This GRE quant practice question is a solid geometry problem solving question. A classic example of equating volume of water in a vessel getting displaced to the volume of the object immersed in the vessel. Volumes of cylinder and sphere.

Question 2 : A cylindrical vessel is filled with water up to some height. If a sphere of diameter 8 cm is dropped into the cylinder, the water level rises by half of the initial level. Instead, if a sphere of diameter 16 cm is dropped, the water level rises to a height h_{2}. What percentage of this new height h_{2} is the initial level of water?

- 33\\frac{1}{3})%
- 40%
- 20%
- 25%
- 16 \\frac{1}{6})%

@ INR 3000

Let the radius of the cylinder be 'r'.

Let the initial height up to which water is filled be 'h'.

Let the height up to which water is filled after the sphere of diameter 8 cm is dropped be h_{1}.

Height of water displaced after sphere 1 is dropped = h_{1} - h

The height of water in the vessel after sphere of diameter 16 cm is dropped = h_{2}

Height of water displaced after sphere 2 is dropped = h_{2} - h.

According to the question, the water level rises by 50% after sphere 1 is dropped into the cylinder.

Therefore, height of water displaced after sphere 1 is dropped, h_{1} - h = 0.5h

The water is stored in a cylindrical vessel. So, it will take the shape of the vessel that holds it.

The volume of water displaced = π r^{2} (height of water displaced) = π r^{2} (0.5h)

The volume of water displaced is the same as the volume of sphere 1 = \\frac{4}{3}) × π × (4)^{3}

Equating the two volumes we get π r^{2} (0.5h) = \\frac{4}{3}) × π × (4)^{3}**...... (1)**

The height of water in the vessel after sphere of diameter 16 cm is dropped = h_{2}

Height of water displaced after sphere 2 is dropped = h_{2} - h.

Volume of water displaced after sphere 2 is dropped = π r^{2} (height of water displaced)

Volume of water displaced = π r^{2} (h2 - h)

The volume of water displaced is the same as the volume of sphere 2 = \\frac{4}{3}) × π × 8^{3}

Equating the two volumes we get π r^{2} (h2 - h) = \\frac{4}{3}) × π × 8^{3}**...... (2)**

Divide equation **(2)** by equation **(1)**: \\frac{πr^2 (h_2−h)}{πr^2(0.5h)}) = \\frac{\frac{4}{3} π8^3}{\frac{4}{3} π4^3})

\\frac{(h_2−h)}{(0.5h)}) = 8 → (h2 - h) = 4h

h_{2} = 5h

Or h = \\frac{1}{5}) h_{2} → i.e., **h = 20% of h _{2}**

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