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Question: In a car lift, compressed air exerts a force \[{F_1}\] on a small piston having a radius of \[5cm\] ...

In a car lift, compressed air exerts a force F1{F_1} on a small piston having a radius of 5cm5cm . This pressure is transmitted to the second piston of a radius 15cm15cm . If the mass of the car to be lifted is 1350  kg1350\;kg . What is F1{F_1}?
A)14.7×103NA)14.7 \times {10^3}N
B)1.47×103NB)1.47 \times {10^3}N
C)2.47×103NC)2.47 \times {10^3}N
D)24.7×103ND)24.7 \times {10^3}N

Explanation

Solution

The pressure is known as the amount of force exerted per unit area. Dimension of pressure is M L1  T2M{\text{ }}{L^{ - 1}}\;{T^{ - 2}}. SI unit of pressure is N/m2N/{m^2} . The principle was invented by the French scientist called Blaise Pascal. Pascal's law defines that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and the walls of the container. According to Pascal's law, a pressure exerted on a piston creates an equal rise in pressure on another piston in a hydraulic system.

Formula used:
F1  πr12=F2πr22\dfrac{{{F_1}}}{{\;\pi {r_1}^2}} = \dfrac{{{F_2}}}{{\pi {r^2}_2}}

Complete step by step answer:
Given: r1=5cm{r_1} = 5cm, r2=5cm{r_2} = 5cm
F=1350kgF = 1350kg
From pascal’s law,
P1=P2{P_1} = {P_2}
Where P1={P_1} = Pressure in the piston 11
P2={P_2} = Pressure in piston 22
F1A1=F2A2\dfrac{{{F_1}}}{{{A_1}}} = \dfrac{{{F_2}}}{{{A_2}}}
F1={F_1} = Force exerted on the piston 11
F2={F_2} = Force exerted on the piston 22
Put area formula, A=πr2A = \pi {r^2}
F1  πr12=F2πr22\dfrac{{{F_1}}}{{\;\pi {r_1}^2}} = \dfrac{{{F_2}}}{{\pi {r^2}_2}}
We have to find F1{F_1}
π\pi is canceled on L.H.S and R.H.S
So, bringing denominator in L.H.S to R.H.S
F1=F2r12r22\Rightarrow {F_1} = \dfrac{{{F_2}{r_1}^2}}{{{r_2}^2}}
By applying all Values in the formula,
F1=1350×9.8×(5×102)2(15×102)2\Rightarrow {F_1} = \dfrac{{1350 \times 9.8 \times {{(5 \times {{10}^{ - 2}})}^2}}}{{{{\left( {15 \times {{10}^{ - 2}}} \right)}^2}}}
F1=1470N\Rightarrow {F_1} = 1470N
F1=1.47×103N\Rightarrow {F_1} = 1.47 \times {10^3}N

Hence, the correct answer is in option (B)1.47×103N(B) \Rightarrow 1.47 \times {10^3}N.

Note: Pressure can be calculated as
P=FAP = \dfrac{F}{A}
Where P =P{\text{ }} = Pressure
F =F{\text{ }} = Force
A =A{\text{ }} = Area
Pascal's law is used in our day to day life in a vehicle braking system and new electronic parking system which is applicable in cars are directly moved to the next floor. And it is used in Hydraulic Jacks. It is used in lifting heavy objects. A hydraulic jack consists of two cylinders.