Question

a cylindrical vessel of 96 cm is filled upto brim it has four holes 1,2,3,4 which are respectively at the heights of 20cm,30cm,43cm and 70.cm from bottom horizontal floor the water falling at longer horizontal distance is from​

Answer 1

$\underline\mathfrak{Question \: }$

A cylindrical vessel of 96 cm is filled upto brim it has four holes 1,2,3,4 which are respectively at the heights of 20cm,30cm,43cm and 70.cm from bottom horizontal floor the water falling at longer horizontal distance is from.

$\underline\mathfrak{Answer}$

The maximum horizontal distance from the vessel comes from hole numbers 3 and 4

Now

$v = \sqrt{2gh}$

where

h is the height of the hole from the top.

$x = vt = \sqrt{2gh} \sqrt{ \frac{2(h - h)}{g} } = 2 \sqrt{h(h - h)}$

Answer 2

Given:

Height of the cylindrical vessel = 96 cm

Height of hole 1 from the bottom = 20 cm

Height of hole 2 from the bottom = 30 cm

Height of hole 3 from the bottom = 43 cm

Height of hole 4 from the bottom = 70 cm

To find:

From which floor the water has longer horizontal length.

Solution:

The vertical distance travelled by water from hole at distance h from the top:

96 - h = 1/2*gt²

t = √[2(96-h)/g]

Horizontal distance in this time will be:

x = vt

Where v = √2gh

So x = √(2gh) * √[2(96-h)/g] = 2√[h(96-h)]

Height of hole 1 from the brim =96- 20 cm = 76 cm

Height of hole 2 from the brim =96- 30 cm = 66 cm

Height of hole 3 from the brim =96- 43 cm =53 cm

Height of hole 4 from the brim =96- 70 cm = 26 cm

The maximum horizontal distance will be for hole 3 which will be x = 2√(43*53) = 95.48 cm

Therefore water coming out from hole 3 will have max horizontal distance.