A) 4
B) 2
C) 1
D) ½
Correct Answer: C
Solution :
\[{{F}_{bottom}}=\rho g\,\,LBh\] \[{{F}_{side}}=\frac{1}{2}Ag\,L{{h}^{2}}\] \[\frac{{{F}_{bottom}}}{{{F}_{side}}}=\frac{2B}{h}=\frac{2B}{2B}=1\]You need to login to perform this action.
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