As the drawing illustrates, a pond full of water has the shape of an inverted cone with the tip sliced off and has a depth of $3 m$ . The atmospheric pressure above the pond is $1.0 \times 10^{5} P a$ . The circular top surface (radius $= R_{2}$ ) and circular bottom surface (radius $= R_{1}$ ) of the pond are both parallel to the ground. The mgnitude of the force acting on the top surface is the same as the magnitude of the force acting on the bottom surface. Obtain $\frac{R_{2}}{R_{1}}$ .

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Solution

Formula Used: $P = P_{0} + ρ g h$
Given,
depth $d = 30 m P_{0} = 1.0 \times 10^{5} P a$
As we know,
$F_{A} = F_{B} P_{A} A_{A} = P_{B} A_{B} P_{0} π R_{2}^{2} = (P_{0} + ρ g h) π R_{1}^{2} 10^{5} π R_{2}^{2} = (10^{5} + 30 \times 10^{3} \times 10) π R_{1}^{2} 10^{5} π R_{2}^{2} = 4 \times 10^{5} R_{1}^{2} \frac{R_{2}}{R_{1}} = 2$

Final answer:(2)

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Viscous Force

Standard IX Physics

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Formula Used: P=P0+ρgh Given, depthd=30 mP0=1.0×105 Pa As we know, FA=FBPAAA=PBABP0πR22=(P0+ρgh)πR21105πR22=(105+30×103×10)πR21105πR22=4×105R21R2R1=2 Final answer:(2)