Question : A hydraulic automobile lift is designed to lift cars with maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear?
(a) 6.92×105 N/m2
(b) 9.63×109 N/m2
(c) 7.82×107 N/m2
(d) 13.76×1011 N/m2
Doubt by Jaskirat & Priyanshi
Solution :
Mass of car (m) = 3000 kg
Weight of the Car = Force Exerted by the Car on the Piston (F2)
= Mass (m) × Acceleration due to gravity (g)
= 3000×9.8
= 29400 N
Weight of the Car = Force Exerted by the Car on the Piston (F2)
= Mass (m) × Acceleration due to gravity (g)
= 3000×9.8
= 29400 N
Area of cross section of piston carrying load (A2)
= 425 cm2
= 425 ×10-4 m2
As per the Pascal's law, pressure will remain same at both the pistons.
P=F1/A1=F2/A2
P=F2/A2
P=29400 / (425 ×10-4)
P=[29400×104]/425
P=[2940×105]/425
P=29400 / (425 ×10-4)
P=[29400×104]/425
P=[2940×105]/425
P=6.917×105
P=6.92×105 N/m2
P=6.92×105 N/m2
Hence, (a) 6.92×105 N/m2, would be the correct option.