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C

- Hướng dẫn giải

Đáp án C

Ta có: $T=2\pi \sqrt{\frac{\mathcal{l}}{g}}\Rightarrow g=\frac{4{\pi }^2\mathcal{l}}{T^2}\Rightarrow \overline{g}=\frac{4{\pi }^2\overline{\mathcal{l}}}{{\overline{T}}^2}=\frac{4{\pi }^2.1,19}{2,{20}^2}=9,7\left(m/s^2\right)$

Vì $g=\frac{4{\pi }^2\mathcal{l}}{T^2}\Rightarrow \delta g=\delta \mathcal{l}+\delta T^2\iff \frac{\Delta g}{\overline{g}}=\frac{\Delta \mathcal{l}}{\overline{\mathcal{l}}}+2\frac{\Delta T}{\overline{T}}\Rightarrow \Delta g=\overline{g}\left(\frac{\Delta \mathcal{l}}{\overline{\mathcal{l}}}+2\frac{\Delta T}{\overline{T}}\right)$

Thay số ta có: $\Delta g=9,7\left(\frac{1}{119}+2\frac{0,01}{2,20}\right)\approx 0,2\Rightarrow g=\left(9,7\pm 0,2\right)\left(m/s^2\right)$

Hoặc có thể giải theo cách khác nhau sau:

Ta có: $T=2\pi \sqrt{\frac{\mathcal{l}}{g}}\Rightarrow g=\frac{4{\pi }^2\mathcal{l}}{T^2}\Rightarrow \overline{g}=\frac{4{\pi }^2\overline{\mathcal{l}}}{\overline{T^2}}=\frac{4{\pi }^2.1,19}{2,{20}^2}=9,7\left(m/s^2\right)$

Lấy ln hai vế $g=\frac{4{\pi }^2\mathcal{l}}{T^2}$ Ta được: $\ln \left(g\right)=\ln \left(4{\pi }^2\right)+\ln \left(\mathcal{l}\right)+\ln \left(\frac{1}{T}\right)+\ln \left(\frac{1}{T}\right)$

$\iff \frac{\Delta g}{\overline{g}}=\frac{\Delta \mathcal{l}}{\overline{\mathcal{l}}}+\frac{\Delta T}{\overline{T}}+\frac{\Delta T}{\overline{T}}\Rightarrow \Delta g=\overline{g}\left(\frac{\Delta \mathcal{l}}{\overline{\mathcal{l}}}+2\frac{\Delta T}{\overline{T}}\right)$

Thay số ta có: $\Delta g=9,7.\left(\frac{1}{119}+2\frac{0,01}{2,20}\right)\approx 0,2\Rightarrow g=\left(9,7\pm 0,2\right)\left(m/s^2\right)$