Question

A solid cone of height $H$ and base radius $H/2$ floats in a liquid of density $\rho$. It is hanging from the ceiling with the help of string. The force by the fluid on the curved surface of the cone is ($P_0=$ atmospheric pressure)

• A. $\pi H^2(\frac{p_0}{4}+\frac{\rho gH}{3})$
• B. $\frac{\pi H^4}{2}\cdot \rho g$
• C. $\frac{\pi H^2}{4}(\frac{p_0}{4}+\rho gH)$
• D. $\frac{\pi H^2}{4}\left(p_0+\rho gH\right)$

Answer 1

The correct option is B $\frac{\pi H^4}{2}\cdot \rho g$

Any vertical component of pressure $P_Y$is called by an equal amount of force acting on the bottom.

Now, $P_X$ is same as pressure $P$, but acting on a plane surface of height $H$.

The net pressure can be assumed to be acting at $A$, from $\frac{H}{3}$ from the bottom.

The net pressure is hence area of the pressure $\Delta le$, i.e., $P_{net}=\frac{1}{2}\times \left(P_{Xmax}\right)\times H=\frac{1}{2}\times \left(\rho gH\right)\times H=\frac{\rho g{H}^2}{2}$

Now total force acting on the surface $F=P\times A_{curved}$

i.e., $F=P\left(A_{curved}\cos \theta \right)=P_{net}{A}_{plane}[\because A_{plane}=A_{curved}\cos \theta ]=P_{net}\times (2\pi \times \frac{H}{2}\times H)=\frac{\rho g{H}^2}{2}\times H^2\pi =\frac{\pi H^4}{2}\cdot \rho g$