Question

A fixed cylindrical vessel is filled with water up to height H. A hole is bored in the wall at a depth h from the free surface of water. For maximum horizontal range h is equal to:
$\left( A \right)H$
$\left( B \right)\dfrac{{3H}}{4}$
$\left( C \right)\dfrac{{3H}}{4}$
$\left( D \right)\dfrac{H}{4}$

Answer 1

Hint: When an incompressible fluid flows steadily, the sum of pressure energy, kinetic energy and potential energy per unit volume of the fluid remains constant. Range is the distance covered by the projectile horizontally. Use the above statements to find the horizontal range.

Formula used:
${P_{atm}} + \rho gH + \dfrac{1}{2}\rho {V^2} = $Constant
Where ${P_{atm}}$ is the pressure, $g$ is the acceleration due to gravity,$\rho $ is the density

Complete step by step answer:
A body projected in air at an angle other than ninety degrees is called projectile. Time of flight is the time taken by body to reach the projection.
Range is the distance covered by the projectile horizontally.
The vertical displacement during time of ascent is the maximum height. In common assumptions air resistance and friction are neglected.
The energy possessed by a body due its motion is termed as Kinetic energy.
It is determined that the body can do work against a restraining force action on the body before the body comes to rest.
When an incompressible fluid flows steadily, the sum of pressure energy, kinetic energy and potential energy per unit volume of the fluid remains constant. This is Bernoulli's theorem.
It relates the pressure difference between two points to change the potential energy and kinetic energy in the pipe.
Consider the velocity at the hole be $v$.
Applying the Bernoulli’s theorem on the top most layer and the hole
$ \Rightarrow {P_{atm}} + \rho gH + \dfrac{1}{2}\rho = {P_{atm}} + \rho gh + \dfrac{1}{2}\rho {V^2}$
Where ${P_{atm}}$is the pressure, $g$is the acceleration due to gravity,$\rho $is the density,
$\Rightarrow V = \sqrt {2g\left( {H - h} \right)} $
Time taken to reach the ground $t = \sqrt {\dfrac{{2h}}{g}} $
Then the horizontal range, $R = V \times t$
$ \Rightarrow \sqrt {2gH - 2gh} \times \sqrt {\dfrac{{2h}}{g}} $
Differentiating the range and equating it to zero gives the maximum horizontal range then we can say that
$\Rightarrow h = \dfrac{H}{2}$

Hence option C is the correct option.

Note: Potential energy is the stored energy inside the body, when the body is deformed.
When the body retains its original state, the potential energy stored in the body will get converted to kinetic energy.
Potential energy is the energy stored due to its location or configuration. The energy possessed by a body due its motion is termed as Kinetic energy.