Cylindrical tank, having a cross-sectional area S, stands fixed on a horizontal surface (Figure). In its wall is a small hole, cross-sectional area of which is $S_{1}$ , is located at a distance $h_{1}$ from the surface of water in the tank and the $h_{2}$ from the bottom. Cross-sectional area of the jet emanating from the holes at the place of its falling on the horizontal surface is $S_{2}$ . $h_{1} + h_{2} = c o n s t a n t . (S > > S_{1})$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$S_{2} = S_{1} \sqrt{\frac{h_{2}}{h_{1} + h_{2}}}$

$S_{2}$ will increase if $h_{1}$ is increased

$S_{2}$ will increase if $h_{2}$ is increased

$S_{2} = S_{1} \sqrt{\frac{h_{1}}{h_{1} + h_{2}}}$

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Applying Bernoulli’s theorem

$\frac{P_{0} + ρ g h_{1}}{ρ g} = \frac{P_{0}}{ρ g} + \frac{v_{1}^{2}}{2 g} = \frac{P_{0}}{ρ g} + \frac{v_{2}^{2}}{2 g} - h_{2} v_{1}^{2} = 2 g h_{1} v_{2}^{2} = 2 g (h_{1} + h_{2}) S_{1} v_{1} = S_{2} v_{2} \Rightarrow S_{2} = S_{1} \sqrt{\frac{h_{1}}{h_{1} + h_{2}}} = S_{1} \sqrt{\frac{1}{1 + h_{2} / h_{1}}}$