A piston of cross-section area 100 , cm^2 is | vedclass

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(C) According to Pascal's Law,the pressure exerted on the fluid is transmitted equally in all directions. Therefore,$P_1 = P_2$.This implies $\frac{F_

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$2 	imes 10^4 \, cm^2$

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(C) According to Pascal's Law,the pressure exerted on the fluid is transmitted equally in all directions. Therefore,$P_1 = P_2$.This implies $\frac{F_1}{A_1} = \frac{F_2}{A_2}$.Given: $A_1 = 100 \, cm^2 = 10^2 \, cm^2$,$F_1 = 10^7 \, dynes$,$m = 2000 \, kg = 2 	imes 10^6 \, g$,and $g \approx 980 \, cm/s^2 \approx 10^3 \, cm/s^2$.Force $F_2 = m 	imes g = 2000 	imes 10^3 	imes 10^3 \, dynes = 2 	imes 10^9 \, dynes$.Substituting the values: $\frac{10^7}{10^2} = \frac{2 	imes 10^9}{A_2}$.$10^5 = \frac{2 	imes 10^9}{A_2}$.$A_2 = \frac{2 	imes 10^9}{10^5} = 2 	imes 10^4 \, cm^2$.

$A$ piston of cross-section area $100 \, cm^2$ is used in a hydraulic press to exert a force of $10^7 \, dynes$ on the water. The cross-sectional area of the other piston which supports an object having a mass $2000 \, kg$ is:

Similar Questions

$A$ hydraulic automobile lift is designed to lift vehicles of mass $5000\,kg$. The area of cross-section of the cylinder carrying the load is $250\,cm^2$. The maximum pressure the smaller piston would have to bear is [Assume $g = 10\,m/s^2$].

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