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The three vessels shown in figure below have same base area.Equal volume of a liquid are passed in the three vessels.The force on the base will be

$\begin{array}{1 1}(A)\;\text{Maximum in vessel A}\\(B)\;\text{Maximum in vessel B}\\(C)\;\text{Maximum in vessel C}\\(D)\;\text{Equal in all vessels}\end{array} $

1 Answer

$\therefore F_C=\omega+N_2$
$F_A=\omega-N_2$
$F_B=\omega$
$F_A < F_B < F_C$
Hence (C) is the correct answer.
answered May 9, 2014 by sreemathi.v
 

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