Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0cm and 3.0 cm respectively.
(a) Find the force on the larger piston when a force of 10N is applied to the smaller piston.
(b) The smaller piston is paused in through 6.0 cm, much does the larger piston move out?

Explanation

Radius of smaller piston, `r_(1)=(1.0cm)/(2) =1/2 xx 10^(-2)m`
Radius of larger piston, `r_(2) =(3.0cm)/(2)=3/2 xx 10^(-2) m`
Force applied to smaller piston, `F_(1)=10N`
Since, the pressure is transmitted undiminished through water. So, `F_(1)/F_(2)=A_(1)/A_(2)`
`rArr F_(1)/F_(2)=r_(1)^(2)/r_(2)^(2)`
`rArr F_(2)=(r_(2)^(2)F_(1))/(r_(1)^(2))`
`rArr F_(2)=r_(2)^(2)/r_(1)^(2) F_(1)=((3/2 xx 10^(-2))/(1/2 xx 10^(-2)))^(2) xx 10=90N`
(b) Distance to which smaller piston is pushed, `L_(1)=6cm =6 xx 10^(-2)m`
As the water is incompressibe. So,
Volume covered by inward movement of smaller piston=Volume covered by outward movement of larger piston.
`rArr L_(1)A_(1)=L_(2)A_(2)`
`rArr L_(1) xx (pi r_(1)^(2))=L_(2) xx (pi r_(2)^(2))`
`rArr L_(2)=r_(1)^(2)/r_(2)^(2) L_(1)=(1/2 xx 10^(-2))/(3/2 xx 10^(-2))^(2) xx 6 xx 10^(-2)`
=0.67cm