A cylinder containing water up to a height of $25 cm$ has a hole of cross-section $\frac{1}{4}c{m^2}$ in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begins to flow out
Diffcult
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(c)Let $ A =$ The area of cross section of the hole
$v =$ Initial velocity of efflux
$d = $ Density of water,
Initial volume of water flowing out per second $ = Av$
Initial mass of water flowing out per second $ = Avd$
Rate of change of momentum $= Adv2$
Initial downward force on the flowing out water $= Adv2$
So equal amount of reaction acts upwards on the cylinder.
Initial upward reaction =$Ad{v^2}$ [As $v = \sqrt {2gh} $]
$\therefore $ Initial decrease in weight $ = Ad\,(2gh)$
$ = 2Adgh$$ = 2 \times \left( {\frac{1}{4}} \right) \times 1 \times 980 \times 25 = 12.5$ $gm-wt.$
$v =$ Initial velocity of efflux
$d = $ Density of water,
Initial volume of water flowing out per second $ = Av$
Initial mass of water flowing out per second $ = Avd$
Rate of change of momentum $= Adv2$
Initial downward force on the flowing out water $= Adv2$
So equal amount of reaction acts upwards on the cylinder.
Initial upward reaction =$Ad{v^2}$ [As $v = \sqrt {2gh} $]
$\therefore $ Initial decrease in weight $ = Ad\,(2gh)$
$ = 2Adgh$$ = 2 \times \left( {\frac{1}{4}} \right) \times 1 \times 980 \times 25 = 12.5$ $gm-wt.$
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