To estimate 'g' from $$g=4$$$$π$$$$2LT2$$, error in measurement of L is ±$$2$$% and error in measurement of T is ±$$3$$%. The error in estimated 'g' will be:

$$g=4$$$$π$$$$2lT2$$ Δ$$ll=2$$%$$=$$±$$2$$×$$10$$−$$2$$ Δ$$TT=$$±$$3$$%$$=$$±$$3$$×$$10$$−$$2$$⇒ Δ$$gg=$$Δ$$ll+2$$ΔTT $$=2$$×$$10$$−$$2+2$$×$$3$$×$$10$$−$$2$$ $$=8$$×$$10$$−$$2=$$±$$8$$%

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Accuracy; precision of instruments and errors in measurements

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Question

To estimate 'g' $(from g = 4 π^{2} \frac{L}{T^{2}})$ , error in measurement of L is $\pm$ 2% and error in measurement of T is $\pm$ 3%. The error in estimated 'g' will be:

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Solution

$\begin{array}{l} g = 4 π^{2} \frac{l}{T^{2}} \\ \frac{Δl}{l} = 2 % = \pm 2 \times 10^{- 2} \\ \frac{ΔT}{T} = \pm 3 % = \pm 3 \times 10^{- 2} \\ \Rightarrow \frac{Δg}{g} = \frac{Δl}{l} + \frac{2 ΔT}{T} \\ = 2 \times 10^{- 2} + 2 \times 3 \times 10^{- 2} \\ = 8 \times 10^{- 2} = \pm 8 % \end{array}$

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