If T=2πlg, then relative errors in T and I are in the ratio

A
$1 / 2$
B
$2$
C
$1 / 2 π$
D
none of these

Solution:

We have,

$\begin{array}{l} T = 2 π \sqrt{\frac{l}{g}} \\ \Rightarrow \log T = \log 2 π + \frac{1}{2} \log l - \frac{1}{2} \log 8 \\ \Rightarrow \frac{1}{T} d T = 0 + \frac{1}{2 l} d l - 0 \\ \Rightarrow \frac{d T}{T} = \frac{1}{2} \frac{d l}{l} \\ \Rightarrow \frac{Δ T}{T} = \frac{1}{2} \frac{Δ l}{l} [∵ d T ≅ Δ T and d l ≅ Δ l] \\ \Rightarrow \frac{Δ T}{T} \frac{Δ l}{l} = \frac{1}{2} \\ \Rightarrow l \end{array}$

If $T = 2 π \sqrt{\frac{l}{g}},$ then relative errors in $T$ and $I$ are in the ratio

Solution:

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