Question

A water tank which is on ground has an arrangement to maintain a constant water level of depth 60cm. Through a hole on its vertical wall at a depth of 20 cm from the free surface water comes out and reaches the ground at a certain distance. To have the same horizontal range another hole can be made at a depth of.
A. 30cm
B. 10cm
C. 40cm
D. 50cm

Answer 1

First find the speed of the water coming from the first hole using the Bernoulli’s theorem. Then use the suitable kinematic equation and calculate the range of the water that is under projectile motion. Then find a relation between the height of the second hole and speed of water at this hole by using the kinematic equations. Then again use Bernoulli’s theorem for the two heights and find the height.
Formula used:
$\rho g{{h}_{1}}+\dfrac{1}{2}\rho v_{1}^{2}=\rho g{{h}_{2}}+\dfrac{1}{2}\rho v_{2}^{2}$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
$v=u+at$
$2as={{v}^{2}}-{{u}^{2}}$

Complete answer:
Let us find the speed of the water at the depth of 60cm by applying Bernoulli's theorem at the surface of water and the depth of 60cm.
i.e. $\rho g{{h}_{1}}+\dfrac{1}{2}\rho v_{1}^{2}=\rho g{{h}_{2}}+\dfrac{1}{2}\rho v_{2}^{2}$ … (i),
Here, $\rho$ is the density of water,
${{h}_{1}}=60cm=0.6m$ is height of the surface
${{h}_{2}}=60-20=40cm=0.4m$, (height of the hole)
${{v}_{1}}=0$ (velocity of the water at the surface)
${{v}_{2}}$ is the speed of the water at the depth of 20cm.
And $g=10m{{s}^{-2}}$ (acceleration due to gravity).
Substitute the known values in equation (i).
$\Rightarrow \rho (10)(0.6)+\dfrac{1}{2}\rho (0)=\rho (10)(0.4)+\dfrac{1}{2}\rho v_{2}^{2}$
$\Rightarrow 6=4+\dfrac{1}{2}v_{2}^{2}$
$\Rightarrow v_{2}^{2}=2(2)$
$\Rightarrow {{v}_{2}}=2m{{s}^{-1}}$.
This means that the horizontal speed of the water at the depth of 20cm is $2m{{s}^{-1}}$.
The water comes down from a height of 40cm = 0.4m. Let us find the time taken to reach the ground by using the formula $s=ut+\dfrac{1}{2}a{{t}^{2}}$ for the vertical motion.
In this case, u = 0, s = 0.4m and a = g = 10$m{{s}^{-2}}$,
$\Rightarrow 0.4=(0)t+\dfrac{1}{2}(10){{t}^{2}}$
$\Rightarrow 0.8=10{{t}^{2}}$
$\Rightarrow {{t}^{2}}=0.08$
$\Rightarrow t=0.2\sqrt{2}s$.
The horizontal velocity of the water will be constant because the acceleration is only in the vertical direction.
Therefore, the horizontal range will be $d={{v}_{2}}t=2\times 0.2\sqrt{2}=0.4\sqrt{2}m$
Let the second hole be height of h and the speed of the water through this hole be v’ and the time taken to reach the ground be t’
Therefore, $d=v't'=0.4\sqrt{2}m$.
$\Rightarrow t'=\dfrac{0.4\sqrt{2}}{v'}$
Use the formula $v=u+at$ for the vertical motion.
Here, u=0, a = g = 10$m{{s}^{-2}}$ and t = t’.
$\Rightarrow v=0+10t'$
$\Rightarrow v=10t'$
$\Rightarrow v=10\left( \dfrac{0.4\sqrt{2}}{v'} \right)=\dfrac{4\sqrt{2}}{v'}$
Now use the formula $2as={{v}^{2}}-{{u}^{2}}$.
$\Rightarrow 2(10)(h)={{\left( \dfrac{4\sqrt{2}}{v'} \right)}^{2}}-0$
$\Rightarrow 20h=\dfrac{32}{v{{'}^{2}}}$ ….. (ii).
Now apply Bernoulli's theorem for the height h and 40cm.
$\Rightarrow \rho (10)(0.4)+\dfrac{1}{2}\rho {{(2)}^{2}}=\rho (10)(h)+\dfrac{1}{2}\rho v{{'}^{2}}$
$\Rightarrow 4+\dfrac{4}{2}=10h+\dfrac{1}{2}v{{'}^{2}}$
$\Rightarrow 6=10h+\dfrac{1}{2}v{{'}^{2}}$
$\Rightarrow 12-20h=+v{{'}^{2}}$
Substitute this value in (ii)
$\Rightarrow 20h=\dfrac{32}{12-20h}$
$\Rightarrow 240h-400{{h}^{2}}=32$
$\Rightarrow 400{{h}^{2}}-240h+32=0$
$\Rightarrow 25{{h}^{2}}-15h+2=0$
Now use the quadratic formula to find h.
$\Rightarrow h=\dfrac{-(-15)\pm \sqrt{{{15}^{2}}-4(25)(2)}}{2(25)}$
$\Rightarrow h=\dfrac{15\pm \sqrt{225-200}}{50}$
$\Rightarrow h=\dfrac{15\pm \sqrt{25}}{50}$
$\Rightarrow h=\dfrac{15\pm 5}{50}$
$\Rightarrow h=\dfrac{20}{50}m$ or $h=\dfrac{10}{50}m$
$\Rightarrow h=0.4m$ or $h=0.2m$
This means that there can be two heights for the same range. We know that the height of the first hole is 0.4m. Therefore, the height of the second hole is 0.2m
This means that the depth of the second hole is $0.6-0.2=0.4m=40cm$.

So, the correct answer is “Option C”.

Note:
We can also find the relation between the velocity of efflux (velocity at which the water comes out of the hole) and the height of the hole by using the relation between the maximum height of a projectile and its horizontal range.