A tank of height $15 \mathrm{~m}$ and cross-section area $10 \mathrm{~m}^2$ is filled with water. There is a small hole of cross-section area $a$ which is much smaller than the container, located at a height of $12 \mathrm{~m}$ from the base of the container. How much force should be applied with a piston at the top level, so that the water coming out of the hole hits the ground at a distance of $16 \mathrm{~m}$ ? (Take, density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ )

• Correct answer
  A

  $233 \mathrm{kN}$
• B

  $200 \mathrm{kN}$
• C

  $320 \mathrm{kN}$
• D

  $400 \mathrm{kN}$

Solution

As, range of water jet $=16 \mathrm{~m}$. If velocity of efflux is $v$, then $v \times \sqrt{\frac{2 H}{g}}=16$, where $H=12 \mathrm{~m}$ Also, $v=\sqrt{2 g h+\frac{2\left(p-p_a\right)}{\rho}}, \rho=1000 \mathrm{~kg}-\mathrm{m}^{-3}$ and $h=3 \mathrm{~m}$ So combining these, we get $$ \begin{aligned} & \left(2 g h+\frac{2\left(p-p_a\right)}{\rho}\right)\left(\frac{2 H}{g}\right)=256 \\ & \Rightarrow \quad 2 \times 10 \times 3+\frac{2 \Delta p}{1000}=\frac{2560}{2 \times 12} \Rightarrow \Delta p=233 \mathrm{k} \frac{\mathrm{N}}{\mathrm{m}^2} \end{aligned} $$ So, force applied $=\Delta p \times A=233 \mathrm{kN}$

Hence, the correct answer is option: 'A'.