Buoyancy and Floatation Questions and Answers

Answer: D

The statement that the upward buoyant force on a body immersed in a fluid equals the weight of the fluid displaced is Archimedes' principle. Newton's third law is about action and reaction, Bernoulli's principle is about pressure and speed in flow, and Pascal's principle is about transmission of pressure in a confined fluid, so only Archimedes' principle fits the statement.

The same principle gives the condition for floating: a body floats when the weight of fluid it displaces equals its own weight, so the buoyant force balances its weight.

Answer: B

The specific gravity is 0.6.

A floating body displaces its own weight of water. With the 10 m length horizontal, the block's cross-section is 2 m by 2 m and it sinks 1.2 m of its 2 m height, so the submerged fraction of its volume is 1.2/2:

\[ \begin{aligned} SG &= \frac{\rho_{\text{block}}}{\rho_{\text{water}}} = \frac{V_{\text{sub}}}{V} \\ &= \frac{1.2}{2} = 0.6 \end{aligned} \]

0.8 and 0.4 do not correspond to the given depth, and 0.5 would put the waterline at exactly mid-height, 1.0 m. The length of 10 m cancels and does not affect the answer.

Answer: D

The centre of gravity must be below the center of buoyancy.

For a fully immersed body the centre of buoyancy, the centroid of the displaced volume, stays fixed relative to the body as it tilts, so the metacentre coincides with it. A small tilt then creates a restoring couple only if the weight acts below the buoyant force, that is if G lies directly below B.

With G above B the couple overturns the body. The metacentre conditions (G below or above M) apply to floating bodies, whose waterplane area lets B shift as they heel.

Answer: A

\(\overline{MG}\) is the metacentric height and \(G\) should lie below \(M\).

When a floating body tilts slightly, the centre of buoyancy shifts and the buoyant force acts along a line that crosses the body's centre line at the metacentre M. The metacentric height is the distance GM from the centre of gravity to the metacentre.

If M is above G, the weight and buoyancy form a couple that rotates the body back upright: stable equilibrium. If M were below G, the couple would increase the tilt.

MB is the metacentric radius (\(I/V\)), not the metacentric height. Whether B lies above or below M does not decide stability; for most floating bodies B is below G and still stable, because what matters is the position of G relative to M.

Answer: B

The centre of buoyancy is 0.62 m above the base.

A floating body displaces its own weight of liquid. With the box 2 m deep, the depth immersed \(h\) satisfies:

\[ \begin{aligned} 0.64 \times 2 &= 1.025 \times h \\ h &= \frac{1.28}{1.025} = 1.249\ \text{m} \end{aligned} \]

The centre of buoyancy is the centroid of the displaced volume, which for a rectangular section is half the immersed depth: \(1.249/2 = 0.62\) m above the base. 1.24 m is the immersed depth itself, not its centroid, and 0.25 m and 0.84 m match no step of the working. The width and length cancel out because both sides of the balance use the same plan area.

Answer: A

The buoyant force passes through the centroid of the displaced volume.

The buoyant force equals the weight of fluid displaced and acts through the centre of gravity of that displaced fluid, which for a uniform fluid is the centroid of the displaced volume. This point is the centre of buoyancy.

The body's weight acts through its centre of gravity, a different point, and the stability of a floating body depends on how the two are placed. The metacentre is the point where the line of buoyancy cuts the body's axis after a small tilt. The centroid of the whole body's volume coincides with the centre of buoyancy only when the body is fully submerged.