$\begin{array}{1 1}(A)\;1.04\% \\(B)\;1 \% \\(C)\;9.8\% \\(D)\;2 \% \end{array} $

$g= 4 \pi ^2 \large\frac{l}{T^2}$

$\bigg(\large\frac{\Delta g}{g}\bigg)_{max} $$ \times 100 = \large\frac{\Delta l}{l} $$ \times 100 +\large\frac{2 \Delta T}{T} $$ \times 100$

$T+\Delta T = \large\frac{t+ \Delta t}{n}$

$T= \large\frac{t}{n}$

$\large\frac{T+\Delta T}{T} = \frac{t+\Delta t}{t} $

$\qquad= \large\frac{\Delta l}{l}$$ \times 100+ \large\frac{2 \Delta t}{t} $$ \times 100$

$\qquad= \large\frac{0.1}{100 }$$ \times 100 +2 \times \large\frac{0.1}{21.2} $$ \times 100$

$\qquad= 0.1 \% + 0.94 \%$

$\qquad= 1.04 \%$

Hence A is the correct answer.

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