A cylindrical vessel with its axis vertical is filled with a liquid of negligible viscosity to a height z and in its side is a small orifice at a height y, both z and y being measured from the bore. If the jet of liquid strikes the horizontal plane through the base at a distance x from bore, the maximum value of x will be

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$\frac{z}{2}$

$\frac{y}{2}$

answer is B.

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Detailed Solution

see fig 8

We know that the velocity of efflux
$V = \sqrt{2 g (z - y)} The time t talen by the jet to reach the ground is given by y = \frac{1}{2} {gt}^{2} or t = \sqrt{(\frac{2 y}{g})} x = vt = \sqrt{{2 g (z - y)}} \times \sqrt{(2 y / g)} = 2 \sqrt{(yz - y^{2})} Now, \frac{dx}{dy} = 2 \times \frac{1}{2} {(yz - y^{2})}^{- 1 / 2} (z - 2 y) The maximum value of x will be when z - 2 y = 0 or z = 2 y ∴ (x)_{max} = 2 \sqrt{(y \times 2 y - y^{2})} = 2 y = z$