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Question: A piston of cross - section area \(100 \mathrm{~cm}^{2}\) is used in a hydraulic press to exert a fo...

A piston of cross - section area 100 cm2100 \mathrm{~cm}^{2} is used in a hydraulic press to exert a force of 10510^{5} dynes on the water. The cross-sectional area of the other piston which supports an object having a mass 1000 kg1000 \mathrm{~kg} is (g=10 m/s2)\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)

  1. 100 cm2100 \mathrm{~cm}^{2}

  2. 109 cm210^{9} \mathrm{~cm}^{2}

  3. 1×106 cm21 \times 10^{6} \mathrm{~cm}^{2}

  4. 2×1010 cm22 \times 10^{10} \mathrm{~cm}^{2}

Explanation

Solution

Calculate Cross Sectional area by using pascal law. Calculate the force acting on the Piston1, Piston 2. By the help of the pascal law we know that P1=P2⇒F1 A1=F2 A2⇒A2=F2 A1 F1\mathrm{P}_{1}=\mathrm{P}_{2} \Rightarrow \dfrac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\dfrac{\mathrm{F}_{2}}{\mathrm{~A}_{2}} \Rightarrow \mathrm{A}_{2}=\dfrac{\mathrm{F}_{2} \mathrm{~A}_{1}}{\mathrm{~F}_{1}} By using this equation we can take out the cross-sectional area of the other piston.

Complete solution:

The law of Pascal is a principle given by Blaise Pascal in fluid mechanics that states that a pressure change is transmitted throughout the fluid at any point in a confined incompressible fluid, so that the same change occurs everywhere. The pressure is equal in all directions at any point in the fluid.

The pressure raised using one piston is communicated everywhere in the fluid by Pascal's law. The same pressure, therefore, acts on the other piston.

Cross-sectional area of piston 1 is A1=100 cm2=0.01 m2\mathrm{A}_{1}=100 \mathrm{~cm}^{2}=0.01 \mathrm{~m}^{2}

Force on piston 1 is F1=105\mathrm{F}_{1}=10^{5} dyne =1 N=1 \mathrm{~N}

Force on piston 2 is F2=1000 kgwt=1000×10=1×104 N\mathrm{F}_{2}=1000 \mathrm{~kg} \mathrm{wt}=1000 \times 10=1 \times 10^{4} \mathrm{~N}

Let the cross-sectional area of Piston 2 be A2=A\mathrm{A}_{2}=\mathrm{A}.

By Pascal's law,

P1=P2\mathrm{P}_{1}=\mathrm{P}_{2}

⇒F1 A1=F2 A2\Rightarrow\dfrac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\dfrac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}

⇒A2=F2 A1 F1\Rightarrow \mathrm{A}_{2}=\dfrac{\mathrm{F}_{2} \mathrm{~A}_{1}}{\mathrm{~F}_{1}}

⇒A2=A=1×104×0.011\Rightarrow {A_2}=A= \dfrac {1\times {10^4} \times {0.01}}{1}

⇒A=1×104×10−2m2\Rightarrow {A}= {1\times {10^4}\times {10^{-2}}} {m^2}

⇒A=1×102m2=102×104cm2\Rightarrow {A}= 1\times{10^2} {m^2}= {10^2}\times{10^4} {{cm}^2}

⇒A=106cm2=1×106cm2\Rightarrow {A}= {10^6} {{cm}^2}=1\times{10^6} {{cm}^2}

The cross-sectional area of the other piston which supports an object having a mass 1000 kg1000 \mathrm{~kg} is 1 !!×!! 106 !! !! cm2\text{1 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{6}}}\text{ }\\!\\!~\\!\\!\text{ c}{{\text{m}}^{\text{2}}}.

Therefore, Correct option is (3).

Note:
What makes pressure so important in fluids is the principle of Pascal, an experimentally verified fact. We often know more about pressure than other physical quantities in fluids because a change in pressure is transmitted undiminished in an enclosed fluid. If at any point of the incompressible liquid some pressure is applied, then the same pressure is transmitted to all points of the liquid and to the walls of the container.