Answer
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Hint The least count of a given stopwatch is said to be as 0.2 seconds. Now, a bob is allowed to undergo 20 oscillations and time is observed. We calculate actual value by subtracting Obtained value with least count and error as ratio of least count and obtained value. Find percentile of error using error factor.
Complete Step By Step Solution
Least count of an instrument is defined as the least and smallest accurate value that can be measured by the given instrument is called the least count of an instrument. The least count is never constant for any instrument and varies accordingly due to manufacturing defects and other errors that occur during processing.
Now in our given question, it is given as a stopwatch that has a least count of \[0.2s\]. This means that the least and accurate value that can be measured using this stopwatch is \[0.2s\]. In a given experiment, a bob is left to oscillate 20 times and it is observed that it takes 25 seconds to complete 20 oscillations. Now the actual value the bob will take to oscillate 20 times is given as the difference between obtained time and least count of the instrument.
\[actual = observed - LC\]
\[ \Rightarrow actual = 25 - 0.2s = 24.8s\]
Now error in time is calculated as the ratio of change in time to the observed value. This means that it is the ratio of least count of the instrument to the observed value.
\[ \Rightarrow \dfrac{{\Delta T}}{T} = \pm (\dfrac{{LC}}{{Observed}})\]
Substituting the values and converting it to percentile we get
\[ \Rightarrow (\dfrac{{\Delta T}}{T} \times 100) = \pm (\dfrac{{0.2}}{{25}} \times 100) = \pm 0.8\% \]
Thus, Option (c) is the right answer for the given question.
Note Oscillation of a body or particle is defined as the repetitive swing motion of the body when it is attached to a single support point . It is the variation with respect to time of its physical state and motion from one end to another.
Complete Step By Step Solution
Least count of an instrument is defined as the least and smallest accurate value that can be measured by the given instrument is called the least count of an instrument. The least count is never constant for any instrument and varies accordingly due to manufacturing defects and other errors that occur during processing.
Now in our given question, it is given as a stopwatch that has a least count of \[0.2s\]. This means that the least and accurate value that can be measured using this stopwatch is \[0.2s\]. In a given experiment, a bob is left to oscillate 20 times and it is observed that it takes 25 seconds to complete 20 oscillations. Now the actual value the bob will take to oscillate 20 times is given as the difference between obtained time and least count of the instrument.
\[actual = observed - LC\]
\[ \Rightarrow actual = 25 - 0.2s = 24.8s\]
Now error in time is calculated as the ratio of change in time to the observed value. This means that it is the ratio of least count of the instrument to the observed value.
\[ \Rightarrow \dfrac{{\Delta T}}{T} = \pm (\dfrac{{LC}}{{Observed}})\]
Substituting the values and converting it to percentile we get
\[ \Rightarrow (\dfrac{{\Delta T}}{T} \times 100) = \pm (\dfrac{{0.2}}{{25}} \times 100) = \pm 0.8\% \]
Thus, Option (c) is the right answer for the given question.
Note Oscillation of a body or particle is defined as the repetitive swing motion of the body when it is attached to a single support point . It is the variation with respect to time of its physical state and motion from one end to another.
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