Let M and m grams be the mass of boat + rock in it. By law of floatation(M+m) grams of water will be displaced. This has a volume of (M+m)cc. That means the level of water in pool has risen up by (M+m)cc with the boat + rock when it is floating. If the rocks in the boat are thrown not into the pool but outside, the weight left is only M grams the weight of the boat. This will have displaced M cc of water only i.e less water than in the case of loaded boat. But if the rocks are put in the water in the pool and not outside, then the water in the pool will rise not only by M cc on account of empty boat but also by a volume of the stone put into it. This volume will be md where m is the mass of rock and d density of rock.<\/p>\n
Now md is < m. Thus the total rise will be (M+md)which is < M + m.<\/p>\n","protected":false},"excerpt":{"rendered":"
Let M and m grams be the mass of boat + rock in it. By law of floatation(M+m) grams of water will be displaced. This has a volume of (M+m)cc. That means the level of water in pool has risen up by (M+m)cc with the boat + rock when it is floating. If the rocks […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[372],"tags":[],"class_list":["post-32019","post","type-post","status-publish","format-standard","hentry","category-physics-general-science"],"acf":[],"yoast_head":"\n