A) \[{{e}_{2}}-{{e}_{1}}\]
B) \[{{e}_{1}}+2{{e}_{2}}\]
C) \[{{e}_{1}}+{{e}_{2}}\]
D) \[{{e}_{1}}-2{{e}_{2}}\]
Correct Answer: B
Solution :
[b]: From the relation, \[h=ut+\frac{1}{2}g{{t}^{2}}\]\[h=\frac{1}{2}g{{t}^{2}}\]\[\Rightarrow \]\[g=\frac{2h}{{{t}^{2}}}\](\[\because \]\[u=0,\]body initially at rest) Taking natural logarithm on both sides, we get In g = In h - 2 In t Differentiating, \[\frac{\Delta g}{g}=\frac{\Delta h}{h}-2\frac{\Delta t}{t}\] For maximum permissible error, or\[{{\left( \frac{\Delta g}{g}\times 100 \right)}_{\max }}=\left( \frac{\Delta h}{h}\times 100 \right)+2\times \left( \frac{\Delta t}{t}\times 100 \right)\] According to problem \[\frac{\Delta g}{h}\times 100={{e}_{1}}\]and\[\frac{\Delta t}{t}\times 100={{e}_{2}}\] Therefore, \[{{\left( \frac{\Delta t}{g}\times 100 \right)}_{\max }}={{e}_{1}}+2{{e}_{2}}\]
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