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Question

(C) The fraction of the thickness of the ice block that remains above the water surface is given by $x = 1 - (ho_{	ext{ice}} / ho_{	ext{water}})

Vedclass · Accepted Answer

$0.9$

Vedclass · Answer

(C) The fraction of the thickness of the ice block that remains above the water surface is given by $x = 1 - (ho_{	ext{ice}} / ho_{	ext{water}})$.Taking the density of ice $ho_{	ext{ice}} \approx 0.9 \, 	ext{g/cm}^3$ and density of water $ho_{	ext{water}} = 1 \, 	ext{g/cm}^3$,we get $x = 1 - 0.9 = 0.1$.The thickness of the ice block is $H = 8 \, m$.The height of the ice block above the water level is $h = H 	imes x = 8 	imes 0.1 = 0.8 \, m$.Since the person is standing on the ice,the distance to the water surface is the height of the ice above the water.Therefore,the minimum length of rope required is approximately $0.8 \, m$.Comparing this with the given options,the nearest value is $0.9 \, m$.

$A$ very large block of ice of the size of a volleyball court and of uniform thickness of $8 \,m$ is floating on water. $A$ person standing near its edge wishes to fetch a bucketful of water using a rope. The smallest length of rope required for this is about ............... $m$.

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