0
$\begingroup$

enter image description here

In this question (b) , I was not able to understand why the upward thrust on the flat surface of the hemisphere will be piR^3, because the "2/3 pi R^3" amount of water has been occupied by the hemisphere itself... and hence the upward thrust on the flat surface should have been equal to the sum of vertical component of force exerted on the curved surface area of hemisphere and the weight of the hemisphere..... So if anyone could clear this doubt it will be helpfull..

Could also tell me why there will be no side thrust on the hemisphere?

Improve this question
$\endgroup$
5
  • $\begingroup$ There is a side thrust but by symmetry the net side thrust is zero. $\endgroup$
    – Farcher
    Commented Dec 21, 2023 at 9:04
  • $\begingroup$ Ohk got it.... could you also like answer the (b) part please? $\endgroup$
    – Adhway
    Commented Dec 21, 2023 at 10:22
  • $\begingroup$ The only downward force which is exerted is due to the volume of water which is equal to the volume of a cylinder of radius $R$ and height $R$, ($\pi R^2 \times R$) minus the volume of a hemisphere of radius $R$, ($\frac 12 \times \frac 43 \pi R^3$). $\endgroup$
    – Farcher
    Commented Dec 21, 2023 at 10:45
  • $\begingroup$ umm I am talking about the (b) part.... why will the upthrust be equal to be pi R^3... $\endgroup$
    – Adhway
    Commented Dec 21, 2023 at 11:24
  • $\begingroup$ So that the net force on the hemisphere is zero. $\endgroup$
    – Farcher
    Commented Dec 21, 2023 at 11:29

0

Reset to default