A hydraulic automobile life has input and output pistons with diameters of 10 cm and 30 cm. The life is used to hold up a car with a weight of `1.44xx 10^(4) N` .
What is the force on the input piston ?

Options

A. `1.6xx10^(3)N`
B. `1.5xx10^(3)N`
C. `1.4 xx10^(3)N`
D. `1.8xx10^(3)N`

Answer

To solve the problem of finding the force on the input piston of a hydraulic lift, we can use the principle of hydraulic systems, which states that the pressure applied on one piston is transmitted equally to the other piston.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Diameter of input piston, \( d_1 = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Diameter of output piston, \( d_2 = 30 \, \text{cm} = 0.3 \, \text{m} \)
- Weight of the car (force on output piston), \( F_2 = 1.44 \times 10^4 \, \text{N} \)

2. **Calculate the Areas of the Pistons:**
The area \( A \) of a circle is given by the formula:
\[ A = \frac{\pi d^2}{4} \]
- Area of input piston, \( A_1 \):
\[ A_1 = \frac{\pi (d_1)^2}{4} = \frac{\pi (0.1)^2}{4} = \frac{\pi \times 0.01}{4} = \frac{\pi}{400} \, \text{m}^2 \]
- Area of output piston, \( A_2 \):
\[ A_2 = \frac{\pi (d_2)^2}{4} = \frac{\pi (0.3)^2}{4} = \frac{\pi \times 0.09}{4} = \frac{\pi}{44.44} \, \text{m}^2 \]

3. **Apply the Hydraulic Principle:**
According to the principle of hydraulics:
\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]
Rearranging this gives:
\[ F_1 = F_2 \times \frac{A_1}{A_2} \]

4. **Substitute the Areas:**
Since \( A_1 = \frac{\pi}{400} \) and \( A_2 = \frac{\pi}{44.44} \):
\[ F_1 = F_2 \times \frac{\frac{\pi}{400}}{\frac{\pi}{44.44}} = F_2 \times \frac{44.44}{400} \]
Now substitute \( F_2 = 1.44 \times 10^4 \, \text{N} \):
\[ F_1 = 1.44 \times 10^4 \times \frac{44.44}{400} \]

5. **Calculate \( F_1 \):**
\[ F_1 = 1.44 \times 10^4 \times 0.1111 = 1.44 \times 10^4 \times 0.1111 \approx 1600 \, \text{N} \]

### Final Answer:
The force on the input piston \( F_1 \) is approximately \( 1600 \, \text{N} \).