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Question

A rectangular block is floating in a liquid. The distance of the metacentre from the point of buoyance is equal to the ratio of

The correct answer is

Moment of inertia of plan to the volume of liquid displaced

Understanding the stability of a floating object, like a rectangular block in a liquid, involves several key concepts from fluid mechanics. One important aspect is the relationship between the metacentre and the point of buoyancy. The question asks about the specific ratio that defines the distance of the metacentre from the point of buoyance.

Floating Block Stability Concepts

To address this question, let's first clarify some fundamental terms related to a floating body:

  • Point of Buoyancy (B): This is the center of gravity of the volume of liquid displaced by the floating block. The buoyant force acts vertically upwards through this point.
  • Metacentre (M): When a floating body tilts or heels through a small angle, the center of buoyancy shifts to a new position (B'). The vertical line passing through this new center of buoyancy (B') intersects the original vertical line passing through the initial center of buoyancy (B) at a point known as the metacentre (M). The position of the metacentre is crucial for determining the stability of the floating object.

Distance of Metacentre from Point of Buoyancy (BM)

The distance between the metacentre (M) and the point of buoyancy (B) is a critical parameter in naval architecture and fluid mechanics. This distance, often denoted as \(\text{BM}\), is determined by the moment of inertia of the waterplane area and the volume of liquid displaced by the floating block. The relationship is given by the formula:

\(\text{BM} = \frac{I}{V_d}\)

Where:

  • \(\text{BM}\) represents the distance of the metacentre from the point of buoyancy.
  • \(I\) represents the moment of inertia of the area of the waterplane (or 'plan') about its longitudinal axis. The waterplane is the area of the floating object's cross-section at the liquid surface.
  • \(V_d\) represents the volume of liquid displaced by the floating block. This is the volume of the submerged part of the block.

Analyzing the Options

Let's evaluate the given options based on the formula \(\text{BM} = \frac{I}{V_d}\):

  • Option 1: Area of plan to the volume of liquid displaced
    This option is incorrect because the formula involves the moment of inertia of the plan, not just its area.
  • Option 2: volume of liquid displaced to Moment of inertia of plan
    This option represents the inverse of the correct ratio (\(\frac{V_d}{I}\)), so it is incorrect.
  • Option 3: Cross-section of the block to the volume of liquid displaced
    This option is incorrect as it refers to the 'cross-section' of the block, which is not necessarily the waterplane area, and again, it mentions area instead of moment of inertia.
  • Option 4: Moment of inertia of plan to the volume of liquid displaced
    This option perfectly matches the derived formula \(\text{BM} = \frac{I}{V_d}\). It correctly states that the distance is the ratio of the moment of inertia of the waterplane (plan) to the volume of liquid displaced.

Therefore, the distance of the metacentre from the point of buoyance is equal to the ratio of the moment of inertia of the plan to the volume of liquid displaced.

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Important Questions from Buoyancy and Floatation

  1. If two objects are weighed in water and both of them lose the same weight, then the two objects must have identical

  2. A wooden cube of side 0.2 m is floating in the water. The density of wood is 600 kg/m3. Then the volume of water displaced by the wooden block is

  3. The stability of a floating body is governed mainly by the ______.

  4. Buoyant force for a floating body passes through:

  5. A can has a total volume of 1200 cm3 and a mass of 200 g. How many grams of lead shots of density 11.4 g/cm3 could it carry without sinking in water? (density of water : 1 g/cm3)

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