Question:
In measuring \(g = \frac{4\pi^2 l}{T^2}\), errors in \(l\) and \(T\) are given. Which gives most accurate result?
In measuring \(g = \frac{4\pi^2 l}{T^2}\), errors in \(l\) and \(T\) are given. Which gives most accurate result?
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Increase oscillation count to reduce time measurement error.
Updated On: Apr 23, 2026
- \(\Delta l=5\,mm,\ \Delta T=0.2s,\ n=10,\ A=5mm\)
- \(\Delta l=5\,mm,\ \Delta T=0.2s,\ n=20,\ A=5mm\)
- \(\Delta l=5\,mm,\ \Delta T=0.1s,\ n=10,\ A=1mm\)
- \(\Delta l=1\,mm,\ \Delta T=0.1s,\ n=50,\ A=1mm\)
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The Correct Option is D
Solution and Explanation
Concept:
\[
\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}
\]
Step 1: Minimise error
• Smaller \(\Delta l\) better
• Smaller \(\Delta T\) better
• Larger \(n\) $\Rightarrow$ reduces timing error
• Small amplitude $\Rightarrow$ SHM validity
Step 2: Best choice
Option (D) has: \[ \text{minimum errors + maximum oscillations} \] Conclusion: (D)
Step 1: Minimise error
• Smaller \(\Delta l\) better
• Smaller \(\Delta T\) better
• Larger \(n\) $\Rightarrow$ reduces timing error
• Small amplitude $\Rightarrow$ SHM validity
Step 2: Best choice
Option (D) has: \[ \text{minimum errors + maximum oscillations} \] Conclusion: (D)
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