Question

a piston of cross sectional area 100cm square is used in a hydraulic press to exert a force of 10 power 7 dynes on the water. the cross sectional area of the other piston which supports a truck of mass 2000 kg is
a) 9.8 into 10 square

b)9. 8 into 10 cube

c) 1.96 into 10 cube

d) 1.96 into 10 power 4​

Answer 1

Answer

Area of the piston which supports the truck = $\sf{1.96 \times 10^4 \;cm^2}$

Explanation

The Cross-sectional area of the first piston, $A_1$ = 100 $cm^2$ = 0.01 $m^{2}$

Force by the first piston, $F_1$ = $10^7$ dynes = 100 N

Mass of the truck supported by the second piston, $m$ = 2000 kg

so,

Let the force on the second piston = $F_2$

then, taking acceleration due to gravity, $g$ = 9.8 $ms^{-2}$

$\implies F_2 = m\;g$

$\implies F_2 = 2000\times 9.8 \;\;N$

Letting the area of the second piston = $A_2$

Now,

According to Pascal's law

On applying any external force on any confined incompressible fluid, it is transmitted undiminished and equally in all direction, resulting in the same pressure difference everywhere.

So, for the first and second piston

$\implies P_1 = P_2$

$\implies \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$

$\implies \dfrac{100\;N}{0.01\;m^2} = \dfrac{2000\times 9.8\;N}{A_2}$

$\implies A_2 =\dfrac{2000\times 9.8\;N\times 0.01\;m^2}{100\;N}$

$\implies A_2 = 2000\times 9.8 \times 10^{-4}\;m^2$

$\implies A_2 = 19.6\times 10^3\times 10^{-4} \;m^2$

$\implies A_2 = 1.96 \;m^2$

$\implies A_2 = 1.96 \times 10^4 \;cm^2$

Therefore,

Area of the piston which supports the truck is $\sf{1.96 \times 10^4 \;cm^2}$

Hence, Option (d) is correct.

Answer 2

$\Large\bf{\color{cyan}GiVeN,}$

ƇƛƧЄ - 1

• Cross-sectional area of a piston is 100 cm².

$\longmapsto\:\:\bf\blue{A_1\:=\:100\:cm^2\:=\:0.01\:m^2}$

• Piston used in a hydraulic pressure and exert 10⁷ dyne force on the water.

$\longmapsto\:\:\bf\pink{F_1\:=\:10^7\:dyne\:=\:100\:N}$

$\Big[\:\bf{{NOTE}\:\longrightarrow\:\purple{1\:dyne\:=\:10^{-5}\:N}\:\atop{\color{olive}1\:cm^2\:=\:10^{-4}\:m^2}}\:\Big]$

ƇƛƧЄ - 2

• Cross-section area of other piston = ?

$\longmapsto\:\:\bf\red{A_2\:=\:?}$

• The other piston which supports a truck of mass 2000 kg.

$\longmapsto\:\:\bf{F_2\:=\:m\:g\:}$

$\longmapsto\:\:\bf{F_2\:=\:2000\times{9.8}\:}$

$\longmapsto\:\:\bf{\color{coral}F_2\:=\:19600\:N\:}$

$\bf{\color{indigo}We\:know\:that,}$

☆ By Pascal's law, pressure applied to a piston fluid will transmitted without a change in magnitude to every point of the fluid and to the walls of the container. The pressure at point in the fluid is equal in all directions. And the same pressure acts on the other piston also, if need.

$\bf\pink{So,}$

$\bf\green{According\:to\:Pascal's\:law,}$

$:\implies\:\:\bf{\color{peru}(Pressure)_1\:=\:(Pressure)_2\:}$

$:\implies\:\:\bf{\dfrac{F_1}{A_1}\:=\:\dfrac{F_2}{A_2}\:}$

$:\implies\:\:\bf{\dfrac{100}{0.01}\:=\:\dfrac{19600}{A_2}\:}$

$:\implies\:\:\bf{10000\:=\:\dfrac{19600}{A_2}\:}$

$:\implies\:\:\bf{A_2\:=\:\dfrac{19600}{10000}\:}$

$:\implies\:\:\bf{A_2\:=\:1.96\:m^2}$

$:\implies\:\:\bf{\color{olive}A_2\:=\:1.96\times{10^4}\:cm^2}$

$\Large\bold\therefore$ The cross-sectional area of the other piston is 1.96 × 10⁴ cm².