A square gate of size $1 \mathrm{~m} \times 1 \mathrm{~m}$ is hinged at its mid-point. A fluid of density $\rho$ fills the space to the left of the gate. The force F required to hold the gate stationary is

• A

  $\frac{\rho g}{3}$
• B

  $\frac{\rho g}{2}$
• Correct answer
  C

  $\frac{\rho g}{6}$
• D

  $\frac{\rho g}{8}$

Solution

The net force acting on the gate element of width dy at a depth y from the surface of the fluid, is $$ \begin{aligned} \mathrm{dy} &=\left(\mathrm{p}_{0}+\rho_{\mathrm{g}} \mathrm{y}-\mathrm{p}_{0}\right) \times 1 \mathrm{dy} \\ &=\rho \mathrm{gydy} \end{aligned} $$ Torque about the hinge is $$ \mathrm{d} \tau=\operatorname{pgydy} \times\left(\frac{1}{2}-\mathrm{y}\right) $$ Net torque experience. by the gate is $$ \begin{aligned} \tau_{\text {net }} &=\int \mathrm{d} \tau+\mathrm{F} \times \frac{1}{2} \\ &=\int_{0}^{1} \rho g \mathrm{ydy}\left(\frac{1}{2}-\mathrm{y}\right)+\mathrm{F} \times \frac{1}{2}=0 \\ \Rightarrow & \mathrm{F}=\frac{\rho \mathrm{g}}{6} \end{aligned} $$ i.e., The force $F$ required to hold the gate stationary is $\frac{p g}{a}$

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