The correct option is
A P-1; Q-3; R-2; S-4
Buoyant force on ball =
√2(2mg) Force on ball w.r.t. vessel, where
Fp is peudo force on ball in frame of vessel.
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/561371/original_3.2.png)
Since, force on ball in frame of vessel is along upward normal to surface of liquid,ball will rise.
(P) - 1, 3, 4
Buoyant force on ball
=vρ(g+g)=4mg ![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/561373/original_3.3.png)
Force on ball w.r.t vessel
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/561378/original_3.4.png)
Since, force on ball in frame of vessel is along upward normal to surface of liquid, ball will rise.
(Q) - 3, 4
Buoyant force on ball
=vρ(g−g)=0 Force on ball w.r.t. vessel is absent.
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/561382/original_3.5.png)
Since, force on ball in frame of vessel is zero, ball will remain stationary.
(R) - 2
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/561696/original_q.PNG)
Since, acceleration of vessel is
g/2 downwards. Therefore,
Buoyant force on ball
=vρ√(g/2)2+(g)2+2(g)(g/2)cos120∘=√3mg Since, force on ball in frame of vessel is along upward normal to surface of liquid, ball will rise.
(S) - 3, 4