The least count of a stop watch is \[0.2s\]. The time of 20 oscillations of a pendulum is measured to be \[25s\]. The percentage error in the measurement of time will be:
(A) \[16\% \]
(B) \[1.8\% \]
(C) \[0.8\% \]
(D) \[0.1\% \]
Answer
Verified
122.4k+ views
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.
Recently Updated Pages
The ratio of the diameters of two metallic rods of class 11 physics JEE_Main
What is the difference between Conduction and conv class 11 physics JEE_Main
Mark the correct statements about the friction between class 11 physics JEE_Main
Find the acceleration of the wedge towards the right class 11 physics JEE_Main
A standing wave is formed by the superposition of two class 11 physics JEE_Main
Derive an expression for work done by the gas in an class 11 physics JEE_Main
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
Class 11 JEE Main Physics Mock Test 2025
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line