Thanks, I think I'm getting somewhere. here's the graph i got.
Though I'm a bit doubtful about the result of data number 4, g can't be that high. Do i have to draw a best fit line?? everything else seems to be almost in a straight line. Or do i just have to calculate the average?
ok i did what you told .i got , $$ g = \frac {\sqrt{2} \omega^2 R \left( h + \sqrt{h^2 + d^2} \right)} {2 d} $$
now what do i do? They have given me data of "d" and "T"(from which i can get ## \omega ##). i need to plot a linear equation in the graph and estimate its slope. pls help.
yeah i realized that. ## \tan 2 \theta = \frac d h ##. And i also know, ## \tan \theta = \frac {\omega^2 R} {g \sqrt{2}} ## . I can relate these two with $$ \tan 2 \theta = \frac {2 \tan \theta} {1 - {\tan^2 \theta}} $$ . But that would give me a complex equation. i need to get a linear equation...
(I'm new here so i cant really write with latex, sorry)
So I balanced the forces at point P.(otherwise the height of water column would not remain same). we have centripetal acceleration (w^2)*x towards +ve x axis and gravity g at -ve y axis ( taking point P as centre). So the two forces...