Question
A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is $528{dm}^{3}(dm$ : decimeter). The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at 70 cm below the top of water level, then horizontal range of water falling on the ground in the beginning is $\_\_\_\_$ cm.
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Solution
Sol. [Key Concept: Efflux speed (Torricelli) $v = \sqrt{2g\ h}$ at depth $h$; then the horizontal range $R = v\sqrt{\frac{2H}{g}}$ using the fall height $H$.]
$${\ v = \pi r^{2}h \Rightarrow 528 \times 10^{- 3} = \pi(0.4)^{2}hh = \frac{528 \times 10^{- 3} \times 7}{(0.4)^{2} \times 22} = 1.05\text{ }m = 105\text{ }cm }{\rightarrow t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times (105 + 35) \times 10^{- 2}}{10}} }{= \sqrt{\frac{2(140) \times 10^{- 2}}{10}}\sqrt{28} \times 10^{- 1}\sec }{R = ut = \sqrt{2 \times 10 \times 70 \times 10^{- 2}} \times \sqrt{28} \times 10^{- 1} }{= (\sqrt{14})(\sqrt{28}) \times 10^{- 1} = 14\sqrt{2} \times 10^{- 1}\text{ }m }{R = 140\sqrt{2}\text{ }cm }$$
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