question_answer

A container is filled with a liquid whose density varies as \[\rho =(10-2y)\,kg/{{m}^{3}}\] with height.   If the area of bottom surface of container is \[200\,\text{c}{{\text{m}}^{2}}\] and height of liquid is 2 m, then the force exerted by container on the liquid is (Take \[g=10\,m/{{s}^{2}},\] neglect atmospheric pressure)

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 \[=0.32\times 10=3.2\,N\]