# A closed conical vessel is filled with water fully and is placed with its vertex down. Thewater is let out at a constant speed. After 21 minutes it was found that the height of thewater column is half of the original height. How much more time in minutes does itrequire to empty the vessel?

AP GAUTHAM
21 Points
7 years ago
3 more minutes.
poojahariharan
20 Points
7 years ago
can you explain with steps pls
AP GAUTHAM
21 Points
7 years ago
let the height of cone be ‘h’ and radius be ‘r’
Volume of water is (1/3)r2h.
After 21min height of water becomes ‘h/2’.Applying simiar trianles concept in the cone, we get new radius as r/2.
Hence remaining volume is (1/3)(r/2)2h/2=(1/3)r2h/8.
Then amount of water drained out is 7(1/3)r2h/8.
Then time taken to empty the vessel =21×[(1/3)r2h/8]/[7 (1/3)r2h/8]=3min.

I’m sorry due to problems ,browser is showing “pi” as .
AP GAUTHAM
21 Points
7 years ago
I’m sorry it’s not showing “pi” at all.
poojahariharan
20 Points
7 years ago
thnk u
Rushang Patel
17 Points
4 years ago
Let R be the radius and h be the total height of the inverted cone.Then volume of water inside cone = 13πR2h .......(1)Volume of water inside cone after 21 min = volume of cone ABCNow, let r be the radius of cone ABC.Now, height of cone ABC = h2So, volume of cone ABC = 13πr2(h2) = 16πr2h ........(2)In ∆AOD and ∆AO`B, we have∠AOD = ∠AO`B [90° each]∠OAD = ∠O`AB (common)⇒∆AOD ~ ∆AO`B (AA)⇒AOAO`=ODO`B (corresponding sides of similar ∆`s are proportional)⇒hh/2=Rr⇒R = 2rNow, from (1) put R = 2r, we getvolume of cone ADE = 43πr2h Volume of water flows out in 21 min =43πr2h−16πr2h = 76πr2hSo, (76πr2h) cubic units of water flows out in 21 minutes⇒(16πr2h) cubic units of water flows out in (21 × 67πr2 × 16πr2h ) = 3 min
Anurag
13 Points
4 years ago
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