A simple-pendulum experiment is performed to find the value of acceleration due to…
Physics · JEE Main · NTA Exams — Physics and Measurement
A simple-pendulum experiment is performed to find the value of acceleration due to gravity by using the expression \(g=4 \pi^{2} \frac{L}{T^{2}}\) where L = 1 m. If ΔL is the error in the measurement of length L, ΔT is the least count of stop watch, then for which of the following values of amplitude (A), ΔL and ΔT, the error in the value of g is minimum?
- A = 0.4 m, ΔL = 0.5 m, ΔT = 0.1 s
- A = 0.4 m, ΔL = 0.1 m, ΔT = 0.05 s
- A = 0.5 m, ΔL = 0.5 m, ΔT = 0.1 s
- A = 0.5 m, ΔL = 0.1 m, ΔT = 0.05 s
Answer
(D) A = 0.5 m, ΔL = 0.1 m, ΔT = 0.05 s
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