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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Fluid Mechanics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Fluid Mechanics

A rectangular tank of dimensions 3 m × 2 m × 1 m is filled with...

A rectangular tank of dimensions 3 m×2 m×1 m3 \text{ m} \times 2 \text{ m} \times 1 \text{ m} is filled with water. Consider a vertical side of area 2 m×1 m2 \text{ m} \times 1 \text{ m} and a horizontal strip of width δx\delta x metre within this side, located at a depth of xx metre from the water surface. Neglecting atmospheric pressure and using g=10 m s−2g = 10 \text{ m s}^{-2}, determine the following values respectively:
(i) the total force exerted by the water on the bottom surface of the tank,
(ii) the force by the water on the horizontal strip,
(iii) the torque of the force on the strip about the bottom edge of this side,
(iv) the total force by the water on this side, and
(v) the total torque by the water on the side about the bottom edge.

Options

  1. A60000 N60000 \text{ N}; 10000xδx N10000 x \delta x \text{ N}; 10000x(1−x)δx N-m10000 x(1-x) \delta x \text{ N-m}; 5000 N5000 \text{ N}; 50003 N-m\dfrac{5000}{3} \text{ N-m}
  2. B60000 N60000 \text{ N}; 20000xδx N20000 x \delta x \text{ N}; 20000x(1−x)δx N-m20000 x(1-x) \delta x \text{ N-m}; 10000 N10000 \text{ N}; 100003 N-m\dfrac{10000}{3} \text{ N-m}
  3. C60000 N60000 \text{ N}; 20000xδx N20000 x \delta x \text{ N}; 20000(1−x)δx N-m20000 (1-x) \delta x \text{ N-m}; 10000 N10000 \text{ N}; 100003 N-m\dfrac{10000}{3} \text{ N-m}
  4. D30000 N30000 \text{ N}; 20000xδx N20000 x \delta x \text{ N}; 20000x(1−x)δx N-m20000 x(1-x) \delta x \text{ N-m}; 20000 N20000 \text{ N}; 200003 N-m\dfrac{20000}{3} \text{ N-m}

Correct answer

B. 60000 N60000 \text{ N}; 20000xδx N20000 x \delta x \text{ N}; 20000x(1−x)δx N-m20000 x(1-x) \delta x \text{ N-m}; 10000 N10000 \text{ N}; 100003 N-m\dfrac{10000}{3} \text{ N-m}

Step-by-step solution

The density of water is ρ=1000 kg m−3\rho = 1000 \text{ kg m}^{-3} and g=10 m s−2g = 10 \text{ m s}^{-2}.

The pressure at the bottom of the tank is P=ρgH=1000×10×1=10000 N m−2P = \rho g H = 1000 \times 10 \times 1 = 10000 \text{ N m}^{-2}.
The area of the bottom surface is A=3×2=6 m2A = 3 \times 2 = 6 \text{ m}^2.
The total force exerted by the water on the bottom surface is Fbottom=P×A=10000×6=60000 NF_{bottom} = P \times A = 10000 \times 6 = 60000 \text{ N}.

The vertical side has dimensions 2 m×1 m2 \text{ m} \times 1 \text{ m}, so its width is 2 m2 \text{ m} and height is 1 m1 \text{ m}.
The pressure at a depth xx is P(x)=ρgx=10000x N m−2P(x) = \rho g x = 10000 x \text{ N m}^{-2}.
The area of the horizontal strip of width δx\delta x is dA=2δx m2dA = 2 \delta x \text{ m}^2.
The force on the strip is dF=P(x)dA=10000x×2δx=20000xδx NdF = P(x) dA = 10000 x \times 2 \delta x = 20000 x \delta x \text{ N}.

The distance of the strip from the bottom edge is (1−x) m(1 - x) \text{ m}.
The torque of the force on the strip about the bottom edge is dτ=dF×(1−x)=20000x(1−x)δx N-md\tau = dF \times (1 - x) = 20000 x(1 - x) \delta x \text{ N-m}.

The total force on the vertical side is obtained by integrating dFdF from x=0x = 0 to x=1x = 1:
Fside=∫0120000x dx=20000[x22]01=10000 NF_{side} = \int_{0}^{1} 20000 x \, dx = 20000 \left[ \dfrac{x^2}{2} \right]_{0}^{1} = 10000 \text{ N}.

The total torque about the bottom edge is obtained by integrating dτd\tau from x=0x = 0 to x=1x = 1:
τside=∫0120000(x−x2) dx=20000[x22−x33]01=20000(12−13)=100003 N-m\tau_{side} = \int_{0}^{1} 20000 (x - x^2) \, dx = 20000 \left[ \dfrac{x^2}{2} - \dfrac{x^3}{3} \right]_{0}^{1} = 20000 \left( \dfrac{1}{2} - \dfrac{1}{3} \right) = \dfrac{10000}{3} \text{ N-m}.
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