Question

hydraulics automobile lift is designed to lift a car with the maximum mark 3000 kg the area of cross section of the piston carrying the load is 425 CM square what is the maximum pressure around the smaller have to bear ​

Answer 1

Given :

Mass of the car = 3000 kg

Area of cross section = 425cm²

To find :

The maximum pressure around the smaller piston.

Solution :

Firstly we have to find the force exerted by the load,

we know that,

F = mg

where,

• F is force exerted
• m is mass
• g is acceleration due to gravity

By substituting the values in the formula,

$\dashrightarrow \sf F = mg$

$\dashrightarrow \sf F = 3000 \times 9.8$

$\dashrightarrow \sf F = 29400 \: N$

Now,

The maximum pressure exerted by the load is,

$\dashrightarrow \sf P = \dfrac{F}{A}$

$\dashrightarrow \sf P = \dfrac{29400}{425 \times {10}^{ - 4} }$

$\dashrightarrow \sf P = 6.917 \times {10}^{5} \: Pa$

Thus, the maximum pressure around the smaller piston is 6.917 × 10⁵ Pa