A) \[{{M}_{e}}\]
B) \[\therefore \]
C) \[{{v}_{e}}=\sqrt{\frac{2G{{M}_{e}}}{{{\operatorname{R}}_{e}}}}\]
D) \[\Omega \]
Correct Answer: A
Solution :
Let height of TV antenna be h and radius of earth is R and R >> h. Let M be receiving station when h<<R then NM may be assumed as tangent to the surface of earth, so \[\Omega \] \[\Omega \] \[\Omega \] \[\Omega \] \[\Omega \] \[1:{{2}^{1/3}}\] \[{{2}^{1/3}}:1\] Since, h << R So, \[2:1\] may be neglected hence, \[1:2\] \[{{m}^{-2}}{{s}^{-1}}\] \[{{m}^{-2}}{{s}^{-1}}\]You need to login to perform this action.
You will be redirected in
3 sec