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Tha area of cross-section of a large tank is $\mathrm{0.5m^{2}}$ . It has a narrow opening near the bottom having area of cross-section $\mathrm{1cm^{2}}$ . A load of $\mathrm{25kg}$ is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is $\mathrm{40cm}$ above the bottom, will be ______ $\mathrm{cms^{-1}}$ .

[ Take $\mathrm{g=10ms^{-2}}$ ]
Option: 1
300

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

Applying Bernoulli's equation at points 1 & 2 we get,

${P_{0}+0+\frac{1}{2} \rho V^{2}=\left(P_{0}+\frac{M g}{A}\right)+\rho g h+\frac{1}{2} \rho v^{2}} \\$

Speed of free surface ( $V$ ) is negligible compared to the speed at the orifice ( $v$ )

${\frac{1}{2} \rho v^{2}=\frac{M g}{A}+\rho g h} \\$

${\frac{10^{3}}{2} \times v^{2}=\frac{25 \times 10}{0.5}+10^{3} \times 10 \times 0.4}$

${\frac{10^{3}}{2} v^{2} =500+4000} \\$

${v^{2} =\frac{4500 \times 2}{10^{3}}=9 \times \frac{10^{3}}{10^{3}}} \\$

${v =3 \mathrm{~m} / \mathrm{s}} \\$

${v =300 \mathrm{~cm} / \mathrm{s}}$

Hence answer is $\mathrm{300 \mathrm{~cm} / \mathrm{s}}$

Posted by

Sanket Gandhi

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