A solid cone of height H and base radius $\frac{H}{2}$ floats in a liquid of density $ρ$ . It is hanging from the ceiling with the help of a string. The force by the fluid on the curved surface of the cone is $(P_{0} = atmospheric pressure)$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$π H^{2} (\frac{P_{0}}{4} + ρ gH)$

$\frac{π H^{2}}{4} (P_{0} + ρ g H)$

$π H^{2} (\frac{P_{0}}{4} + \frac{ρ gH}{3})$

$π H^{2} (\frac{P_{0}}{4} + \frac{ρ gH}{6})$

answer is B.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{1}{3} π {(\frac{H}{2})}^{2} H ρ g = (P_{0} + ρ g H) π {(\frac{H}{2})}^{2} - F_{v}$

$F_{v} = P_{0} π \frac{H^{2}}{4} + ρ g \frac{H^{3}}{4} π - \frac{π H^{3} ρ g}{12}$

$= P_{0} \frac{π H^{2}}{4} + \frac{ρ {gH}^{3} π}{6} = π H^{2} (\frac{P_{0}}{4} + \frac{ρ gH}{6})$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test