A) \[\theta ={{\tan }^{-1}}\left[ \frac{\pi }{2}\left( \frac{{{\rho }_{1}}-{{\rho }_{2}}}{{{\rho }_{1}}+{{\rho }_{2}}} \right) \right]\]
B) \[\theta ={{\tan }^{-1}}\frac{\pi }{2}\left( \frac{{{\rho }_{1}}-{{\rho }_{2}}}{{{\rho }_{1}}+{{\rho }_{2}}} \right)\]
C) \[\theta ={{\tan }^{-1}}\pi \left( \frac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)\]
D) None of above
Correct Answer: D
Solution :
[d] Let us find the pressure at the lowest point 1. Since the liquid has density |
\[{{\rho }_{2}}\] and height |
\[h_{2}^{'}\] on the right hand side of point 1, we have |
\[{{p}_{1}}={{\rho }_{1}}g{{h}_{1}}................(1)\] |
Since two liquid columns of height \[{{h}_{1}}\] and \[{{h}_{2}}\] densities \[{{\rho }_{1}}\] are \[{{\rho }_{2}}\] are situated above point 1, on the left-hand side, we have |
\[{{P}_{2}}={{\rho }_{1}}g{{h}_{2}}+{{\rho }_{2}}gh_{2}^{'}..............(2)\] |
Equating \[{{P}_{1}}\] and \[{{P}_{2}}\], we get |
\[{{\rho }_{1}}{{h}_{2}}+{{\rho }_{2}}h_{2}^{'}={{\rho }_{1}}{{h}_{1}}\] |
Substituting |
\[h_{2}^{'}=R\sin \theta +R\cos \theta ,\,\,{{h}_{2}}=R(1-\cos \theta )\]and \[{{h}_{1}}=R(1-\sin \theta )\] |
\[{{\rho }_{1}}R(1-\cos \theta )+{{\rho }_{2}}R(\sin \theta +\cos \theta )={{\rho }_{1}}R(1-sin\theta )\] |
This gives |
\[\tan \theta =\frac{{{\rho }_{1}}-{{\rho }_{2}}}{{{\rho }_{1}}+{{\rho }_{2}}}\] |
You need to login to perform this action.
You will be redirected in
3 sec