Home
Class 12
PHYSICS
An iceberg is floating in water. The den...

An iceberg is floating in water. The density of ice in the iceberg is 917 kg `m^(-3)` and the density of water is 1024 kg `m^(-3)`. What percentage fraction of the iceberg would be visible?

Text Solution

AI Generated Solution

The correct Answer is:
To determine the percentage fraction of the iceberg that is visible above the water, we can use the principle of buoyancy. Here’s a step-by-step solution: ### Step 1: Understand the buoyancy principle The buoyant force acting on the iceberg must equal the weight of the iceberg for it to float. The buoyant force is equal to the weight of the water displaced by the submerged part of the iceberg. ### Step 2: Write down the equations Let: - \( V \) = total volume of the iceberg - \( V_s \) = volume of the submerged part of the iceberg - \( \rho_{ice} \) = density of ice = 917 kg/m³ - \( \rho_{water} \) = density of water = 1024 kg/m³ - \( g \) = acceleration due to gravity (which will cancel out) The weight of the iceberg can be expressed as: \[ W_{ice} = V \cdot \rho_{ice} \cdot g \] The buoyant force can be expressed as: \[ F_b = V_s \cdot \rho_{water} \cdot g \] ### Step 3: Set the buoyant force equal to the weight of the iceberg Since the iceberg is floating, we have: \[ V_s \cdot \rho_{water} \cdot g = V \cdot \rho_{ice} \cdot g \] ### Step 4: Cancel out \( g \) Since \( g \) appears on both sides, we can cancel it: \[ V_s \cdot \rho_{water} = V \cdot \rho_{ice} \] ### Step 5: Express the volume of the submerged part From the equation above, we can express the volume of the submerged part: \[ V_s = V \cdot \frac{\rho_{ice}}{\rho_{water}} \] ### Step 6: Substitute the known densities Substituting the known values: \[ V_s = V \cdot \frac{917}{1024} \] ### Step 7: Calculate the fraction of the iceberg that is submerged The fraction of the iceberg that is submerged is: \[ \text{Fraction submerged} = \frac{V_s}{V} = \frac{917}{1024} \] ### Step 8: Calculate the fraction that is visible The fraction of the iceberg that is visible above the water is: \[ \text{Fraction visible} = 1 - \text{Fraction submerged} = 1 - \frac{917}{1024} \] Calculating this gives: \[ \text{Fraction visible} = \frac{1024 - 917}{1024} = \frac{107}{1024} \] ### Step 9: Convert to percentage To find the percentage of the iceberg that is visible: \[ \text{Percentage visible} = \left( \frac{107}{1024} \right) \times 100 \approx 10.43\% \] ### Step 10: Round to the nearest whole number Rounding this to the nearest whole number gives approximately 10%. ### Final Answer Thus, the percentage fraction of the iceberg that is visible is approximately **10%**. ---

To determine the percentage fraction of the iceberg that is visible above the water, we can use the principle of buoyancy. Here’s a step-by-step solution: ### Step 1: Understand the buoyancy principle The buoyant force acting on the iceberg must equal the weight of the iceberg for it to float. The buoyant force is equal to the weight of the water displaced by the submerged part of the iceberg. ### Step 2: Write down the equations Let: - \( V \) = total volume of the iceberg ...
Promotional Banner

Topper's Solved these Questions

  • LIQUIDS

    VMC MODULES ENGLISH|Exercise LEVEL -2|50 Videos
  • LIQUIDS

    VMC MODULES ENGLISH|Exercise JEE MAIN (LEVEL - 1 )|31 Videos
  • LIQUIDS

    VMC MODULES ENGLISH|Exercise LEVEL-0 ( Short Answer Type - II )|15 Videos
  • LAWS OF MOTION

    VMC MODULES ENGLISH|Exercise IMPECCABLE|53 Videos
  • MAGNETIC EFFECTS OF CURRENT

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|78 Videos

Similar Questions

Explore conceptually related problems

The density of water is ........ "kg/m"^3 .

An iceberg is floating partially immersed in sea water. The density of sea water is 1.03 g cm^(-3) and that of ice is 0.92 g cm^(-3) . The approximate percentage of total volume of iceberg above the level of sea water is

The mass of 1 litre of water is 1 kg. Find the density of water in kg m^(-3)

The order of magnitude of the density of nuclear matter is 10^(4) kg m^(-3)

The density of iron is 7.8xx 10^3 "kg m"^(-3) What is its relative density?

The density of copper is 8.9 g cm^(-3) . What will be its density in kg m^(-3)

The density of sand is 1500 "kg/m"^3 . What is the mass of 1m^3 of sand ?

The density of ice x cm^(-3) and that of water is y gcm^(-3) . What is the change in volume when mg of ice melts?

The density of sand is 1500 "kg/m"^3 . What is the mass of 10m^3 of sand ?

An iceberg of density 900kg//m^(3) is floating in water of density 1000 kg//m^(3) . The percentage of volume of ice cube outside the water is

VMC MODULES ENGLISH-LIQUIDS-LEVEL -1
  1. A container is partially filled with a liquid of density rho2 A capill...

    Text Solution

    |

  2. A drop of liquid of density rho is floating half-immersed in a liquid ...

    Text Solution

    |

  3. A large open top container of negligible mass and uniform cross sectio...

    Text Solution

    |

  4. A liquid flows through two capillary tubes A and B connected in series...

    Text Solution

    |

  5. A uniform tube is shown in figure, Which is open at one en and closed ...

    Text Solution

    |

  6. If a 5.0 cm long capillary tube with 0.10 mm internal diameter open at...

    Text Solution

    |

  7. A cubical block (of sie 2m) of mass 20 kg slides on inclined plane lub...

    Text Solution

    |

  8. A drop of water of radius 0.0015 mm is falling in air. If the coeffici...

    Text Solution

    |

  9. A metallic sphere of radius 1.0 xx 10^(-3) m and density 1.0 xx 10^(4)...

    Text Solution

    |

  10. The cylindrical tube of a spray pump has a cross-section of 8 cm^2, on...

    Text Solution

    |

  11. An iceberg is floating in water. The density of ice in the iceberg is ...

    Text Solution

    |

  12. A mercury drop of radius 1.0 cm is sprayed into 10^(6) droplets of equ...

    Text Solution

    |

  13. The excess pressure inside a spherical drop of water is four times tha...

    Text Solution

    |

  14. Work done in forming a liquid drop of radius R is W(1) and that of rad...

    Text Solution

    |

  15. The surface tension of a liquid is 5 Newton per metre. If a film is he...

    Text Solution

    |

  16. Two liquids of densities 2 rho " and " rho having their volumes in the...

    Text Solution

    |

  17. A container (see figure) contains a liquid upto a height h. The densit...

    Text Solution

    |

  18. A film of soap solution is formed on a loop frame loop of 6.28 cm long...

    Text Solution

    |

  19. Eight spherical rain drops of the same mass and radius are falling dow...

    Text Solution

    |

  20. A solid uniform ball of volume V floats on the interface of two immisc...

    Text Solution

    |