Answer:
[tex]2\ cm[/tex]
Step-by-step explanation:
[tex]We\ are\ given\ that,\\Radius\ of\ the\ conical\ vessel=5\ cm\\Height\ of\ the\ conical\ vessel=24\ cm\\Hence,\\As\ we\ know\ that,\\Volume\ of\ a\ cone=\frac{1}{3}\pi r^2h\\Hence,\\The\ volume\ of\ the\ conical\ vessel=\frac{1}{3}\pi (5)^224= \frac{1}{3}\pi 25*24=200\pi \\Hence,\\The\ volume\ of\ the\ conical\ vessel=The\ amount\ of\ water\ filled\ in\ it\\Hence,\\The\ amount\ of\ water\ filled\ in\ the\ conical\ vessel=200\pi\\Lets\ consider\ the\ cylindrical\ vessel:\\Volume\ of\ a\ cylinder=\pi r^2h[/tex]
[tex]As\ we\ are\ not\ said\ that\ the\ cylindrical\ vessel\ was\ empty,\ let\ its\ initial\\ height\ be\ h_1\\Radius\ of\ the\ Cylindrical\ vessel=10\ cm\\Initial\ height\ of\ the\ cylindrical\ vessel=h_1\\Initial\ volume\ of\ the\ cylindrical\ vessel=\pi *10^2*h_1=100h_1\pi \\Initial\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi \\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=Final\ amount\ of\ water\\ in\ the\ cylindrical\ vessel +Amount\ of\ water\ in\ conical\ vessel[/tex]
[tex]Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi +200\pi \\Now,\\Let\ the\ final\ height\ be\ h_2\\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=\pi *10^2*h_2=100h_2\pi \\Hence,\\100h_2\pi =100h_1\pi +200\pi \\Hence,\\100h_2=100h_1+200\\100h_2-100h_1=200\\100(h_2-h_1)=200\\h_2-h_1=2\\Hence,\\As\ the\ rise\ in\ level\ is\ the\ difference\ between\ the\ initial\ and\ final\\ heights:\\The\ rise\ in\ level\ of\ water=2\ cm[/tex]
