We can make use of the complying with layout to assist in the resolution:

The deepness of the water in the cone is stood for by \$latex x\$. After that, the distance \$latex r\$ of the cross-section of the water is offered by:

\$\$ tan( 30 ^ {circ} )= frac {r} {x} \$\$

\$\$ r= xtan( 30 ^ {circ} )\$\$

\$\$= frac {x} {sqrt {3}} \$\$

The quantity of water in the cone is offered by:

\$\$ V= frac {1} {3} pi r ^ 2x\$\$

\$\$= frac {1} {3} pi left( frac {x} {sqrt {3}} right) ^ 2x\$\$

\$\$= frac {1} {9} pi x ^ 3\$\$

Then, the by-product of the quantity relative to \$latex x\$ is:

\$\$ dfrac {dV} {dx} pi x ^ 2\$\$

From the concern, we understand that \$latex dfrac {dV} {dt} =5\$. Additionally, by the chain guideline, we have:

\$\$ dfrac {dV} {dt} =dfrac {dV} {dx} dfrac {dx} {dt} \$\$

Then,

\$\$ 5= frac {1} {3} pi x ^ 2 dfrac {dx} {dt} \$\$

\$\$ dfrac {dx} {dt} = frac {15} {pi x ^ 2} \$\$

When \$latex x= 10\$, we have:

\$\$ dfrac {dx} {dt} =frac {15} {pi (10 )^ 2} =frac {3} {20pi} \$\$

The price of adjustment of deepness when the deepness is 10 centimeters is \$latex frac {3} {20pi} \$ cm/s.