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A water hose pipe of cross-sectional area $5 {cm}^{2}$ is used to fill a tank of 120 L. It has been observed that it takes 2 min to fill the tank. Now, a nozzle with an opening of cross-sectional area $1 {cm}^{2}$ is attached to the hose. The nozzle is held so that water is projected horizontally from a point $1 m$ above the ground. The horizontal distance over which the water can be projected is (Take $g = 10 m / s^{2}$ )

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$4.47 m$

$8.64 m$

$8 m$

$3 m$

answer is C.

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Detailed Solution

Volume flow rate $= av = \frac{V}{t}$

$∴ v_{1} = \frac{V}{A_{1} t} = \frac{120 \times 10^{- 3}}{5 \times 10^{- 4} \times 2 \times 60} Now, A_{1} v_{1} = A_{2} v_{2}$

$A_{2}$ is $\frac{1}{5}$ th of $A_{1}$ .

Hence, $v_{2}$ is five times of $v_{1}$ or $10 m / s$ .

$t_{0} = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 1}{10}} = 0.447 s R = v_{2} t_{0} = 4.47 m$

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