Advertisements
Advertisements
प्रश्न
In a hydraulic machine, a force of 2 N is applied on the piston of area of cross section 10 cm2. What force is obtained on its piston of area of cross section 100 cm2 ?
Advertisements
उत्तर
Force on narrow piston , F1 = 2N
Area of cross - section of narrow piston , A1 = 10 cm2
Let Force on wider piston be F2
Area of cross - section of wider piston , A2 = 100 cm2
By the principle of hydraulic machine ,
Pressure on narrow piston = pressure on wider piston
or, `"F"_1/"A"_1 = "F"_2/"A"_2`
or , `2/10 = "F"_2/100`
or , `"F"_2 = 20 "N"`
APPEARS IN
संबंधित प्रश्न
A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25 cm respectively.
66640 Pa pressure is exerted by 0.50 m vertical column of a liquid. If g = 9.8 Nkg−1, calculate the density of the liquid.
The area of cross-sections of the pump plunger and press plunger of a hydraulic press is 0.02 m2 and 8 m2 respectively. If the hydraulic press overcomes a load of 800 kgf, calculate the force acting on the pump plunger.
How does the liquid pressure on a diver change according to the following condition:
When the driver moves to the greater depth?
State two applications of Pascal's law.
Analogy
Pascal’s law: ______:: Surface tensin: ______.
An iceberg of density 700 kg/m3 is floating in the water of density 1000 kg/m3. The percentage of the volume of ice cube outside the water is:
