Consider a water jar of radius \(R\) that has water filled up to height \(H\) and is kept on a stand of height \(h\) (see figure) . Through a hole of radius \(r\) \((r << R)\) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is \(x.\) Then

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Answer & Solution

Correct Answer

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

Detailed explanation

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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