Answer
$(a)\space0.8\space kN$
$(b)\space 2.16\space mm$
$(c)\space 7.04\space J,7.02\space J$
Work Step by Step
Here we use the principle of Pascal's law. Whatever pressure results from the force on the smaller piston is transmitted through the fluid to the larger piston. So $P_{1}=P_{2}$
We know, $P=\frac{F}{A}$
$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}$
$\frac{F}{\pi (\frac{1.04cm}{2})^{2}}=\frac{3.25\times10^{3}N}{\pi (\frac{2.1cm}{2})^{2}}=\gt F=0.8\space kN$
0.8 kN must be applied to the smaller piston.
(b) Here we use the principle of conservation of volume to find the motion.
$V_{1}=V_{2}$
$\pi r_{1}^{2}\times8.8\space mm=\pi r_{1}^{2}L$
$(\frac{1.04cm}{2})^{2}8.8\space mm=(\frac{2.1cm}{2})^{2}L$
$2.16\space mm=L$
(c) Here we use the equation work (W)=FS
Let's plug known values into this equation.
Small piston,
$W_{1}=0.8\space kN\times8.8\times10^{-3}m=7.04\space J$
Large piston
$W_{2}=3.25\space kN\times2.16\times10^{-3}m=7.02\space J$
