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#### Please solve RD Sharma class 12 chapter 13 Differentials Errors and Approximations exercise multiple choice question 1 maths textbook solution

Answer:  A) $1%$

Hint:  Here, we all know the formula of period of a pendulum is

$T=2 \pi \sqrt{\frac{L}{g}}$

Given:

Given,   $\left(\frac{\Delta L}{L}\right) \times 100=2$

[If we let the length of pendulum is $L$]

Solution:

Given $\left(\frac{\Delta L}{L}\right) \times 100=2$......equation 1               (if we let the length of pendulum is L)

we all know the formula of period of a pendulum is $\mathrm{T}=2 \pi \times \mathrm{V}(\mathrm{l} / \mathrm{g})$

By the formula of approximation in derivation, we get

Time period,

$T=2 \pi \sqrt{\frac{L}{g}}$

Taking Log on both sides, we get

$\log T=\log 2 \pi+\frac{1}{2} \log \mathrm{L}-\frac{1}{2} \log g$

Differentiating both sides w.r.t x, we get

So,

$\begin{gathered} \frac{1}{T} \frac{d T}{d L}=\frac{1}{2 L} \\\\ \frac{d T}{d L}=\frac{T}{2 L} \end{gathered}$

$\left(\frac{\Delta T}{T}\right) \times 100=\frac{1}{2} \times\left(\frac{\Delta L}{L}\right) \times 100$

$\left(\frac{\Delta T}{T}\right) \times 100=\frac{1}{2} \times 2$               (from  equation 1 $\left(\frac{\Delta L}{L}\right) \times 100=2$ )

\begin{aligned} &\left(\frac{\Delta T}{T}\right)=\frac{1}{100} \\\\ &\left(\frac{\Delta T}{T}\right)=1 \% \end{aligned}