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Assuming that shear stress at the base of a mountain is equal to the force per unit area due to its weight. Calculate the maximum possible height of a mountain on the earth if breaking stress of a typical rock is $3 \times 10^{8} {Nm}^{- 3}$ and its density is $3 \times 10^{3} kg m^{- 3}$ . (Take g = 10 ${ms}^{- 2}$ )

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16 km

4 km

8 km

10 km

answer is C.

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Detailed Solution

For a mountain of height h and base area A, weight

W = Ah $ρ$ g

So, pressure at the base due to its own weight is

$P = \frac{W}{A} = \frac{Ahρg}{A} = hρg$

The mountain will exist, if

$hρg \leq Breaking stress$

$h \leq \frac{Breaking stress}{ρg}$

$h \leq \frac{30 \times 10^{7}}{3 \times 10^{3} \times 10} = 10^{4} m = 10 km$

$∴ h_{\max} = 10 km$

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