Water is filled in a uniform container of area of cross-section A. A hole of cross section area $a$ (<<A) is made in the container at the height of 20m above the base. Water stream out and hits a small block on surface at some distance from container. The block is moved on the surface in such a way that the stream always hits on the block. The initial velocity of the block (in m/s) is (Given $\frac{a}{A} = \frac{1}{20}$ take $g = 10 m s^{- 2}$ )

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

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

velocity of efflux $v = \sqrt{2 g y}$
Range $x$ $= \sqrt{2 g y} \times \sqrt{\frac{2 h}{g}}$
Question Image
The velocity of block must be $\frac{d x}{d t}$
$v_{b} = \frac{d x}{d t} = \sqrt{\frac{2 h}{g}} \times \sqrt{2 g} \times \frac{1}{2 \sqrt{y}} \frac{d y}{d t}$
$= \frac{\sqrt{h}}{\sqrt{y}} \frac{d y}{d t}$ ---------------(i)
Using equation of continuity
$A \frac{d y}{d t} = a \sqrt{2 g y}$ --------------(ii)
From equation (i) and (ii)
$v_{b} = \sqrt{\frac{h}{y}} \times \frac{a}{A} \sqrt{2 g y}$

$= \sqrt{2 g h} \times \frac{a}{A} = 20 \times \frac{1}{20} = 1 m s^{- 1}$