Answer a-

Height of mountain range -4 Km

For erosion of 2 km of material to attain isostatic equlibrium

the depth of compensation before and after erosion:

ρc g hc + ρm g hm = ρc g h’c + ρm g h’m = ρc g ( hc - he ) + ρm g h’m ------------------------ eq. 1

(here ρc – crustal density, ρm- Mantle density, hc – crustal height, h’c – Crustal height after erosion,

hm – Mantle thickness, h’m – Mantle thickness at depth of compensation, he – eroded height )

The equation can be expressed as

Δ h = h’m - hm ------------------------ eq. 2

ρm (h’m - hm) = ρc he

Δ h = (ρc/ ρm) he ------------------------ eq. 3

Using the value of crustal density - 2800 and mantle density of 3300 and he 2 Km

Δ h = (2800/3300) * 2 km

Δ h = 0.85 * 2 Km

                        Δ h = 1.70 Km

(Hence for erosion of 2 Km the loss in height will be only .30 Km)

                       The new height = 4 – 0.3 Km

                                                = 3.7 Km

Part b-

To bring back the mountain to sea level the required change in height = 4 Km

Change in height of mountain for erosion of 1 Km

Using eq 3 –

                                =1 – ((2800/3300) * 2) Km

                                = 0.15 Km

Erosion needed for 4 Km change in height

                                = 4/.15

                                =26.66 Km