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 :
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.
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