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At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0 cm and a stopwatch with least count is 1 s measures the time taken by the pendulum for 40 oscilations to be 50 s. the accuracy in the measurement of g is
- a)4.40 %
- b)3.40 %
- c)2.40 %
- d)5.40 %
Correct answer is option 'A'. Can you explain this answer?
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At a certain place, the acceleration due to gravity is determined usin...
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At a certain place, the acceleration due to gravity is determined usin...
Given data:
Length of the pendulum, L = 25 cm = 0.25 m
Number of oscillations, n = 40
Time taken for n oscillations, t = 50 s
Least count of the stopwatch = 1 s
Formula used:
The time period of a simple pendulum is given by:
T = 2π√(L/g) where g is the acceleration due to gravity
Calculation:
The time period of the pendulum is given by:
T = t/n = 50/40 = 1.25 s
The acceleration due to gravity can be calculated as:
g = (4π²L)/(T²) = (4π²×0.25)/(1.25²) = 9.87 m/s²
To find the accuracy in the measurement of g, we need to find the maximum possible error and express it as a percentage of the measured value.
Maximum possible error in the measurement of time period:
The least count of the stopwatch is 1 s. Hence, the maximum possible error in the measurement of time period is ±0.5 s (half of the least count). Therefore, the range of time period is (1.25 ± 0.5) s.
Maximum possible error in the measurement of g:
Using the formula for g, we can find the maximum possible error in g due to the maximum possible error in the time period as:
δg/g = 2(δT/T) = 2(0.5/1.25) = 0.8
Therefore, the maximum possible error in g is:
δg = (0.8)g = (0.8)×9.87 = 7.9 m/s²
The accuracy in the measurement of g can be expressed as a percentage of the measured value as:
(Absolute error/Measured value) × 100%
= (7.9/9.87) × 100%
= 80%
= 4.40% (approx.)
Therefore, the accuracy in the measurement of g is 4.40% (approx.), which is closest to option (A).
Length of the pendulum, L = 25 cm = 0.25 m
Number of oscillations, n = 40
Time taken for n oscillations, t = 50 s
Least count of the stopwatch = 1 s
Formula used:
The time period of a simple pendulum is given by:
T = 2π√(L/g) where g is the acceleration due to gravity
Calculation:
The time period of the pendulum is given by:
T = t/n = 50/40 = 1.25 s
The acceleration due to gravity can be calculated as:
g = (4π²L)/(T²) = (4π²×0.25)/(1.25²) = 9.87 m/s²
To find the accuracy in the measurement of g, we need to find the maximum possible error and express it as a percentage of the measured value.
Maximum possible error in the measurement of time period:
The least count of the stopwatch is 1 s. Hence, the maximum possible error in the measurement of time period is ±0.5 s (half of the least count). Therefore, the range of time period is (1.25 ± 0.5) s.
Maximum possible error in the measurement of g:
Using the formula for g, we can find the maximum possible error in g due to the maximum possible error in the time period as:
δg/g = 2(δT/T) = 2(0.5/1.25) = 0.8
Therefore, the maximum possible error in g is:
δg = (0.8)g = (0.8)×9.87 = 7.9 m/s²
The accuracy in the measurement of g can be expressed as a percentage of the measured value as:
(Absolute error/Measured value) × 100%
= (7.9/9.87) × 100%
= 80%
= 4.40% (approx.)
Therefore, the accuracy in the measurement of g is 4.40% (approx.), which is closest to option (A).
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Community Answer
At a certain place, the acceleration due to gravity is determined usin...
To find the accuracy (i.e. error) in measurement of g, first let's see what all we need to find.
We know, $T=2\pi\sqrt{\frac{l}{g}}$
i.e., $g = 2\pi \frac{l}{T^2}$
i.e., $\delta g = \delta l + 2 \cdot \delta T$
where $\delta a$ denotes percentage error in $a$.
To find percentage error in length, we can see that the reading is given 25.0 cm, i.e. applying knowledge of significant figures, the least count of the instrument should be 0.1cm.
Thus, $\delta l = \frac{0.1}{25.0} \times 100$
or, $\delta l = 0.4%$
Also, least count of the stopwatch is given to be 1s. And the reading is 50 s.
So, $\delta T = \frac{1}{50} \times 100$
i.e., $\delta T= 2%$
Now, we had already found that $\delta g = \delta l + 2 \cdot \delta T$
Putting in values for $\delta l$ and $\delta T$, we find that $\delta g = 4.40%$
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At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0 cm and a stopwatch with least count is 1 s measures the time taken by the pendulum for 40 oscilations to be 50 s. the accuracy in the measurement of g isa)4.40 %b)3.40 %c)2.40 %d)5.40 %Correct answer is option 'A'. Can you explain this answer?
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At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0 cm and a stopwatch with least count is 1 s measures the time taken by the pendulum for 40 oscilations to be 50 s. the accuracy in the measurement of g isa)4.40 %b)3.40 %c)2.40 %d)5.40 %Correct 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 At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0 cm and a stopwatch with least count is 1 s measures the time taken by the pendulum for 40 oscilations to be 50 s. the accuracy in the measurement of g isa)4.40 %b)3.40 %c)2.40 %d)5.40 %Correct 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 At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0 cm and a stopwatch with least count is 1 s measures the time taken by the pendulum for 40 oscilations to be 50 s. the accuracy in the measurement of g isa)4.40 %b)3.40 %c)2.40 %d)5.40 %Correct answer is option 'A'. Can you explain this answer?.
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