Openai/6919f112-be58-8003-b6d5-fc3458a0d806

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gpt-5-t-mini

Assistant: Tank Y the inverted cone will empty first.

Tank Y (the inverted cone) will empty first.

Quick intuition: the outflow speed depends only on the water height (Torricelli: v=2ghv=\sqrt{2gh}v=2gh​), so the rate the water level falls depends on how much cross-sectional area the tank has at each height. The cone has much smaller area near the bottom, so for the same outflow the water level must drop faster — overall it empties quicker.

Short derivation (clean):

Let outlet area be aaa, local tank cross-section A(h)A(h)A(h) at height hhh. Volume flow Q=a2ghQ=a\sqrt{2gh}Q=a2gh​. So

dVdt=−Q=−a2gh,dV=A(h) dh\frac{dV}{dt} = -Q = -a\sqrt{2gh},\qquad dV = A(h)\,dhdtdV​=−Q=−a2gh​,dV=A(h)dh so total emptying time

T=∫0HA(h)a2gh dh.T=\int_0^H \frac{A(h)}{a\sqrt{2gh}}\,dh.T=∫0H​a2gh​A(h)​dh. For a cylinder Acyl=A0=V/HA_{\text{cyl}}=A_0=V/HAcyl​=A0​=V/H, so

Tcyl=2Va2gH.T_{\text{cyl}}=\frac{2V}{a\sqrt{2gH}}.Tcyl​=a2gH​2V​. For a cone (vertex at bottom) A(h)=kh2A(h)=k h^2A(h)=kh2 with k=3V/H3k=3V/H^3k=3V/H3, so

Tcone=2kH5/25a2g=6V5a2gH.T_{\text{cone}}=\frac{2kH^{5/2}}{5a\sqrt{2g}}=\frac{6V}{5a\sqrt{2gH}}.Tcone​=5a2g​2kH5/2​=5a2gH​6V​. Therefore

TconeTcyl=6/52=35=0.6,\frac{T_{\text{cone}}}{T_{\text{cyl}}}=\frac{6/5}{2}=\frac{3}{5}=0.6,Tcyl​Tcone​​=26/5​=53​=0.6, so the cone (Tank Y) empties in 60%60\%60% of the time the straight tank (Tank X) does.