A cylindrical tank of height 0.4 m is open at the top and has a diameter 0.16 m, Water is filled in It up to a height of 0.16 m. How long it will take to empty the tank through a hole of radius $5 \times 10^{- 3}$ m at its bottom?

46.26 sec

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4.6 sec

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462.6 sec

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0.46 sec

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Solution

The correct option is A 46.26 sec
$h_{1} = 0.4 c m; h_{2} = 0; R = D / 2 = 0.08 c m; r = 2 \times 10^{- 3} c m$
$t = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{h_{1}} - \sqrt{h_{2}}]$
$t = (\frac{0.08}{5 \times 10^{(} - 3)})^{2} \sqrt{\frac{2}{9.8}} [\sqrt{0.16} - \sqrt{0}]$
$t = 46.26$ s.
hence, it will take 46.26s to empty the tank .

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A cylindrical tank of height 0.4 m is open at the top and has a diameter 0.16 m, Water is filled in It up to a height of 0.16 m. How long it will take to empty the tank through a hole of radius 5×10−3 m at its bottom?

The correct option is A 46.26 sech1=0.4cm;h2=0;R=D/2=0.08cm;r=2×10−3cmt=Aa√2g[√h1−√h2]t=(0.085×10(−3))2√29.8[√0.16−√0]t=46.26s.hence, it will take 46.26s to empty the tank .

A cylindrical tank of height 0.4 m is open at the top and has a diameter 0.16 m, Water is filled in It up to a height of 0.16 m. How long it will take to empty the tank through a hole of radius $5 \times 10^{- 3}$ m at its bottom?