JEE Class main Answered
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Asked by swayamagarwal2114 | 21 Jul, 2022, 17:18: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/d753778374184cf33bc126f17773c50b62d96a515f6ac9.94367818f4.png)
Pressure P is the force acting on unit area .If we consider a fluid element of small size dr , then pressure dp
due to this small element dr is given as
dp = ρ g(r) dr
where ρ is uniform density and g(r) is acceleration due to gravity at distance r
g(r) = G ( m / r2 )
where G is universal gravitation constant and m is mass of fluid in sphere of radius r.
![begin mathsize 14px style g left parenthesis r right parenthesis space equals space G space fraction numerator begin display style 4 over 3 end style pi space r cubed space rho over denominator r squared end fraction space equals space 4 over 3 pi space G space rho space r end style](https://images.topperlearning.com/topper/tinymce/cache/f2bb48536d8ddbafce600b33cfeffccb.png)
Presuure at centre due to full radius R is given as
![begin mathsize 14px style P space equals space integral subscript 0 superscript R rho space g left parenthesis r right parenthesis space d r space equals space 4 over 3 pi space G space rho squared space integral subscript 0 superscript R r space d r space equals space space 2 over 3 pi space G space rho squared R squared end style](https://images.topperlearning.com/topper/tinymce/cache/30a1493e026153c4be4a20fa60159e80.png)
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