A) \[5.8\times {{10}^{4}}K\]
B) \[\text{14 }\times \,\,\text{1}{{0}^{\text{4}}}\text{ K}\]
C) 1 cm
D) 2 cm
Correct Answer: D
Solution :
At the mean position, the speed will be maximum\[=\frac{n\times \frac{1}{2}m{{\upsilon }^{2}}}{t}\] So, \[=\frac{360\times \frac{1}{2}\times 2\times {{10}^{-2}}\times {{(100)}^{2}}}{60}\] and \[g=\frac{Gm}{{{R}^{2}}}\] \[\left( \text{Here}:{{M}_{m}}=\frac{{{M}_{e}}}{9},{{R}_{m}}\frac{{{R}_{e}}}{2} \right)\] or \[\overrightarrow{\text{F}}=\left( \text{2\hat{i}}+\text{4\hat{j}} \right)\] or \[\overrightarrow{S}=\left( \text{3\hat{j}}+\text{5\hat{k}} \right)\text{m}\] or \[{{\upsilon }_{A}}\]
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