Home
Class 10
MATHS
The diameter of a hollow cylindrical ves...

The diameter of a hollow cylindrical vessel is 14 cm .There is some water in it. When a cubical iron piece is fully immeresed in it , the water surface rises by `8(9)/(14)`cm .Find the edge of the cube.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Understand the given information We have a hollow cylindrical vessel with a diameter of 14 cm. When a cubical iron piece is fully immersed in it, the water level rises by \(8 \frac{9}{14}\) cm. We need to find the edge length of the cube. ### Step 2: Calculate the radius of the cylinder The radius \( r \) of the cylinder can be calculated from the diameter: \[ r = \frac{\text{diameter}}{2} = \frac{14 \text{ cm}}{2} = 7 \text{ cm} \] ### Step 3: Convert the rise in water level to an improper fraction The rise in water level is given as \(8 \frac{9}{14}\) cm. We convert this mixed fraction to an improper fraction: \[ 8 \frac{9}{14} = \frac{8 \times 14 + 9}{14} = \frac{112 + 9}{14} = \frac{121}{14} \text{ cm} \] ### Step 4: Calculate the volume of water displaced The volume of water displaced when the cube is immersed is equal to the volume of the cylinder that corresponds to the rise in water level. The volume \( V \) of the cylindrical section can be calculated using the formula: \[ V = \pi r^2 h \] Substituting the values: \[ V = \pi \times (7 \text{ cm})^2 \times \frac{121}{14} \text{ cm} \] \[ V = \pi \times 49 \text{ cm}^2 \times \frac{121}{14} \text{ cm} \] ### Step 5: Substitute the value of \(\pi\) Using \(\pi \approx \frac{22}{7}\): \[ V = \frac{22}{7} \times 49 \times \frac{121}{14} \] ### Step 6: Simplify the expression First, simplify \( \frac{22 \times 49 \times 121}{7 \times 14} \): \[ = \frac{22 \times 49 \times 121}{98} \] Calculating \( 22 \times 49 = 1078 \): \[ = \frac{1078 \times 121}{98} \] Now, simplifying \( \frac{1078 \times 121}{98} \): \[ = \frac{130678}{98} \] ### Step 7: Calculate the volume of the cube The volume of the cube is given by \( a^3 \), where \( a \) is the edge length of the cube. Thus, we have: \[ a^3 = \frac{130678}{98} \] ### Step 8: Find the edge length of the cube To find \( a \), we need to take the cube root of the volume: \[ a = \sqrt[3]{\frac{130678}{98}} \] Calculating this gives: \[ a = 11 \text{ cm} \] ### Conclusion The edge of the cube is \( 11 \text{ cm} \). ---

To solve the problem, we will follow these steps: ### Step 1: Understand the given information We have a hollow cylindrical vessel with a diameter of 14 cm. When a cubical iron piece is fully immersed in it, the water level rises by \(8 \frac{9}{14}\) cm. We need to find the edge length of the cube. ### Step 2: Calculate the radius of the cylinder The radius \( r \) of the cylinder can be calculated from the diameter: \[ ...
Promotional Banner

Topper's Solved these Questions

  • VOLUME AND SURFACE AREA OF SOLIDS

    NAGEEN PRAKASHAN ENGLISH|Exercise Problems From NCERT/exemplar|10 Videos
  • VOLUME AND SURFACE AREA OF SOLIDS

    NAGEEN PRAKASHAN ENGLISH|Exercise Exercise|68 Videos
  • TRIANGLES

    NAGEEN PRAKASHAN ENGLISH|Exercise Revision Exercise Long Questions|1 Videos

Similar Questions

Explore conceptually related problems

A solid metallic ball is dropped in a cylindrical vessel which has some waater .when this ball is fully immersed in water the water surface rises by 2 cm find the volume of ball if the radius of base of cylindrical vessel is 7cm.

There is some water in a cylindrical vessel of radus 6 cm a sphere of radus 2 cm is dropped in to the vessel .Find the increase in the height of water surface.

150 spherical marbles, each of diameter 1.4 cm are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.

There is some water in a cylindrical vessel of diameter 12 cm. A solid metallic spere of redius 4 cm is dropped in to it . Find the increase in height of the water surface if sphere is fully immersed in to the water.

A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 5/9 cm . Find the diameter of the cylindrical vessel.

A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 5/9 cm . Find the diameter of the cylindrical vessel.

Lead spheres of diameter 6 cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18 cm and water rises by 40 cm. Find the number of lead spheres dropped in the water.

When a metal cube is completely submerged in water contained in a cylindrical vessel with diameter 30 cm, the level of water rises by 1 (41)/(99) cm. Find : (i) the length of edge of the cube. (ii) the total surface area of the cube.

A cylindrical tub of radius 16cm contains water to a depth of 30cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9cm. What is the radius of the ball?

The internal and external diameters of a hollow hemispherical vessel are 42 cm and 45 cm respectively. Find its capacity and also its outer curved surface area.