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 $3 \times 10^{8} N m^{- 2}$ and its density is then $3 \times 10^{3} k g m^{- 3}$ .
(Take $g = 10 m s^{- 2}$ )

4 k m

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8 k m

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10 k m

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16 k m

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Solution

The correct option is C $10 k m$
For a mountain of height $h$ and base area $A$ , weight $W - A h ρ g$
So, the pressure at the base due to its own weight is
$P = \frac{W}{A} = \frac{A h ρ g}{A} = h ρ g$
The mountain will exist if $h ρ g \leq$ breaking stress
$∴ h \leq \frac{B r e a k i n g s t r e s s}{ρ g}$
$h \leq \frac{30 \times 10^{7}}{3 \times 10^{3} \times 10} = 10^{4} m = 10 k m$
$∴ h_{m a x} = 10 k m$

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Similar questions

km

3 \times 10^{8} {Nm}^{- 2}

3

30 \times 10^{7} N m^{- 2}

3 \times 10^{3} k g / m^{3}

7.9 \times 10^{9} N m^{- 2}

7.9 \times 10^{3} k g m^{- 3}

g = 10 m s^{- 2}

1.5 m^{2}

10^{3} k g m^{- 3}

g = 10 m s^{- 2}]

7.8 \times 10^{9} N M^{- 2}

= 7.8 \times 10^{3} k g m^{- 3}

g = 10 N k g^{- 1}

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