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Physics-

General

Easy

Question

For the arrangement shown in figure the time interval after which the water jet ceases to cross the wall (area of cross section of tank is A and orifice is ‘a’)

$\frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{\frac{x^{2}}{4 y}})$
$\frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{\frac{4 y}{x^{2}}})$
$\frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{\frac{4 y^{2}}{x}})$
null

The correct answer is: $fraction numerator A over denominator a end fraction square root of fraction numerator 2 over denominator g end fraction end root open parentheses square root of H minus square root of fraction numerator x to the power of 2 end exponent over denominator 4 y end fraction end root close parentheses$

Velocity of efflux = $square root of 2 g h end root$
Find ‘h’ to have range of ejected water $greater or equal than x$
$rightwards double arrow x equals square root of 2 g h end root. square root of fraction numerator 2 y over denominator g end fraction end root rightwards double arrow h equals fraction numerator x to the power of 2 end exponent over denominator 4 y to the power of 2 end exponent end fraction$
time taken by liquid to drain out from H to h is $equals fraction numerator A over denominator a end fraction square root of fraction numerator 2 over denominator g end fraction end root open square brackets square root of H minus square root of h close square brackets$

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