A) 3.2 N
B) 1.6 N
C) 12.5 N
D) 20 N
Correct Answer: A
Solution :
Total mass of liquid, \[dm=\rho \,A\,dy\] \[m=\int\limits_{0}^{2}{(10-2y)(0.02)dy}\] \[=(0.02)[10y-{{y}^{2}}]_{0}^{2}=0.02[20-4]=0.32\,kg\] Force on the bottom of container = mg \[=0.32\times 10=3.2\,N\]You need to login to perform this action.
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