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Find the time taken for height of water level to fall from 'h' to 'h'in a vessel, if water comes out from a small hole at the bottom |
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Answer» SOLUTION :From EQUATION of continity for an incompressible liquid `Av^(1)=asqrt(2gh)` (or) `v^(1)=((a)/(A))sqrt(2gh)` where `v^(1)` is the RATE of descendence of the LEVEL of free surface. So, `v^(1)` can be WRITTEN as: `V^(1)=(-dh)/(dt)` `therefore (-dh)/(dt)=((a)/(A))sqrt(2gh)` (or) `overset(h^(1))underset(h)int(dh)/(sqrt(h))=-((a)/(A))sqrt(2g)overset(t)underset(t=0)intdt` `therefore 2(sqrt(h^(1)-sqrt(h))) =-((a)/(A))(sqrt(2g))t` (or)`t=(2A)/(asqrt(2g))(sqrt(h^(1))-sqrt(h))prop t` |
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