Physics - Practical physics Question with Solution | TestHub

From the relation

$\text{h}=\text{ut}+\frac{1}{2}{\text{gt}}^2$

$\text{h}=\frac{1}{2}{\text{gt}}^2\Rightarrow \text{g}=\frac{2\text{h}}{{\text{t}}^2}\left(\because \text{body initially at rest}\right)$

Taking natural logarithms on both sides, we get

$\ln \left(g\right)=\ln \left(h\right)-2\ln \left(t\right)+\ln \left(2\right)$

$\text{Differentiating,}\frac{\Delta \text{g}}{\text{g}}=\frac{\Delta \text{h}}{\text{h}}-2\frac{\Delta \text{t}}{\text{t}}$

For maximum permissible errors, we add both the errors. So

${\left(\frac{\Delta \text{g}}{\text{g}}\times 100\right)}_{\text{max}}=\left(\frac{\Delta \text{h}}{\text{h}}\times \text{100}\right)+2\times \left(\frac{\Delta \text{t}}{\text{t}}\times \text{100}\right)$


$\frac{\Delta \text{h}}{\text{h}}\times \text{100}={\text{e}}_1\text{and}\frac{\Delta \text{t}}{\text{t}}\times \text{100}={\text{e}}_2$

$\text{Therefore,}{\left(\frac{\Delta \text{g}}{\text{g}}\times \text{100}\right)}_{\text{max}}= {\text{e}}_1+2{\text{e}}_2$