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While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the Pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be?
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While measuring acceleration due to gravity by a simple pendulum, a st...
For finding error in case of g is given by:
Δg/g = Δl/l + 2ΔT/T
⇒ Δg/g x 100 = Δl/l x 100 + 2 x ΔT/T x 100
⇒ % error in g = % error in l + 2 x % error in T
⇒ % error in g = 2 % + 2 x 1 %
= 4 %
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
While measuring acceleration due to gravity by a simple pendulum, a st...
Calculation of Percentage Error in the Measurement of Acceleration due to Gravity
- Given positive error in length of pendulum: 2%
- Given positive error in time period: 1%
Formula for calculating acceleration due to gravity (g) using a simple pendulum:
g = (4 * π^2 * L) / T^2
Calculating the percentage error in length of pendulum (ΔL) and time period (ΔT):
- ΔL = 2%
- ΔT = 1%
Calculating the percentage error in acceleration due to gravity (Δg):
- Using error propagation formula:
Δg/g = sqrt((ΔL/L)^2 + (2ΔT/T)^2)
Substitute the given values:
Δg/g = sqrt((0.02)^2 + 2(0.01)^2)
Δg/g = sqrt(0.0004 + 2(0.0001))
Δg/g = sqrt(0.0004 + 0.0002)
Δg/g = sqrt(0.0006)
Δg/g ≈ 0.0245
Converting to percentage:
Δg/g ≈ 0.0245 * 100%
Δg/g ≈ 2.45%
Therefore, the student's actual percentage error in the measurement of the value of g is approximately 2.45%.
- Given positive error in length of pendulum: 2%
- Given positive error in time period: 1%
Formula for calculating acceleration due to gravity (g) using a simple pendulum:
g = (4 * π^2 * L) / T^2
Calculating the percentage error in length of pendulum (ΔL) and time period (ΔT):
- ΔL = 2%
- ΔT = 1%
Calculating the percentage error in acceleration due to gravity (Δg):
- Using error propagation formula:
Δg/g = sqrt((ΔL/L)^2 + (2ΔT/T)^2)
Substitute the given values:
Δg/g = sqrt((0.02)^2 + 2(0.01)^2)
Δg/g = sqrt(0.0004 + 2(0.0001))
Δg/g = sqrt(0.0004 + 0.0002)
Δg/g = sqrt(0.0006)
Δg/g ≈ 0.0245
Converting to percentage:
Δg/g ≈ 0.0245 * 100%
Δg/g ≈ 2.45%
Therefore, the student's actual percentage error in the measurement of the value of g is approximately 2.45%.
Community Answer
While measuring acceleration due to gravity by a simple pendulum, a st...
4%
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While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the Pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be?
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While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the Pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the Pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the Pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be?.
While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the Pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the Pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the Pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be?.
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