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An object floats in water such that 14th of its volume is above the surface of water.\displaystyle{W}{h}{e}{n}{t}{h}{e}{s}{a}{m}{e}{o}{b}{j}{e}{c}{t}{i}{s}{m}{a}{d}{e}\to{f}{l}{o}{a}{t}\in{s}{a}{m}{p}\le{o}{f}{i}{m}{p}{u}{r}{e}{a}{l}{c}{o}{h}{o}{l},{t}{h}{e}{v}{o}{l}{u}{m}{e}{t}\hat{{r}}{e}{m}{a}\in{s}{o}{u}{t}{s}{i}{d}{e}{t}{h}{e}{l}{i}{q}{u}{i}{d}{i}{s}{110}{o}{f}\to{t}{a}{l}{v}{o}{l}{u}{m}{e}.{T}{h}{e}{r}{e}{l}{a}{t}{i}{v}{e}{d}{e}{n}{s}{i}{t}{y}{o}{f}{t}{h}{e}{s}{a}{m}{p}\le{o}{f}{a}{l}{c}{o}{h}{o}{l}{w}{i}{l}{l}{b}{e}
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According to Archimedes's principle, buoyant force, FB=Volume of the object immersed in liquid × density of liquid ×acceleration due to gravity Let V be the volume of object. For water: FB=3V4×ρw×g Or FB=3V4×ρw×10=152Vρw - (i) For alcohol: FB=9V10×ρa×g=9V10×ρa×10=9Vρa - (ii) From eqn. (i) and (ii), 152Vρw=9Vρa∴ρaρw=152×9=1518=0.84
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Question Text | An object floats in water such that 14th of its volume is above the surface of water.\displaystyle{W}{h}{e}{n}{t}{h}{e}{s}{a}{m}{e}{o}{b}{j}{e}{c}{t}{i}{s}{m}{a}{d}{e}\to{f}{l}{o}{a}{t}\in{s}{a}{m}{p}\le{o}{f}{i}{m}{p}{u}{r}{e}{a}{l}{c}{o}{h}{o}{l},{t}{h}{e}{v}{o}{l}{u}{m}{e}{t}\hat{{r}}{e}{m}{a}\in{s}{o}{u}{t}{s}{i}{d}{e}{t}{h}{e}{l}{i}{q}{u}{i}{d}{i}{s}{110}{o}{f}\to{t}{a}{l}{v}{o}{l}{u}{m}{e}.{T}{h}{e}{r}{e}{l}{a}{t}{i}{v}{e}{d}{e}{n}{s}{i}{t}{y}{o}{f}{t}{h}{e}{s}{a}{m}{p}\le{o}{f}{a}{l}{c}{o}{h}{o}{l}{w}{i}{l}{l}{b}{e} |
Answer Type | Text solution:1 |
Upvotes | 150 |