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Fluid Mechanics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsFluid Mechanics3 Marks
Question
Water is filled in a rectangular tank of size $3m \times 2m \times 1m$.
  1. Find the total force exerted by the water on the bottom surface of the tank.
  2. Consider a vertical side of area $2m \times 1m$. Take a horizontal strip of width $ox$ metre in this side, situated at a depth of $x$ metre from the surface of water. Find the force by the water on this strip.
  3. Find the torque of the force calculated in part.$(b)$ about the bottom edge of this side.
  4. Find the total force by the water on this side.
  5. Find the total torque by the water on the side about.
The bottom edge. Neglect the atmospheric pressure and take $g = 10m/s^2$.
✓

Answer

  1. The dimensions of the tank are,
Length $(L) = 3m,$ Breadth $(B) = 2m,$ Height $(H) = 1m.$
Since the tank is filled, the height of water in the tank $= H = 1m.$
The pressure of water at the bottom $= \rho \text{gH}$
$= 1000 \times 10 \times 1$
$= 10000N/m^2$
Area of the bottom of the tank $= L \times B$
$= 3m \times 2m = 6m^2$
So the force on the bottom $=$ Pressure $\times$ Area
$= 10000 \times 6$
$= 60000N.$
  1. The pressure at $\times m$ below the surface of the water $= \rho\text{gx}$
Area of a horizontal strip $= 2 \times x$
The force on this strip $F =$ Area pressure
$=2\times \text{x}^2\rho\text{gx}$
$= 2\times \text{x}^2\times 1000\times 10$
$= 20000 \text{x}^2\text{N}$
  1. The perpendicular distance of this force from the bottom of edge of this side
$= H - x = 1 - xm$
Hence the torque of the force $=$ Force $\times$ perpendicular distance
$= 20000x \times x \times (1 - x)Nm$
$= 20000 \times x(1- x) \times xNm.$
  1. The total force of the water on this side $=\int\text{Fda}, ($from $x = 0$ to $x = 1m,)$
where $F$ is the pressure at the depth $xm$ and $da =$ area of the strip
$= spg \times x (2 \times dx) \{dx$ is the width of the strip$\}$ 
​​​​​​​$=2\text{pg}\int\text{x}\cdot\text{dx}$
$=2\text{pg}\times\Big[\frac{\text{x}^2}{2}\Big]$
$= pgH^2$
$= 1000 \times 10 \times 1^2$
$= 1000N.$​​​​​​​
  1. Total Force $=\frac{1}{2}\times\text{pg}\times\text{H}\times2$
$= 10000N$
The height of this resultant force from the bottom of edge of the side
will be one$-$third of $H($height of the $C.G$. from the bottom$) =\frac{\text{H}}{3}$
Hence the torque $=10000\times\frac{1}{2}$
$=10000\times\frac{1}{2}$
$=\frac{10000}{3}\text{Nm}$

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