and its length is reduced by $64\%$ . What is the time taken, in seconds, for this modified pendulum to complete $30$ oscillations on the lunar station?

Show answer & explanation

Answer: 36

Answer

The time taken for the modified pendulum to complete 30 oscillations on the lunar station is 36 s.

The initial period on Earth is

T_1 = \frac{40}{50} = 0.8\text{ s}

. The formula for the period of a simple pendulum is

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

. When length decreases by

64\%

, the remaining length ratio is

\frac{L_2}{L_1} = 0.36

. The ratio of gravity is

\frac{g_1}{g_2} = \frac{10}{1.6} = 6.25

. Taking the ratio gives

\frac{T_2}{T_1} = \sqrt{0.36 \times 6.25} = \sqrt{2.25} = 1.5

. Thus, the new period is

T_2 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}

. For

30

oscillations, the total time is

t = 30 \times 1.2\text{ s} = 36\text{ s}

.

Step-by-Step Solution

1

Calculate the initial period of oscillation on Earth

T_1 = 0.8\text{ s}

Period

T_1

is total time divided by the number of oscillations:

T_1 = \frac{40\text{ s}}{50} = 0.8\text{ s}

.

2

Set up the ratio for period under altered length and gravitational field

\frac{T_2}{T_1} = 1.5

Using

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

, we have

\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1} \cdot \frac{g_1}{g_2}} = \sqrt{(1 - 0.64) \cdot \frac{10}{1.6}} = \sqrt{0.36 \cdot 6.25} = \sqrt{2.25} = 1.5

.

3

Determine the new period of oscillation

T_2 = 1.2\text{ s}

T_2 = 1.5 \times T_1 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}

.

4

Calculate the total time required for 30 oscillations

t_2 = 36\text{ s}

Total time

t_2 = N_2 \times T_2 = 30 \times 1.2\text{ s} = 36\text{ s}

.

Key Concept

Period of a simple pendulum and its dependence on length and gravitational acceleration

Question 8Question →

A student records the time taken for a simple pendulum to complete $20$ complete oscillations as $30\text{ s}$ . What is the period of oscillation of the pendulum, in seconds?

Show answer & explanation

Answer: 1.5

Answer

The period of oscillation of the pendulum is

1.5\text{ s}

.

The period of oscillation (

T

) is the time required for one complete cycle. Dividing the total time (

30\text{ s}

) by the number of oscillations (

20

) gives

1.5\text{ s}

.

Step-by-Step Solution

1

Identify the given values for total time and total number of oscillations.

Total time

t = 30\text{ s}

and number of oscillations

N = 20

.

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

2

Divide the total time by the number of oscillations to calculate the period.

T = \frac{30\text{ s}}{20} = 1.5\text{ s}

.

Applying the period formula

T = \frac{t}{N}

yields the duration of one period.

Key Concept

Period of Oscillation

Estimated Time:45s

Question 9Question →

A ticker-tape timer connected to a power supply operates at a frequency of $50\text{ Hz}$ . A continuous strip of paper tape pulled through the timer records a sequence of dots. Calculate the total time interval, in seconds, between the $1\text{st}$ dot and the $21\text{st}$ dot on the tape.

Show answer & explanation

Answer: 0.4

Answer

0.4 s

On a ticker tape, the time interval between consecutive dots represents one period

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

. The total elapsed time for

N

dots corresponds to

(N - 1)

spaces. For 21 dots, there are 20 spaces, giving a total time of

20 \times 0.02\text{ s} = 0.40\text{ s}

.

Step-by-Step Solution

1

Find the number of spaces between dots

20 spaces

Between

N

dots on a ticker-tape, there are

(N - 1)

time intervals.

2

Calculate the period per space

0.02 s

The period

T

is the reciprocal of the operating frequency

f = 50\text{ Hz}

.

3

Multiply the number of spaces by the period per space

0.4 s

Total time duration is the product of the total number of intervals and the time for one interval.

Key Concept

Calculation of time interval using ticker-tape timer frequency and dot count

Question 10Question →

A student measures the time taken for $40$ complete oscillations of a simple pendulum using a digital stopwatch that has a negative zero error of $-0.40\text{ s}$ . If the stopwatch displays a reading of $47.60\text{ s}$ for the oscillations, what will be the correct period of oscillation when the pendulum's length is reduced to one-fourth ( $1/4$ ) of its initial length?

0.60\text{ s}

0.59\text{ s}

0.30\text{ s}

2.40\text{ s}

Show answer & explanation

Answer: $0.60\text{ s}$

Answer

The correct period of oscillation for the shortened pendulum is

0.60\text{ s}

.

The correct answer is obtained by first adjusting the measured time for negative zero error (

47.60\text{ s} - (-0.40\text{ s}) = 48.00\text{ s}

), calculating the initial period per oscillation (

48.00 / 40 = 1.20\text{ s}

), and then taking into account that period varies with the square root of pendulum length. Reducing length to

1/4

reduces period by a factor of

\sqrt{4} = 2

, yielding

0.60\text{ s}

.

Step-by-Step Solution

1

Correct the recorded total time for instrument zero error

True total time

t = \text{Uncorrected Time} - \text{Zero Error} = 47.60\text{ s} - (-0.40\text{ s}) = 48.00\text{ s}

Negative zero error means the timer reads less than the actual elapsed time, so the absolute value of the error must be added back.

2

Calculate the initial period of oscillation

Initial period

T_1 = \frac{t}{N} = \frac{48.00\text{ s}}{40} = 1.20\text{ s}

Period is defined as the time per single oscillation.

3

Apply the pendulum scaling relationship for length and period

New period

T_2 = T_1 \times \sqrt{\frac{L_2}{L_1}} = 1.20\text{ s} \times \sqrt{\frac{1}{4}} = 1.20\text{ s} \times 0.5 = 0.60\text{ s}

The period of a simple pendulum is proportional to

\sqrt{L}

, so quartering the length reduces the period to half of its original value.

Key Concept

Measurement of time using a stopwatch with zero error correction combined with simple pendulum period dependence on length.

Estimated Time:2m 0s

Question 11Question →

An experimenter uses a digital stopwatch to record the duration of $40$ complete oscillations of a simple pendulum. The stopwatch displays a reading of $50.8\text{ s}$ , but it has a known positive zero error of $+0.8\text{ s}$ . What is the correct period of oscillation of the pendulum?

1.25\text{ s}

1.29\text{ s}

1.27\text{ s}

0.80\text{ s}

Show answer & explanation

Answer: $1.25\text{ s}$

Answer

The correct period of oscillation is

1.25\text{ s}

.

To find the true duration of the oscillations, the positive zero error of

+0.8\text{ s}

must be subtracted from the stopwatch display of

50.8\text{ s}

, giving an actual elapsed time of

50.0\text{ s}

. Dividing this corrected time by the

40

complete oscillations gives

T = 50.0\text{ s} / 40 = 1.25\text{ s}

.

Step-by-Step Solution

1

Calculate the true time interval by correcting for the instrument zero error.

t_{\text{actual}} = 50.8\text{ s} - 0.8\text{ s} = 50.0\text{ s}

A positive zero error means the timer reads higher than the true time, so the error value must be subtracted.

2

Calculate the period of a single oscillation by dividing the total corrected time by the number of oscillations.

T = \frac{t_{\text{actual}}}{N} = \frac{50.0\text{ s}}{40} = 1.25\text{ s}

The period

T

is defined as the time required to complete one full oscillation cycle.

Key Concept

Measurement of Time with Zero Error Correction

Estimated Time:1m 30s

Question 12Question →

A ticker-tape timer operates at an alternating current frequency of $50\text{ Hz}$ . During a mechanics experiment, a paper tape pulled through the device records a section containing $11$ consecutive dots. What is the total time elapsed for this section of the tape?

0.20\text{ s}

0.22\text{ s}

0.55\text{ s}

550\text{ s}

Show answer & explanation

Answer: $0.20\text{ s}$

Answer

The total time elapsed for this section of the tape is

0.20\text{ s}

.

The period of a

50\text{ Hz}

ticker-tape timer is

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

. A sequence of

11

consecutive dots contains

10

time intervals (

11 - 1 = 10

). Multiplying

10

intervals by

0.02\text{ s}

yields

0.20\text{ s}

.

Step-by-Step Solution

1

Determine the time interval (period) between consecutive dots recorded by the timer.

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

The ticker-tape timer makes one dot every period

T

of the operating frequency.

2

Calculate the number of intervals (spaces) between the first and last dot in the sequence.

\text{Number of intervals } n = N - 1 = 11 - 1 = 10

Time elapses between dots, so

N

dots define

N-1

spaces.

3

Multiply the number of intervals by the time period of one interval to get total time elapsed.

t = 10 \times 0.02\text{ s} = 0.20\text{ s}

Total duration is the product of the number of spaces and the duration of a single space.

Key Concept

Ticker-Tape Timer Period and Time Interval Calculation

Measurement of Time

Answer

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