A) \[2.5\times {{10}^{-7}}:1\]
B) \[1:2.5\times {{10}^{-4}}\]
C) \[1:1\]
D) \[4\times {{10}^{-5}}:1\]
Correct Answer: A
Solution :
: Angular momentum about own axis\[{{J}_{A}}=I\omega \] \[\therefore \]\[JA=\frac{2}{5}M{{R}^{2}}\times \frac{2\pi }{T}\]where\[T=1\]day or \[{{J}_{A}}=\frac{2}{5}M{{R}^{2}}\times \frac{2\pi }{1}\] ................ (i) Angular momentum of earth in its orbit around sun. \[{{J}_{0}}=M{{r}^{2}}{{\omega }_{1}}\] or \[{{J}_{0}}=M{{r}^{2}}\times \frac{2\pi }{365}\] \[\therefore \] \[\frac{{{J}_{A}}}{{{J}_{0}}}=\frac{2}{5}M{{r}^{2}}\times 2\pi \times \frac{360}{M{{r}^{2}}\times 2\pi }\] \[=\frac{2\times 365}{5}{{\left( \frac{R}{r} \right)}^{2}}=146\times {{\left( \frac{6\times {{10}^{6}}}{1.5\times {{10}^{11}}} \right)}^{2}}\] \[=2.336\times {{10}^{-7}}\]or \[\frac{{{J}_{A}}}{{{J}_{0}}}\approx \frac{2.5\times {{10}^{-7}}}{1}\]
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