Question

A very large block of ice of the size of a volleyball court and of uniform thickness of \[8\,{\text{m}}\] 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
(A) \[3.6\,{\text{m}}\]
(B) \[1.8\,{\text{m}}\]
(C) \[0.9\,{\text{m}}\]
(D) \[0.4\,{\text{m}}\]

Answer 1

Hint:Use the concept of Archimedes’ principle. Use the equation for the buoyant force exerted on the object in the liquid. Also use the equation for density of an object. As the ice block is floating on the water, the buoyant force on the block is equal to the weight of the block submerged in the water.

Formulae used:
The buoyant force \[F\] on an object floating on the liquid is
\[F = \rho Vg\] …… (1)
Here, \[\rho \] is the density of the liquid, \[V\] is the volume of the object and \[g\] is acceleration due to gravity.
The density \[\rho \] of an object is
\[\rho = \dfrac{M}{V}\] …… (2)
Here, \[M\] is the mass of the object and \[V\] is the volume of the object.

Complete step by step answer:
We have given that a large block of ice of mass \[{M_i}\] having size equal to the volleyball court with thickness \[8\,{\text{m}}\] is floating on the water.
\[{h_i} = 8\,{\text{m}}\]
Since the block of ice is floating on the water, the upward buoyant force on the ice block is equal to the weight of the ice block submerged in the water in the downward direction.
\[F = {M_i}g\] …… (3)
Rewrite equation (2) for the mass of the ice block.
\[{M_i} = {\rho _i}{V_i}\]
Here, \[{\rho _i}\] and \[{V_i}\] are the density and total volume of the ice block respectively.
The volume \[{V_i}\] of the ice block submerged in the water is equal to the volume \[{V_w}\] of the water displaced by the block.
Substitute \[{V_w}\] for \[{V_i}\] in the above equation.
\[{M_i} = {\rho _i}{V_w}\]
Substitute for \[F\] and \[{\rho _i}{V_w}\] for \[{M_i}\] in equation (3).
\[{\rho _w}{V_i}g = {\rho _i}{V_w}g\]
\[ \Rightarrow {\rho _w}{V_i} = {\rho _i}{V_w}\]
The volume of the ice block is given by multiplication of its length, breadth and thickness which is also true for the volume of the water displaced by the ice. Hence, the volume of the ice and the water displaced are proportional to their thickness.
\[V \propto h\]
Also the density of ice is 0.9 times the density of the water.
\[{\rho _i} = 0.9{\rho _w}\]
Hence, the above equation becomes
\[ \Rightarrow {\rho _w}{h_i} = 0.9{\rho _w}{h_w}\]
\[ \Rightarrow {h_i} = 0.9{h_w}\]
Substitute \[8\,{\text{m}}\] for \[{h_i}\] in the above equation.
\[ \Rightarrow {h_i} = 0.9\left( {8\,{\text{m}}} \right)\]
\[ \Rightarrow {h_i} = 7.2\,{\text{m}}\]
Hence, the height of the ice in the water is \[7.2\,{\text{m}}\].
The height \[h\] of the rope that should be used is the difference between the total thickness \[{h_w}\] of the ice block and the thickness \[{h_i}\] of the ice block submerged in water.
\[h = {h_w} - {h_i}\]
\[ \Rightarrow h = \left( {8\,{\text{m}}} \right) - \left( {7.2\,{\text{m}}} \right)\]
\[ \therefore h = 0.8\,{\text{m}}\]
Therefore, the minimum length of the rope should be \[0.8\,{\text{m}}\]. So, the length of the rope can be a little greater than \[0.8\,{\text{m}}\].

Hence, the correct option is C.

Note: One can also solve the same question without using Archimedes’ principle. The density of the ice is 0.9 times the density of the water. So, the thickness of the block submerged in the water is also 0.9 times the total thickness of the ice block. Finally, the minimum length of the rope required is equal to the difference between the total thickness and submerged thickness of the ice block.