Total pressure on a 1m × 1m gate immersed vertically at a depth of 2 m below the free water surface will be A. 1000 kg B. 4000 kg C. 2000 kg D. 2500 kg

[amp_mcq option1=”1000 kg” option2=”4000 kg” option3=”2000 kg” option4=”2500 kg” correct=”option1″]

The correct answer is $\boxed{\text{C}}$.

The pressure at a depth of $h$ below the free surface of water is given by the equation $P = \rho g h$, where $\rho$ is the density of water and $g$ is the acceleration due to gravity. The density of water is approximately $1000 \frac{kg}{m^3}$ and the acceleration due to gravity is approximately $9.8 \frac{m}{s^2}$. Therefore, the pressure on a 1m × 1m gate immersed vertically at a depth of 2 m below the free water surface is $P = (1000 \frac{kg}{m^3})(9.8 \frac{m}{s^2})(2 m) = 19600 \frac{N}{m^2}$. This is equivalent to a force of $19600 \frac{N}{m^2} \times 1 m \times 1 m = 19600 N$ on the gate.

Option A is incorrect because it is the weight of the gate. The weight of the gate is given by the equation $W = mg$, where $m$ is the mass of the gate and $g$ is the acceleration due to gravity. The mass of the gate is given by the equation $m = \rho V$, where $\rho$ is the density of the gate and $V$ is the volume of the gate. The density of the gate is approximately $2700 \frac{kg}{m^3}$ and the volume of the gate is given by the equation $V = LWH$, where $L$ is the length of the gate, $W$ is the width of the gate, and $H$ is the height of the gate. The length, width, and height of the gate are not given, so the weight of the gate cannot be calculated.

Option B is incorrect because it is the pressure at a depth of 1 m below the free water surface. The pressure at a depth of 1 m below the free water surface is given by the equation $P = \rho g h = (1000 \frac{kg}{m^3})(9.8 \frac{m}{s^2})(1 m) = 9800 \frac{N}{m^2}$.

Option D is incorrect because it is the pressure at a depth of 2 m below the free water surface times 2. The pressure at a depth of 2 m below the free water surface is given by the equation $P = \rho g h = (1000 \frac{kg}{m^3})(9.8 \frac{m}{s^2})(2 m) = 19600 \frac{N}{m^2}$. Therefore, the pressure at a depth of 2 m below the free water surface times 2 is $19600 \frac{N}{m^2} \times 2 = 39200 \frac{N}{m^2}$.