A piston of cross-section area 100 cm² is used in a hydraulic press to exert a force of 10⁷ dynes on the water. The cross-sectional area of the other piston which supports an object having a mass 2000 kg. is

The correct option is 2 x 10⁴cm².

To find the cross-sectional area of the other piston, we can use Pascal's Law, which states that any change in pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid.

Therefore, the pressure ($P$) exerted by both pistons must be equal:
$P=\frac{F_1}{A_1}=\frac{F_2}{A_2}$
where:

• $F_1$ is the force on the first piston,
• $A_1$ is the cross-sectional area of the first piston,
• $F_2$ is the force on the second piston (equal to the weight of the supported object), and
• $A_2$ is the cross-sectional area of the second piston.

Step 1: Identify the given values and convert them to CGS units
Area of the first piston, $A_1=100{\text{cm}}^2$
Force on the first piston, $F_1={10}^7\text{dynes}$
Mass of the object on the second piston, $m=2000\text{kg}=2000\times 1000\text{g}=2\times {10}^6\text{g}$
Acceleration due to gravity, $g\approx 10{\text{m/s}}^2=1000{\text{cm/s}}^2$

Step 2: Calculate the weight force $F_2$ of the object
The force acting on the second piston is:
$F_2=m\times g$
$F_2=(2\times {10}^6\text{g})\times \left(1000{\text{cm/s}}^2\right)=2\times {10}^9\text{dynes}$

Step 3: Solve for the area $A_2$ using Pascal's Law
$\frac{F_1}{A_1}=\frac{F_2}{A_2}$
Rearranging the equation to isolate $A_2$:
$A_2=\frac{F_2\times A_1}{F_1}$
Substitute the values into the equation:
$A_2=\frac{(2\times {10}^9\text{dynes})\times 100{\text{cm}}^2}{{10}^7\text{dynes}}$
$A_2=\frac{2\times {10}^{11}}{{10}^7}=2\times {10}^4{\text{cm}}^2$