NEET · Physics · STD 11 - 13. oscillations
Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as :
(Take \(\pi^2=9.8\), and \(g=9.8 m / s ^2\) )
- A 0.75 m
- B 1.5 m
- C 2 m
- D 1 m
Answer & Solution
Correct Answer
(D) 1 m
Step-by-step Solution
Detailed explanation
(D) 1 m
Time period \(T=\frac{60}{30}=2 s\).
The formula for the time period of a simple pendulum is \(T=2 \pi \sqrt{\frac{L}{g}}\).
Squaring both sides, we get \(T^2=4 \pi^2 \frac{L}{g}\).
Substituting the given values \(T=2 s, \pi^2=9.8\), and \(g=9.8 m / s ^2\) :
\((2)^2=4 \times 9.8 \times \frac{L}{9.8}\)
\(4=4 L\)
\(L=1 m\)
Time period \(T=\frac{60}{30}=2 s\).
The formula for the time period of a simple pendulum is \(T=2 \pi \sqrt{\frac{L}{g}}\).
Squaring both sides, we get \(T^2=4 \pi^2 \frac{L}{g}\).
Substituting the given values \(T=2 s, \pi^2=9.8\), and \(g=9.8 m / s ^2\) :
\((2)^2=4 \times 9.8 \times \frac{L}{9.8}\)
\(4=4 L\)
\(L=1 m\)
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