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(D) Given, the rate of flow of water, $Q = 12 \pi 	ext{ litre/minute}$.Converting this to $SI$ units $(m^3/s)$:$Q = \frac{12 \pi 	imes 10^{-3} 	ext

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(D) Given, the rate of flow of water, $Q = 12 \pi 	ext{ litre/minute}$.Converting this to $SI$ units $(m^3/s)$:$Q = \frac{12 \pi 	imes 10^{-3} 	ext{ m}^3}{60 	ext{ s}} = 0.2 \pi 	imes 10^{-3} 	ext{ m}^3/s = 2 \pi 	imes 10^{-4} 	ext{ m}^3/s$.The diameter of the pipe is $d = 2 	ext{ cm} = 0.02 	ext{ m}$, so the radius is $r = 0.01 	ext{ m} = 10^{-2} 	ext{ m}$.The cross-sectional area $A$ is given by $A = \pi r^2 = \pi (10^{-2})^2 = \pi 	imes 10^{-4} 	ext{ m}^2$.Using the equation of continuity, $Q = A 	imes v$, where $v$ is the velocity of water:$2 \pi 	imes 10^{-4} = (\pi 	imes 10^{-4}) 	imes v$.Solving for $v$:$v = \frac{2 \pi 	imes 10^{-4}}{\pi 	imes 10^{-4}} = 2 	ext{ m/s}$.

Water flows through a horizontal pipe of variable cross-section at the rate of $12 \pi$ litre per minute. The velocity of the water at the point where the diameter of the pipe becomes $2 \text{ cm}$ is (in $\text{ m/s}$)

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