In a brama press, a force of 10kg is applied on a piston of area of crossection 5cm². Then what is the upperthrust exerted by the puston of the area of crossection of 50cm²?
Answers
Answer:
In a typical hydraulic press, a force of 20 N is exerted on a small piston of area 0.050 m2m2. What is the force exerted by a large piston on load if it has an area of 0.50 m2m2 ?
A) 200 N
B) 100 N
C) 50 N
D) 10 N
Answer
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Hint
As, here are two pistons one is small and other is large we have to calculate the force exerted by the large piston. For this we know that the hydraulic pressure is the ratio of force per unit area. We write the hydraulic pressure for both small and large piston. After equating both hydraulic pressure and substitute the given values then we will get the required value of the force by the large piston.
Complete step by step solution
Here, it is given that; Force exerted on small piston is F1=20NF1=20N
And the area on which force is exerted by small piston is A1=0.050m2A1=0.050m2
As, hydraulic pressure is the ratio of applied force to the area on which force is applied
Therefore, for small piston hydraulic pressure is P1=200.050P1=200.050 …………………. (1)
Let the force exerted by the large piston on load is F2F2.
And area on which force is exerted by the large piston is A2=0.50m2A2=0.50m2
Therefore, hydraulic pressure for a large piston is P2=F20.50P2=F20.50 ……………………. (2)
As the, both hydraulic pressure is equal so equating the equation (1) and (2), we get
⇒200.05=F20.5⇒200.05=F20.5
⇒F2=200N⇒F2=200N
Thus, the force exerted by the large piston is 200 N.
Hence, (A) option is correct.