A piston of cross-sectional area `100 cm^(2)` is used in a hydraulic pressure to exert a force of `10^(7)` dyne on the water. The cross-sectional area of the other piston which support a truck of mass 2000 kg is

Options

A. `9.8xx10^(2) cm^(2)`
B. `9.8xx10^(3) cm^(2)`
C. `1.96xx10^(3) cm^(2)`
D. `1.96xx10^(4) cm^(2)`

Explanation

To solve the problem, we will use Pascal's law, which states that the pressure applied to a confined fluid is transmitted undiminished in every direction throughout the fluid. This means that the force exerted by one piston can be used to find the force and area of another piston.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Cross-sectional area of the first piston, \( A_1 = 100 \, \text{cm}^2 \)
- Force exerted by the first piston, \( F_1 = 10^7 \, \text{dyne} \)
- Mass of the truck, \( m = 2000 \, \text{kg} \)

2. **Convert Units:**
- Convert the force from dyne to Newton:
\[ 1 \, \text{dyne} = 10^{-5} \, \text{N} \]
\[ F_1 = 10^7 \, \text{dyne} = 10^7 \times 10^{-5} \, \text{N} = 10^2 \, \text{N} = 100 \, \text{N} \]

3. **Calculate the Weight of the Truck:**
- The weight \( F_2 \) of the truck can be calculated using the formula:
\[ F_2 = m \cdot g \]
where \( g \approx 9.8 \, \text{m/s}^2 \).
\[ F_2 = 2000 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19600 \, \text{N} \]

4. **Apply Pascal's Law:**
- According to Pascal's law, the pressure exerted by both pistons is equal:
\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]
- Rearranging to find \( A_2 \):
\[ A_2 = \frac{F_2 \cdot A_1}{F_1} \]

5. **Substitute the Values:**
- Substitute \( F_1 = 100 \, \text{N} \), \( F_2 = 19600 \, \text{N} \), and \( A_1 = 100 \, \text{cm}^2 \):
\[ A_2 = \frac{19600 \, \text{N} \cdot 100 \, \text{cm}^2}{100 \, \text{N}} = 19600 \, \text{cm}^2 \]

6. **Final Result:**
- The cross-sectional area of the piston supporting the truck is:
\[ A_2 = 19600 \, \text{cm}^2 \]