A) 6 km
B) 19 km
C) 10 km
D) 15 km
Correct Answer: C
Solution :
| At the bottom of a mountain of height h, the force per unit area due to the weight of the mountain is hg where, \[\rho \] the density of the material of the mountain and \[g\] is the acceleration due to gravity. The m material at the bottom experiences this force in the direction and the sides of the mountain are free. Therefore, this is not a case of pressure of bulk compression. There is a shear component approximately \[hg\] itself. Now, the elastic limit for a typical rock is \[30\times {{10}^{7}}N{{m}^{-2}}\]. Equating this to hg with |
| \[r=3\times {{10}^{3}} \operatorname{kg}\,{{m}^{-3}}\] gives\[hg=30\times {{10}^{7}}N{{m}^{-2}}\] Or \[h=30\times {{10}^{7}}N{{m}^{-2}}/\left( 3\times {{10}^{3}}kg{{m}^{-3}}\times 10m{{s}^{-2}} \right)\]\[=10km\] |
| Which is more than the height of Mt. Everest |
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