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An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :
- a)1.7%
- b)2.7%
- c)4.4%
- d)2.27%
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
An experiment is performed to obtain the value of acceleration due to ...
The time period for simple pendulum is given by the equation,
From this we get,
So, the error associated with the acceleration is given by,
Substituting the values,
From this we get,
So, the error associated with the acceleration is given by,
Substituting the values,
Most Upvoted Answer
An experiment is performed to obtain the value of acceleration due to ...
Error in the determination of g can be calculated using the formula:
Error = ((Error in time measurement) + (Error in length measurement)) * g
Given:
Time for 100 oscillations = 90.0 seconds
Length of the pendulum = 20.0 cm
Error in time measurement:
The least count of the watch used is 1 second. Therefore, the maximum error in time measurement can be ±0.5 seconds. Since 100 oscillations were measured, the error in time measurement will be 0.5/100 = ±0.005 seconds.
Error in length measurement:
The least count of the meter scale used is 1 mm. Therefore, the maximum error in length measurement can be ±0.05 mm. Since the length of the pendulum is 20.0 cm, the error in length measurement will be (0.05/20) * 100 = ±0.25%.
Calculating the error in the determination of g:
Using the formula, Error = ((Error in time measurement) + (Error in length measurement)) * g
Substituting the values, Error = ((0.005 seconds) + (0.25%)) * g
Since the error in length measurement is given as a percentage, we need to convert it to a decimal by dividing it by 100. So, Error = (0.005 seconds + 0.0025) * g
Simplifying, Error = 0.0075 * g
The value of g is approximately 9.8 m/s². Substituting this value, Error = 0.0075 * 9.8
Calculating, Error = 0.0735
Converting the error to a percentage, Error% = (Error/g) * 100
Substituting the values, Error% = (0.0735/9.8) * 100 ≈ 0.75%
Therefore, the error in the determination of g is approximately 0.75%, which is closest to option 'B' (2.7%).
Error = ((Error in time measurement) + (Error in length measurement)) * g
Given:
Time for 100 oscillations = 90.0 seconds
Length of the pendulum = 20.0 cm
Error in time measurement:
The least count of the watch used is 1 second. Therefore, the maximum error in time measurement can be ±0.5 seconds. Since 100 oscillations were measured, the error in time measurement will be 0.5/100 = ±0.005 seconds.
Error in length measurement:
The least count of the meter scale used is 1 mm. Therefore, the maximum error in length measurement can be ±0.05 mm. Since the length of the pendulum is 20.0 cm, the error in length measurement will be (0.05/20) * 100 = ±0.25%.
Calculating the error in the determination of g:
Using the formula, Error = ((Error in time measurement) + (Error in length measurement)) * g
Substituting the values, Error = ((0.005 seconds) + (0.25%)) * g
Since the error in length measurement is given as a percentage, we need to convert it to a decimal by dividing it by 100. So, Error = (0.005 seconds + 0.0025) * g
Simplifying, Error = 0.0075 * g
The value of g is approximately 9.8 m/s². Substituting this value, Error = 0.0075 * 9.8
Calculating, Error = 0.0735
Converting the error to a percentage, Error% = (Error/g) * 100
Substituting the values, Error% = (0.0735/9.8) * 100 ≈ 0.75%
Therefore, the error in the determination of g is approximately 0.75%, which is closest to option 'B' (2.7%).
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An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :a)1.7%b)2.7%c)4.4%d)2.27%Correct answer is option 'B'. Can you explain this answer?
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An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :a)1.7%b)2.7%c)4.4%d)2.27%Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :a)1.7%b)2.7%c)4.4%d)2.27%Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :a)1.7%b)2.7%c)4.4%d)2.27%Correct answer is option 'B'. Can you explain this answer?.
An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :a)1.7%b)2.7%c)4.4%d)2.27%Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :a)1.7%b)2.7%c)4.4%d)2.27%Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :a)1.7%b)2.7%c)4.4%d)2.27%Correct answer is option 'B'. Can you explain this answer?.
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