Discussion
In an experiment, the acceleration due to gravity is determined by measuring the time period of a simple pendulum. If there is an error of 1% in the measurement of time period, the error in the value of g is :
Answer: A
The error in g is 2%.
For a simple pendulum \(T = 2\pi\sqrt{L/g}\), so \(g = 4\pi^{2}L/T^{2}\). The relative error in a quantity raised to a power is that power times the relative error:
\[ \begin{aligned} \frac{\Delta g}{g} &= 2\frac{\Delta T}{T} \\ &= 2 \times 1\% = 2\% \end{aligned} \]
The option 1% ignores the square, and 0.5% treats g as depending on the square root of T.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development