Answer:
[tex]\frac{dh}{dt} = 0.05 m/min[/tex]
Explanation:
As we know that the dimensions of the base of the tank is given as
[tex]L = 2 m[/tex]
[tex]W = 3 m[/tex]
now the base area is given as
[tex]A = 2m \times 3 m[/tex]
[tex]A = 6 m^2[/tex]
now we know that volume of the liquid filled in the tank is given as
[tex]V = Area \times height[/tex]
[tex]V = 6 \times h[/tex]
now we will differentiate it with respect to time both sides
[tex]\frac{dV}{dt} = 6 \times \frac{dh}{dt}[/tex]
here we know that
[tex]\frac{dV}{dt} = 0.3 m^3/min[/tex]
now we have
[tex]0.3 m^3/min = (6 m^2) \frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = 0.05 m/min[/tex]
