A piston of cross-sectional area $100 c m^{2}$ is used in a hydraulic press to exert a force of $10^{7}$ dyne on water. The cross-sectional area of the other piston which supports a truck of mass 200 kg is in $m^{2} .$ [Take g=10ms^-2]

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Detailed Solution

$Cross-sectional area of piston 1 is A_{1} = 100 {cm}^{2} = 0.01 m^{2}$

$Force on piston 1 is F_{1} = 10^{7} dyne = 100 N$

$Force on piston 2 is F_{2} = 2000 kg wt = 2000 \times 10 = 2 \times 10^{4} N$

$P_{1} = P_{2} (Pascal's law); \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$

$\Rightarrow A = \frac{2 \times 10^{4} \times 0.01}{100} = 2 m^{2}$

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