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A cubical box of wine has a small spout located in one of the bottom corners. When the box is full and placed on a level surface, opening the spout results in a flow of wine with a initial speed of \mathrm{v_0}  (see figure). When the box is half empty, someone tilts it at \mathrm{45^{\circ}} so that the spout is at the lowest point (see figure). When the spout is opened the wine will flow out with a speed of

Option: 1

v_0


Option: 2

v_0 / 2


Option: 3

\mathrm{v}_0 / \sqrt{2}


Option: 4

v_0 / \sqrt[4]{2}


Answers (1)

best_answer

Initial

\mathrm{\begin{aligned} & \mathrm{a} \rho g=\frac{1}{2} \rho \mathrm{V}^2 \\ & \mathrm{~V}_0=\sqrt{2 g \mathrm{a}} \\ & \text { Now } \mathrm{V}=\sqrt{2 \frac{\mathrm{a}}{\sqrt{2}}}=\frac{\mathrm{V}_0}{\sqrt[4]{2}} \end{aligned}}

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Rakesh

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