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 follow these steps:

### Step 1: Understand the given data
We have:
- Diameter of the input piston (d1) = 10 cm
- Diameter of the output piston (d2) = 30 cm
- Weight of the car (F2) = 1.44 × 10^4 N

### Step 2: Calculate the areas of the pistons
The area (A) of a circle is given by the formula:
\[ A = \frac{\pi}{4} d^2 \]

**For the input piston (A1):**
\[ A1 = \frac{\pi}{4} (10 \, \text{cm})^2 = \frac{\pi}{4} (100 \, \text{cm}^2) = 25\pi \, \text{cm}^2 \]

**For the output piston (A2):**
\[ A2 = \frac{\pi}{4} (30 \, \text{cm})^2 = \frac{\pi}{4} (900 \, \text{cm}^2) = 225\pi \, \text{cm}^2 \]

### Step 3: Apply Pascal's Law
According to Pascal's Law, the pressure applied on the input piston is equal to the pressure on the output piston:
\[ \frac{F1}{A1} = \frac{F2}{A2} \]

Where:
- F1 = force on the input piston
- F2 = force on the output piston (weight of the car)

### Step 4: Rearrange the equation to find F1
From the equation above, we can express F1 as:
\[ F1 = F2 \times \frac{A1}{A2} \]

### Step 5: Substitute the values
Substituting the known values:
\[ F1 = (1.44 \times 10^4 \, \text{N}) \times \frac{25\pi}{225\pi} \]

The \(\pi\) cancels out:
\[ F1 = (1.44 \times 10^4 \, \text{N}) \times \frac{25}{225} \]

### Step 6: Simplify the fraction
\[ \frac{25}{225} = \frac{1}{9} \]

So,
\[ F1 = (1.44 \times 10^4 \, \text{N}) \times \frac{1}{9} \]

### Step 7: Calculate F1
\[ F1 = \frac{1.44 \times 10^4}{9} \]

Calculating this gives:
\[ F1 \approx 1.6 \times 10^3 \, \text{N} \]

### Final Answer
The force on the input piston (F1) is approximately **1.6 × 10^3 N**.

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