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A tube of length L is shown in the figure. The radius of the cross section at point (1) is 2 cm and at point (2) is 1 cm , respectively. If the velocity of water entering at point (1) is 2 m/s, then the velocity of water leaving point (2) will be
For a tube with varying cross-sectional areas, the product of the area ($$A$$) and the velocity ($$v$$) is constant.
$$A_1 v_1 = A_2 v_2$$
$$A = \pi r^2$$
$$\pi r_1^2 v_1 = \pi r_2^2 v_2$$
$$r_1^2 v_1 = r_2^2 v_2$$
$$(2)^2 \cdot 2 = (1)^2 \cdot v_2$$
$$4 \cdot 2 = 1 \cdot v_2$$
$$v_2 = 8 \text{ m/s}$$
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