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The accuracy in the measurement of time for 20 oscillations of pendulum for the following measurements, t1 = 39.6 s, t2 = 39.9 s and t3 = 39.5 s is:
- a)± 0.005 s
- b)± 0.3 s
- c)± 0.5 s
- d)± 0.4 s
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The accuracy in the measurement of time for 20 oscillations of pendulu...
Given,
t1 = 39.6 s
t2 = 39.9s
t3 = 39.5 s
Least count of measuring instrument = 0.1 s
[As measurements have only one decimal place] Percision in the measurement = Least count of the measuring instrument =0.1 s
Mean value of time for 20 oscillations
t1 = 39.6 s
t2 = 39.9s
t3 = 39.5 s
Least count of measuring instrument = 0.1 s
[As measurements have only one decimal place] Percision in the measurement = Least count of the measuring instrument =0.1 s
Mean value of time for 20 oscillations
Absolute errors in the measurements
Δt1 = t - t1 = 39.7 - 39.6 = 0.1s
Δt2 = t - t2 = 39.7 - 39.9 = -0.2 s
Δt3 = t - t3 = 39.7 - 39.5 = 0.2 s
Δt1 = t - t1 = 39.7 - 39.6 = 0.1s
Δt2 = t - t2 = 39.7 - 39.9 = -0.2 s
Δt3 = t - t3 = 39.7 - 39.5 = 0.2 s
Mean absolute error
(rounding-off upto decimal place)∴ Accuracy of measurement = ±0.2 s
Most Upvoted Answer
The accuracy in the measurement of time for 20 oscillations of pendulu...
0.141 s
b) 0.03 s
c) 0.007 s
d) 0.003 s
We can use the formula for the period of a pendulum, T=2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.
Assuming the length of the pendulum is constant, we can find the average value of g using the three measurements of time:
t1= 39.6 s
t2 = 39.9 s
t3 = 39.5 s
The period T for each measurement is:
T1 = t1/20 = 1.98 s
T2 = t2/20 = 1.995 s
T3 = t3/20 = 1.975 s
Using T=2π√(L/g), we can write:
g = (4π^2L)/(T^2)
The average value of g is:
g_avg = (g1 + g2 + g3)/3
g1 = (4π^2L)/(T1^2) = (4π^2L)/(1.98^2) = 9.81 m/s^2
g2 = (4π^2L)/(T2^2) = (4π^2L)/(1.995^2) = 9.80 m/s^2
g3 = (4π^2L)/(T3^2) = (4π^2L)/(1.975^2) = 9.82 m/s^2
g_avg = (9.81 + 9.80 + 9.82)/3 = 9.81 m/s^2
The standard deviation of the measurements can be calculated using the formula:
σ = √[(∑(xi - x_avg)^2)/n]
where xi is the i-th measurement, x_avg is the average value of the measurements, and n is the number of measurements.
Using this formula, we can find:
σ = √[((1.98-1.975)^2 + (1.995-1.975)^2 + (1.975-1.975)^2)/3] = 0.007 s
Therefore, the accuracy in the measurement of time for 20 oscillations of pendulum is 0.007 s. The answer is (c).
b) 0.03 s
c) 0.007 s
d) 0.003 s
We can use the formula for the period of a pendulum, T=2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.
Assuming the length of the pendulum is constant, we can find the average value of g using the three measurements of time:
t1= 39.6 s
t2 = 39.9 s
t3 = 39.5 s
The period T for each measurement is:
T1 = t1/20 = 1.98 s
T2 = t2/20 = 1.995 s
T3 = t3/20 = 1.975 s
Using T=2π√(L/g), we can write:
g = (4π^2L)/(T^2)
The average value of g is:
g_avg = (g1 + g2 + g3)/3
g1 = (4π^2L)/(T1^2) = (4π^2L)/(1.98^2) = 9.81 m/s^2
g2 = (4π^2L)/(T2^2) = (4π^2L)/(1.995^2) = 9.80 m/s^2
g3 = (4π^2L)/(T3^2) = (4π^2L)/(1.975^2) = 9.82 m/s^2
g_avg = (9.81 + 9.80 + 9.82)/3 = 9.81 m/s^2
The standard deviation of the measurements can be calculated using the formula:
σ = √[(∑(xi - x_avg)^2)/n]
where xi is the i-th measurement, x_avg is the average value of the measurements, and n is the number of measurements.
Using this formula, we can find:
σ = √[((1.98-1.975)^2 + (1.995-1.975)^2 + (1.975-1.975)^2)/3] = 0.007 s
Therefore, the accuracy in the measurement of time for 20 oscillations of pendulum is 0.007 s. The answer is (c).
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Community Answer
The accuracy in the measurement of time for 20 oscillations of pendulu...
T1 = 39.5 sec
t2 = 39.6 sec
t3 = 39.9 sec
average of all of these will be the true value.
(39.5 + 39.6 + 39.9)/3 = 39.66666 which is approx. 39.7 sec as we have to round it to three significant digits as the measurements also have three significant places.
so the max error can be 0.2 sec in the case of t1 and t2 compared to the true value.
So plus minus 2 is the correct answer.
hope you got it.
t2 = 39.6 sec
t3 = 39.9 sec
average of all of these will be the true value.
(39.5 + 39.6 + 39.9)/3 = 39.66666 which is approx. 39.7 sec as we have to round it to three significant digits as the measurements also have three significant places.
so the max error can be 0.2 sec in the case of t1 and t2 compared to the true value.
So plus minus 2 is the correct answer.
hope you got it.
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The accuracy in the measurement of time for 20 oscillations of pendulum for the following measurements, t1= 39.6 s, t2 = 39.9 s and t3 = 39.5 s is:a)± 0.005 sb)± 0.3 sc)± 0.5 sd)± 0.4 sCorrect answer is option 'A'. Can you explain this answer?
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The accuracy in the measurement of time for 20 oscillations of pendulum for the following measurements, t1= 39.6 s, t2 = 39.9 s and t3 = 39.5 s is:a)± 0.005 sb)± 0.3 sc)± 0.5 sd)± 0.4 sCorrect answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The accuracy in the measurement of time for 20 oscillations of pendulum for the following measurements, t1= 39.6 s, t2 = 39.9 s and t3 = 39.5 s is:a)± 0.005 sb)± 0.3 sc)± 0.5 sd)± 0.4 sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The accuracy in the measurement of time for 20 oscillations of pendulum for the following measurements, t1= 39.6 s, t2 = 39.9 s and t3 = 39.5 s is:a)± 0.005 sb)± 0.3 sc)± 0.5 sd)± 0.4 sCorrect answer is option 'A'. Can you explain this answer?.
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