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Take a differential area dA at distance r from the center.
Here, We take a cylindrical element of uniform cross section dA to Calculate Pressure.
dF= -(GMr/R^3) x dm = -(GMr/R^3) x dA x dr x density = -[(GMr/R^3) x dr] x density x dA
dF1= -density x dA x ∫_R^r(GMr/R^3) x dr (Here, dF1 is net force acting on surface at r radius)
= (GM/2R^3) x density x dA x [ -r^2+R^2]
Pressure= dF1/dA = (GM/2R^3) x density x [ R^2-r^2] =(GM/2R^3) x (M/V) x [ R^2-r^2] =(GM/2R^3) x (3M/4piR^3) x [ R^2-r^2] =(3GM^2/8piR^6) x [ R^2-r^2] =(3GM^2/8piR^4) x [1-r^2/R^2]
this is a really good question guys
ReplyDeletei would lyk u guys to try it before seeing the sol
so therefore i will be posting it on 11th nov
The answer is 3GM^2/8piR^4 x [1-r^2/R^2].
ReplyDeleteTake a differential area dA at distance r from the center.
ReplyDeleteHere, We take a cylindrical element of uniform cross section dA to Calculate Pressure.
dF= -(GMr/R^3) x dm
= -(GMr/R^3) x dA x dr x density
= -[(GMr/R^3) x dr] x density x dA
dF1= -density x dA x ∫_R^r(GMr/R^3) x dr (Here, dF1 is net force acting on surface at r radius)
= (GM/2R^3) x density x dA x [ -r^2+R^2]
Pressure= dF1/dA
= (GM/2R^3) x density x [ R^2-r^2]
=(GM/2R^3) x (M/V) x [ R^2-r^2]
=(GM/2R^3) x (3M/4piR^3) x [ R^2-r^2]
=(3GM^2/8piR^6) x [ R^2-r^2]
=(3GM^2/8piR^4) x [1-r^2/R^2]