A crane can lift up `10,000 kg` of coal in `1` hour form a mine of `180 m` depth. If the efficiency of the crane is `80 %`, its input power must be `(g = 10 ms^(-2))`.

Options

A. 5 kW
B. 6.25 kW
C. 50 kW
D. 62.5 kW

Explanation

To solve the problem step by step, we will calculate the input power required by the crane to lift the coal, taking into account its efficiency.

### Step 1: Calculate the Work Done (W)
The work done against gravity to lift the coal can be calculated using the formula:
\[ W = mgh \]
Where:
- \( m = 10,000 \, \text{kg} \) (mass of coal)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 180 \, \text{m} \) (height)

Substituting the values:
\[ W = 10,000 \, \text{kg} \times 10 \, \text{m/s}^2 \times 180 \, \text{m} \]
\[ W = 10,000 \times 10 \times 180 \]
\[ W = 18,000,000 \, \text{J} \, \text{(or 18 MJ)} \]

### Step 2: Calculate the Output Power (P_out)
The output power can be calculated using the formula:
\[ P_{\text{out}} = \frac{W}{T} \]
Where \( T \) is the time taken in seconds.
Given that the time is 1 hour:
\[ T = 1 \, \text{hour} = 3600 \, \text{seconds} \]

Substituting the values:
\[ P_{\text{out}} = \frac{18,000,000 \, \text{J}}{3600 \, \text{s}} \]
\[ P_{\text{out}} = 5000 \, \text{W} \, \text{(or 5 kW)} \]

### Step 3: Calculate the Input Power (P_in)
Given the efficiency (\( \eta \)) of the crane is 80%, we can relate the input power to the output power using the formula:
\[ \eta = \frac{P_{\text{out}}}{P_{\text{in}}} \]
Rearranging gives:
\[ P_{\text{in}} = \frac{P_{\text{out}}}{\eta} \]

Substituting the values:
\[ P_{\text{in}} = \frac{5000 \, \text{W}}{0.80} \]
\[ P_{\text{in}} = 6250 \, \text{W} \, \text{(or 6.25 kW)} \]

### Final Answer
The input power required by the crane must be **6.25 kW**.

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