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a hydraulic jack is used to lift a car in which a man stands on one piston and car is put on another one. what should be the minimum weight of the man to lift the car?

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A car is to be lifted by a hydraulic jack which consists of two pistons. The large piston is 80 cm in diameter and the smaller piston is 16 cm in diameter. If W is the weight of the car, then how much smaller force is needed on small pistonto lift the car?

Radius of large piston , R(2)= 40 cm =0.40 m Radius of small piston , R(1) = 8 cm =0.08 m F(2) =W , F(1)= ? Ac cording to Pascal's law (F1)/(pi R(1)^(2)) = (F2)/(pi R(2)^(2)) or, F(1) = F(2) (R(1)^(2))/(R(2)^(2)) = W xx ((0.08)/(0.40))^(2) = 0.04 W. = 4% of the weight of car.

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A car is to be lifted by a hydraulic jack which consists of two pistons. The large piston is 80 cm in diameter and the smaller piston is 16 cm in diameter. If W is the weight of the car, then how much smaller force is needed on small pistonto lift the car?

Updated On: 27-06-2022

Text Solution Solution

R 2 =40cm=0.40m R2=40cm=0.40m

R 1 =8cm=0.08m R1=8cm=0.08m F 2 =W, F 1 =? F2=W,F1=?

Ac cording to Pascal's law

F 1 π R 2 1 = F 2 π R 2 2 F1πR12=F2πR22 or, F 1 = F 2 R 2 1 R 2 2 =W× ( 0.08 0.40 ) 2 =0.04W

F1=F2R12R22=W×(0.080.40)2=0.04W

. =4% =4%

of the weight of car.

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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स्रोत : www.doubtnut.com

In a hydraulic jack as shown, mass of the car is W =800 kg , A 1=10 cm 2, A 2=10 m 2. The minimum force F required to lift the car is: [Take .g =10 m / s 2]B. 0.8 NС. 8 ND. 16 N

In a hydraulic jack as shown, mass of the car is W =800 kg , A 1=10 cm 2, A 2=10 m 2. The minimum force F required to lift the car is: [Take .g =10 m / s 2]B. 0.8 NС. 8 ND. 16 N

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In a hydraulic jack as shown, mass of the car is W =800 kg , A 1=10 cm 2, A 2=10 m 2. The minimum force F required to lift the car is: [Take .g =10 m / s 2]B. 0.8 NС. 8 ND. 16 N

Question

In a hydraulic jack as shown, mass of the car is

W = 800 kg , A 1 = 10 cm 2 , A 2 = 10 m 2 . The minimum force F

required to lift the car is: [Take

g = 10 m/s 2 ]

A 1 N B 0.8 N C 8 N D 16 N Open in App Solution

The correct option is B

0.8 N Given, A 1 = 10 cm 2 A 2 = 10 m 2 = 10 5 cm 2 F 2 = W g = 8000 N

Pressure in the same liquid at the same level should be the same, so we can say that pressure at both the pistons are equal.

F 1 A 1 = F 2 A 2

By substituting values,

F 10 = 8000 10 5 or F = 0.8 N Suggest Corrections 8

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Q. In a hydraulic jack as shown, mass of the car is

W = 800 kg , A 1 = 10 cm 2 , A 2 = 10 m 2 . The minimum force F

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g = 10 m/s 2 ]

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2000 k g with a force of 1000 N

using a car jack. Find the mechanical advantage of the car jack. (Given,

g = 10 m s − 2 )

Q. The press plungers of a Bramah (hydraulic) press is

40 c m 2

in cross-section and is used to lift a load of mass

800 k g

. What minimum force is required to be applied to the pump plunger if its cross-sectional area is

0.02 m 2 ?

Q. A

1 m

long uniform beam of

2 k g

mass is being lifted vertically up by a force F at the

100 c m

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a 1 , a 2 , a 3

of the three blocks shown in figure

( 6 − E 8 )

if a horizontal force of

10 N is applied on ( a ) 2 k g block, ( b ) 3 k g block, ( c ) 7 k g block. Take g = 10 m / s 2 . Take g = 10 m / s 2 .

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स्रोत : byjus.com

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross

Click here👆to get an answer to your question ✍️ A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross - section of the piston carrying the load is 425 cm ^2 . What maximum pressure would the smaller piston have to bear ?

Class 11 >>Physics

>>Mechanical Properties of Fluids

>>Problems on Pascal's Law

>>A hydraulic automobile lift is designed

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm

Question 2

. What maximum pressure would the smaller piston have to bear ?

Pressure on the piston P=

Medium Open in App Solution Verified by Toppr A F ​ Force F=m×a =3000×9.8 =29400 N

Area of cross section A=425×10

−4 sqm

Therefore the pressure P=

425×10 −4 3000×9.8 ​ =6.92×10 5 Pa.

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