A) 10m
B) 7.1m
C) 5 m
D) 3.2 m
Correct Answer: D
Solution :
Let the velocity at point B is \[{{E}_{eq}}=12V\]. From conservation of total mechanical energy, \[{{V}_{eq}}=\frac{2\times 2}{2+2}=1\Omega \] \[i=\frac{{{E}_{eq}}}{R+{{R}_{eq}}}=\frac{12}{5+1}=\frac{12}{6}=1\Omega \] \[N\sin \theta =m{{\omega }^{2}}r\sin \theta \] \[\Rightarrow \] \[N=m{{\omega }^{2}}r\sin \theta \] \[N\cos \theta =mg\] Now, let maximum height attained by water stream, be (H). \[\frac{1}{\cos \theta }=\frac{{{\omega }^{2}}r}{g}\] \[\omega =\sqrt{\frac{g}{r\,\,\cos \theta }}\] \[\phi =B.A\] \[\phi =|B||A|\cos \theta \] \[\varepsilon =-\frac{d\phi }{dt}=\frac{d[|B||A|\cos \theta ]}{dt}\]
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