JEE Main 2014 · 12 Apr · Shift 1 · Physics Q9

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PhysicsSingle correct+4 / -1Question 9 of 90

A cylindrical vessel of cross-section A contains water to a height h. There is a hole in the bottom of radius 'a'. The time in which it will be emptied is :

\dfrac{2A}{\pi a^2}\sqrt{\dfrac hg}

\dfrac{\sqrt2A}{\pi a^2}\sqrt{\dfrac hg}

\dfrac{2\sqrt2A}{\pi a^2}\sqrt{\dfrac hg}

\dfrac A{\sqrt2\,\pi a^2}\sqrt{\dfrac hg}

Answer and worked solution

Official answer: B

$-A\dfrac{dy}{dt}=\pi a^2\sqrt{2gy}$ . Integrating from $h$ to 0:
$t=\frac A{\pi a^2}\cdot\frac{2\sqrt h}{\sqrt{2g}}=\frac{\sqrt2A}{\pi a^2}\sqrt{\frac hg}.$

Topics:Bernoulli's principle Continuity equation

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 9