A cone of volume V plugs a circular hole in a water-tank base, with half the cone height protruding below the hole. Water depth equals cone height h. If hydrostatic force on the cone is ρVg/n, determine n. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A cone of volume V plugs a circular hole in a water-tank base, with half the cone height protruding below the hole. Water depth equals cone height h. If hydrostatic force on the cone is ρVg/n, determine n. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Let the cone’s height be h and top radius R; cone volume is V = πR²h/3. Its upper half is exposed to water, with radius reducing from R to R/2 at the tank base. Step 2 — Imagine closing the exposed frustum at the hole plane. The upward hydrostatic buoyancy on this closed frustum equals ρg × (7V/8). Step 3 — The imagined bottom disk of radius R/2 at depth h would experience upward force ρghπ(R/2)² = 3ρgV/4, but it is not wetted in the actual setup. Step 4 — Actual net force on the cone…