# Soil Mechanics Questions and Answers – Consolidation Equation in Polar Coordinates

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This set of Soil Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Consolidation Equation in Polar Coordinates”.

1. The transformation from Cartesian to plane coordinates in x-direction is given by ______
a) x=rsinθ
b) x=rcosθ
c) x=rcos2θ
d) x=rsin2θ

Explanation: The transformation from Cartesian to plane coordinates in x-direction is given by,
x=rcosθ. Where, x=coordinate in Cartesian form
r=coordinate in polar form.
r2=x2+y2 and θ=tan-1(y/x).

2. The transformation from Cartesian to plane coordinates in y-direction is given by ______
a) y=rsinθ
b) y=rcosθ
c) y=rcos2θ
d) y=rsin2θ

Explanation: The transformation from Cartesian to plane coordinates in y-direction is given by,
y=rsinθ. Where, y=coordinate in Cartesian form
r=coordinate in polar form.
r2=x2+y2 and θ=tan-1(y/x).

3. The transformation from Cartesian to plane coordinates in x-direction is given by ______
a) z=rsinθ
b) z=rcosθ
c) z=z
d) z=r2sinθcosθ

Explanation: In the z-direction, since it does not make any angle with the axis, the z-direction in both the Cartesian as well as the polar coordinates is the same.
∴ z=z.
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4. In polar form the term, $$\frac{∂r}{∂x}$$ is given by______
a) $$\frac{∂r}{∂x}=sinθ$$
b) $$\frac{∂r}{∂x}=cosθsinθ$$
c) $$\frac{∂r}{∂x}=cosθ$$
d) $$\frac{∂r}{∂x}=sin2θ$$

Explanation: Since we know, r2=x2+y2 —————-(1)
∴ differentiating (1) with respect to x, we get,
$$\frac{∂r}{∂x}=\frac{x}{r}=\frac{rcosθ}{r}=cosθ.$$

5. In polar form the term, $$\frac{∂r}{∂y}$$ is given by______
a) $$\frac{∂r}{∂y}=sinθ$$
b) $$\frac{∂r}{∂y}=cosθsinθ$$
c) $$\frac{∂r}{∂y}=cosθ$$
d) $$\frac{∂r}{∂y}=sin2θ$$

Explanation: Since we know, r2=x2+y2 —————-(1)
∴ differentiating (1) with respect to y, we get,
$$\frac{∂r}{∂y}=\frac{y}{r}=\frac{rsinθ}{r}=sinθ.$$

6. In polar form the term, $$\frac{∂θ}{∂x}$$ is given by______
a) $$\frac{∂θ}{∂x}=\frac{sinθ}{r}$$
b) $$\frac{∂θ}{∂x}=-cosθsinθ$$
c) $$\frac{∂θ}{∂x}=-\frac{cosθ}{r}$$
d) $$\frac{∂θ}{∂x}=\frac{-sinθ}{r}$$

Explanation: Since we know,
θ=tan-1(y/x) —————-(1)
∴ differentiating (1) with respect to x, we get,
$$\frac{∂θ}{∂x}=\frac{-sinθ}{r}.$$

7. In polar form the term, $$\frac{∂θ}{∂y}$$ is given by______
a) $$\frac{∂θ}{∂y}=\frac{sinθ}{r}$$
b) $$\frac{∂θ}{∂y}=cosθsinθ$$
c) $$\frac{∂θ}{∂y}=\frac{cosθ}{r}$$
d) $$\frac{∂θ}{∂y}=\frac{sin2θ}{r}$$

Explanation: Since we know,
θ=tan-1(y/x) —————-(1)
∴ differentiating (1) with respect to y, we get,
$$\frac{∂θ}{∂y}=\frac{cosθ}{r}.$$

8. The partial differentiation of excess hydrostatic pressure \overline{u} as a function of r and θ with respect to x is given by _______
a) $$\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ$$
b) $$\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r} \frac{∂\overline{u}}{∂θ} cosθ$$
c) $$\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} sinθ-\frac{1}{r} \frac{∂\overline{u}}{∂θ} sinθ$$
d) $$\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} sinθ-\frac{1}{r} \frac{∂\overline{u}}{∂θ} cosθ$$

Explanation: Partially differentiating the excess hydrostatic pressure \overline{u} with respect to x,
$$\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} \frac{∂r}{∂x}+\frac{∂\overline{u}}{∂θ}\frac{∂θ}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ.$$
∴ $$\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ.$$

9. The term $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}$$ in terms of r and θ is given by _______
a) $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}-\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}$$
b) $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}+\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}$$
c) $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}-\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}$$
d) $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}+\frac{1}{c^2}\frac{∂^2 \overline{u}}{∂θ^2}$$

Explanation: The second order of the partial differentiation of excess hydrostatic pressure u with respect to x is,
$$\frac{∂^2 \overline{u}}{∂x^2}=(\frac{∂}{∂r} cosθ-\frac{1}{r} sinθ \frac{∂}{∂θ})(\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} θsinθ) ————-(1)$$
also, the second order of the partial differentiation of excess hydrostatic pressure \overline{u} with respect to y is,
$$\frac{∂^2 \overline{u}}{∂y^2}=(\frac{∂}{∂r} cosθ-\frac{1}{r} sinθ \frac{∂}{∂θ})(\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} θsinθ)$$ ———-(2)
$$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}+\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}.$$

10. In case of radial symmetry, $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}$$ is_________
a) $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}$$
b) $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}$$
c) $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=-\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}$$
d) $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=-\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}$$

Explanation: The term $$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}$$ in terms of r and θ is given by,
$$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}+\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}.$$
In case of radial symmetry, \overline{u} is independent of θ and hence we get,
$$\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}$$

11. In three dimensional consolidation of sand drain, having radial symmetry, the governing consolidation equation is _______
a) $$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})-C_{vz}\frac{∂^2 \overline{u}}{∂z^2}$$
b) $$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}$$
c) $$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}-\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}$$
d) $$\frac{∂\overline{u}}{∂t}=C_{vz} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}$$

Explanation: In three dimensional consolidation of sand drain, for the case of radial symmetry,
Cvx=Cvy =Cvr
∴ the governing consolidation equation is,
$$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}.$$

12. The radial flow part of governing consolidation equation of three dimensional consolidation having radial symmetry is _______
a) $$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})$$
b) $$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}$$
c) $$\frac{∂\overline{u}}{∂t}=C_{vz} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}$$
d) $$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})$$

Explanation: In three dimensional consolidation of sand drain, having radial symmetry, the governing consolidation equation is,
$$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2},$$ in this the radial flow part of equation is,
$$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r}).$$

13. The one dimensional flow part of governing consolidation equation of three dimensional consolidation having radial symmetry is _______
a) $$\frac{∂\overline{u}}{∂t}=C_{vr} \frac{∂\overline{u}}{∂r^2}$$
b) $$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{∂\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}$$
c) $$\frac{∂\overline{u}}{∂t}=C_{vz} (\frac{∂\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}$$
d) $$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{∂\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})$$

Explanation: In three dimensional consolidation of sand drain, having radial symmetry, the governing consolidation equation is,
$$\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{∂\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}$$, in this the one dimensional flow part of equation is,
$$\frac{∂\overline{u}}{∂t}=C_{vr} \frac{∂\overline{u}}{∂r^2}$$.

14. The equation given by Carillo in 1942 relating the degree of consolidation in one dimensional flow (Uz) and radial flow (Ur) is _______
a) (1-U)=(1-Uz)(1+Ur)
b) (1-U)=(1-Uz)(1-Ur)
c) (1-U)=(1+Uz)(1-Ur)
d) (1-U)=(1+Uz)(1+Ur)

Explanation: The equation given by Carillo in 1942 relating the degree of consolidation in one dimensional flow (Uz) and radial flow (Ur) is,
(1-U)=(1-Uz)(1-Ur). This is the combination of the one dimensional flow and radial flow parts from the governing consolidation equation.

15. The time factor Tv for the vertical flow is given by _______
a) $$T_v=\frac{C_{vz} t}{H^2}$$
b) $$T_v=\frac{-C_{rz} t}{H^2}$$
c) $$T_v=\frac{C_{vz}}{H^2}$$
d) $$T_v=\frac{C_{vz} t}{H}$$

Explanation: The time factor Tv for the vertical flow is given by,
$$T_v=\frac{C_{vz} t}{H^2},$$
Where Cvz is the coefficient of consolidation in z-direction
t=elapsed time
H=average drainage path.

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