Consider a water jar of radius R that h | vedclass

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Question

According to Bernoulli's principle,

$\frac{1}{2}\rho v_1^2 + \rho gh = \frac{1}{2}\rho v_2^2$

$v_1^2 + 2gh = v_2^2$

$2gH + 2gh = v_2^2\,\

Vedclass · Accepted Answer

$x\, = \,r{\left( {\frac{H}{{H + h}}} \right)^{\frac{1}{4}}}$

Vedclass · Answer

According to Bernoulli's principle,

$\frac{1}{2}\rho v_1^2 + \rho gh = \frac{1}{2}\rho v_2^2$

$v_1^2 + 2gh = v_2^2$

$2gH + 2gh = v_2^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

${a_1}{v_1} = {a_2}{v_2}$

$\pi {r^2}\sqrt {2gh}  = \pi {x^2}{v_2}$

$\frac{{{r^2}}}{{{x^2}}}\sqrt {2gh}  = {v_2}$

Substituting the value of $v_2$ in equation $(i)$

$2gH + 2gh = \frac{{{r^4}}}{{{x^4}}}2gh\,\,or,\,\,x = r{\left[ {\frac{H}{{H + h}}} \right]^{\frac{1}{4}}}$

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