There are two holes one each along the opposite sides of a wide rectangular tank. The cross section of each hole is $0.01 m^{2}$ and the vertical distance between the holes is one meter. The tank is filled with water. The net force on the tank in Newton when water flows out of the holes is: (Density of water $1000 k g / m^{3}$ )

100

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200

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300

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400

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Solution

The correct option is A $200$
force of the tank is $\frac{d}{d t} (m v)$
$= \frac{d m}{d t} . v$
$= \frac{d (s A h)}{d t} . v$
$= s A v^{2}$
$= 1000 \times 0.01 \times (\sqrt{2 g h})^{2}$
$= 1000 \times 0.01 α 2 \times 10 \times 1$
$= 200. N$

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The correct option is A 200force of the tank is ddt(mv)=dmdt.v=d(sAh)dt.v=sAv2=1000×0.01×(√2gh)2=1000×0.01α2×10×1=200.N

The correct option is A $200$
force of the tank is $\frac{d}{d t} (m v)$
$= \frac{d m}{d t} . v$
$= \frac{d (s A h)}{d t} . v$
$= s A v^{2}$
$= 1000 \times 0.01 \times (\sqrt{2 g h})^{2}$
$= 1000 \times 0.01 α 2 \times 10 \times 1$
$= 200. N$