Question

A student pushes down with a force of 2N on the plunger of syringe A. Syringe B has a plunger with an area that is 3 times the area of plunger A. What is the force produced by plunger B?

Answer 1

Explanation:

Since pressure is transmitted undiminished throughout the water

∴

A

1

F

1

=

A

2

F

2

where F

1

and F

2

are the forces on the smaller and on the larger pistons respectively and A

1

and A

2

are the respective areas.

∴F

2

=

A

1

A

2

F

1

=

π(D

1

/2)

2

π(D

2

/2)

2

F

1

=(

D

1

D

2

)

2

F

1

=

(1×10

−2

m)

2

(3×10

−2

m)

2

×10N=90N

Answer 2

Answer:

PRESSURE ON BOTH THE PLUNGERS WILL BE SAME

$\frac{F1}{A1}=\frac{F2}{A2} \\\frac{2}{A1} =\frac{F2}{3A1} \\F2=6N$