A ticker-tape timer operates at an alternating current frequency of 50\text{ Hz}. During a… — JAMB UTME

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?

$0.20\text{ s}$ Answer
B
$0.22\text{ s}$
C
$0.55\text{ s}$
D
$550\text{ s}$

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

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.

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.

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

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