There are two holes one each along the opposite sides of a wide rectangular tank. The cross section of each hole is 0.01 m2 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 the water flows out of the holes is: (density of water is 1000 kg/m3) - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Given,

$Area of cross section A = 0.01 m^{2} and density of water is 1000 kg / m^{3}$

We know that,

$F_{1} = (ρA) {v_{1}}^{2} and F_{2} = (ρA) {v_{2}}^{2}$

$Net force = F_{2} - F_{1}$

$F_{net} = (ρA) {v_{2}}^{2} - (ρA) {v_{1}}^{2} = ρA ({v_{2}}^{2} - {v_{1}}^{2})$

${v_{1}}^{2} = \sqrt{2 {gh}_{0}} and {v_{2}}^{2} = \sqrt{2 g (h_{0} + 1)}$

$∴ F_{net} = ρA 2 g = 1000 \times 0.01 \times 2 \times 10 = 200 N$

Hence the correct answer is 200.

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