Question

A piston of cross sectional area 100 cm square is used in a hydraulic press to exert a force of 10^7 dyne on the water.The cross-sectional area of the Other Piston which supports an object having a mass of 2000 kg is?

Answer 1

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Formula to be used
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$= > \bf \: \frac{force(f)}{area(a)} = \frac{f1}{a1} = \frac{f2}{a2}$
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Area (A1) = 100 cm²

$\bf \: = \frac{100}{100 \times 100} = 0.01 \: {m}^{2}$

Force(F1) = 10^7 dynes =( 10^7 ×10^-5) N

$\bf \: = 100 \: newton$
Force (F2) = mass × gravity.

$\bf = 2000 \: kg \times 9.8\: m \: {s}^{ - 2} \: ( since \: \: g = 9.8 \: m \: {s}^{ - 2} ) \\ \\ \bf \: = 19600 \: \: newton$
Area (A2) =?

So..
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$\bf = > \frac{f1}{a1} = \frac{f2}{a2} \\ \\ = > \frac{100}{0.01} = \frac{19600}{a2} \\ \\ \bf \: = > a2 = \frac{19600}{10000} = 1.960 \: {m}^{2}$
So..
The required area is

$\bf \: = 1.960 \: {m}^{2} \: \: or \: \: 19.60 \: \times {10}^{3} \: \: cm ^{2}$
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Answer 2

F1/A2= F2/A2

On solving this ahead you reach to A2 = 2000× 10 to power 6 × 100/ 10 to power 7

A2 = 2000× 10 to power 8/ 10 to power 7

Which is then equal to 2000× 10 = 20000 cm or 2 m

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