A) \[9\text{ }m{{s}^{-2}}\]
B) \[8.5\text{ }m{{s}^{-2}}\]
C) \[10m{{s}^{-2}}\]
D) \[9.5\text{ }m{{s}^{-2}}\]
Correct Answer: D
Solution :
Above the surface of earth at height h, acceleration due to gravity \[g=g\frac{1}{{{\left( 1+\frac{h}{R} \right)}^{2}}}\] Given that, \[g=10\,m{{s}^{-2}}\]at surface of earth \[g=10\frac{1}{{{\left( 1+\frac{h}{R} \right)}^{2}}}\] ?(i) Below the surface of earth atdepth h, acceleration due to gravity \[g=g\left( 1-\frac{h}{R} \right)\] \[\therefore \] \[g=10\left( 1-\frac{h}{R} \right)\] ?(ii) From Eq. (i) \[9=10{{\left( 1+\frac{h}{R} \right)}^{-2}}\] By Binomial theorem \[9=10\left( 1-\frac{2h}{R} \right)\] ?(iii) \[\therefore \] \[\frac{9}{10}-1=-\frac{2h}{R}\] or \[-\frac{1}{10}=-\frac{2h}{R}\] \[\frac{R}{20}=h,\] Now, we put the value of h in Eq. (ii), we have \[g\,=10\left( 1-\frac{R/20}{R} \right)\] \[=10\left( 1-\frac{1}{20} \right)\] \[=10\left( \frac{19}{20} \right)\] \[=9.5\,m{{s}^{-2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
