The price of a machine is depreciated at the rate of 10% yearly and the ultimate scrap value was Rs. 6561. Find the effective life of the macine. The price of the machine is Rs. 10,000.

Answer

To find the effective life of the machine, we will use the formula for depreciation. The price of the machine is depreciated at a rate of 10% yearly, and the ultimate scrap value is Rs. 6561. The initial price of the machine is Rs. 10,000.

### Step-by-Step Solution:

1. **Understand the Depreciation Formula**:
The formula for depreciation can be derived from the compound interest formula:
\[ \text{Final Price} = \text{Initial Price} \times \left(1 - \frac{\text{Rate}}{100}\right)^{\text{Time}} \]
Here, the rate of depreciation is 10%, so we will use \(1 - \frac{10}{100} = 0.9\).

2. **Identify the Given Values**:
- Initial Price (P) = Rs. 10,000
- Final Price (S) = Rs. 6,561
- Rate of Depreciation = 10% (which we will use as 0.9 in the formula)

3. **Set Up the Equation**:
Plugging the values into the formula:
\[ 6561 = 10000 \times (0.9)^t \]

4. **Rearrange the Equation**:
To isolate \((0.9)^t\), divide both sides by 10,000:
\[ \frac{6561}{10000} = (0.9)^t \]
This simplifies to:
\[ 0.6561 = (0.9)^t \]

5. **Express 6561 and 10000 in Powers**:
Notice that:
\[ 6561 = 9^4 \quad \text{and} \quad 10000 = 10^4 \]
Therefore, we can rewrite the equation as:
\[ \frac{9^4}{10^4} = (0.9)^t \]
Which can be expressed as:
\[ \left(\frac{9}{10}\right)^4 = (0.9)^t \]

6. **Equate the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ 4 = t \]

7. **Conclusion**:
The effective life of the machine is \(t = 4\) years.

### Final Answer:
The effective life of the machine is **4 years**.