Two syringes of different area of cross sections filled with water are
connected with a tight rubber tube filled with water. Radius of the smaller
piston and larger piston are 2 cm and 4 cm. Find the force exerted on
the larger piston when force of 100 N is applied to the smaller piston.
Answers
Explanation:
Since pressure is transmitted undiminished throughout the water
\therefore \displaystyle \frac{F_1}{A_1} =\frac{F_2}{A_2}∴
A
1
F
1
=
A
2
F
2
where F_1F
1
and F_2F
2
are the forces on the smaller and on the larger pistons respectively and A_1A
1
and A_2A
2
are the respective areas.
\therefore \displaystyle F_2=\frac{A_2}{A_1}F_1 = \frac{\pi (D_2/2)^2}{\pi (D_1/2)^2} F_1 = \left ( \frac{D_2}{D_1} \right )^2 F_1∴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
\displaystyle = \frac {(3 \times 10^{-2}\,m)^2}{(1 \times 10^{-2}\,m)^2} \times 10\,N = 90\,N=
(1×10
−2
m)
2
(3×10
−2
m)
2
×10N=90N
Pressure on the smaller piston = F/A
= 100/(πr²)
= 100/π{2x10^(-2)}²
= 25/π x10⁴
Now pressure on the second piston will be equal to that of first..
Thus pressure in second piston = F/A
25/π x 10⁴. = F/(πR²)
F = 25/π x πR² x 10⁴
F = 25R² x 10⁴
F = 25 x {4x10^(-2)}² x 10⁴
F = 400 N
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