Home
Class 12
PHYSICS
A unit of area, often used in measuring ...

A unit of area, often used in measuring land areas, is the hectare defined as `10^4 m^2` An open-pit coal mine consumes 75 hectares of land, down to a depth of 26 m, each year. What volume of earth, in a cubic kilometre, is removed in this time?

A

0.01

B

0.02

C

0.03

D

0.04

Text Solution

AI Generated Solution

The correct Answer is:
To find the volume of earth removed by the open-pit coal mine in cubic kilometers, we can follow these steps: ### Step 1: Convert hectares to square meters 1 hectare is defined as \(10^4 \, \text{m}^2\). Therefore, for 75 hectares: \[ \text{Area} = 75 \, \text{hectares} \times 10^4 \, \text{m}^2/\text{hectare} = 75 \times 10^4 \, \text{m}^2 \] ### Step 2: Calculate the volume in cubic meters The volume \(V\) can be calculated using the formula: \[ V = \text{Area} \times \text{Depth} \] Given that the depth is 26 m: \[ V = (75 \times 10^4 \, \text{m}^2) \times (26 \, \text{m}) = 1950 \times 10^4 \, \text{m}^3 \] ### Step 3: Convert cubic meters to cubic kilometers To convert cubic meters to cubic kilometers, we use the conversion factor: \[ 1 \, \text{km}^3 = 10^9 \, \text{m}^3 \] Thus, to convert \(1950 \times 10^4 \, \text{m}^3\) to cubic kilometers: \[ V = \frac{1950 \times 10^4 \, \text{m}^3}{10^9 \, \text{m}^3/\text{km}^3} = \frac{1950 \times 10^4}{10^9} \, \text{km}^3 = 1950 \times 10^{-5} \, \text{km}^3 = 0.195 \, \text{km}^3 \] ### Final Answer The volume of earth removed each year is approximately: \[ 0.195 \, \text{km}^3 \]
Promotional Banner

Similar Questions

Explore conceptually related problems

In a tyre of "Ferrari" car of Mr. Obama, a tube having a volume of 12.3 litres is filled with air at a pressure of 4 atm at 300 K.Due to travelling, the temperature of the tube and air inside it raised to 360 K and pressure reduced to 3.6 atm in 20 minutes. If the porosity (number of pores per unit area) of the tube material is 5xx10^3 pores/ cm^2 and each pore can transfer air from inside to outside of the tube with the rate of 6.023xx10^8 molecules per minute.Calculate the total surface area (m^2) of the tube.(R=0.082 Lt-atm/mole-K) Give your answer divide by 100.