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A spying plane flying over the sea has to make emergency landing due to some technical problem. The operator saw a huge ice block floating in the sea water is 15//17. If this plane weighs 360 kgwt and lands on this block, it is found that the ice block. |
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Answer» Solution :Given the fraction of the ice block inside the SEA water `=(15)/(17)` `:.` ratio of the density of the ice block `(P_(i))` to the density of the sea water `(p_(w))=(15)/(17)` Let the density of ice block be `p_(i)=15X` Then density of the sea water `p_(w)=17X` Mass of the block kept on the ice block = 360 kg Let 'M' be the mass of the ice block. `:.M+360` = mass of water displaced = (volume of water displaced) `p_(w)` But the volume of water displaced = volume of the ice block = `(M)/(15x)` `:.M+360=((M)/(15x))(17x)=(17M)/(15)` `:.(17M)/(15)-M=360` `rArrM=(360xx15)/(2)=2700kg` `:.` mass of the ice block 2700 kg |
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