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Assuming that the atmosphere has the same density anywhere as at sea level (\rho = 1.3 kg/m3)
and g to the constaht (g=10 m/s2). What should be the appropriate height of the atmosphere ( \rho_0= 1. 01 x 105 N/m2)?

Option: 1

6 km


Option: 2

8 km


Option: 3

12 km


Option: 4

18 km


Answers (1)

best_answer

1 atmosphere =  1. 01x 105 N/m2 = \rho gh

\therefore h = \frac{1. 01\times 10^{5} }{1.3\times 10} = 7789 m \approx 8 km

Posted by

Divya Prakash Singh

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