Question 1 |
A 5 \mathrm{~cm} long metal rod A B with initially at a uniform temperature of T_{0}{ }^{\circ} \mathrm{C}. Thereafter, temperature at both the ends are maintained at 0^{\circ} \mathrm{C}. Neglecting the heat transfer from the lateral surface of the rod, the heat transfer in the rod is governed by the one-dimensional diffusion equation \frac{\partial T}{\partial t}=D \frac{\partial^{2} T}{\partial x^{2}}, where D is the thermal diffusively of the metal, given as 1.0 \mathrm{~cm}^{2} / \mathrm{s}.
The temperature distribution in the rod is obtained as
T(x, t)=\sum_{n=13,5 \ldots .}^{\infty} C_{n} \sin \frac{n \pi x}{5} e^{-\beta n^{2} t} where x is in \mathrm{cm} measured from A to B with x=0 at A, t is s, C_{n} are constants in { }^{\circ} \mathrm{C}, \mathrm{T} is in { }^{\circ} \mathrm{C} and \beta is in \mathrm{s}^{-1}.
The value of \beta (in \mathrm{s}^{-1}, rounded off to three decimal places) is ___
The temperature distribution in the rod is obtained as
T(x, t)=\sum_{n=13,5 \ldots .}^{\infty} C_{n} \sin \frac{n \pi x}{5} e^{-\beta n^{2} t} where x is in \mathrm{cm} measured from A to B with x=0 at A, t is s, C_{n} are constants in { }^{\circ} \mathrm{C}, \mathrm{T} is in { }^{\circ} \mathrm{C} and \beta is in \mathrm{s}^{-1}.
The value of \beta (in \mathrm{s}^{-1}, rounded off to three decimal places) is ___
0.395 | |
0.125 | |
0.254 | |
0.685 |
Question 1 Explanation:
Given: Diffusion equation:
\frac{\partial T}{\partial t}=\mathrm{D} \frac{\partial^{2} \mathrm{~T}}{\partial \mathrm{x}^{2}}
And also,
D=1
\mathrm{T}(0, \mathrm{t})=0
\mathrm{T}(5,5)=0
T(x, 0)=T_{0}
We know, solution of equation is given by,
T(x, t)=\left(C_{1} \cos \alpha x+C_{2} \sin \alpha x\right) C_{3} e^{-\alpha^{2} t} \;\;\; ...(1)
For \mathrm{T}(0, t)=0
\left(C_{1}+0\right) C_{3} e^{-\alpha^{2} t}=0
\Rightarrow \mathrm{C}_{1}=0
For \mathrm{T}(\mathrm{s}, \mathrm{t})=0
\left(C_{2} \sin \alpha 5\right) C_{3} \mathrm{e}^{-25 t}=0
\Rightarrow \sin 5 \alpha=0
\Rightarrow \alpha=\frac{\mathrm{n} \pi}{5}
Now, from equation (1)
T(x, t)=C_{2} C_{3} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2}}{25} t}
=b_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2} t}{25}}
Generalize solution
T(x, t)=\sum_{n=1}^{\infty} b_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2} t}{25}}
For \mathrm{T}(\mathrm{x}, 0)=\mathrm{T}_{0}
\Rightarrow T_{0}=\sum_{n=1}^{\infty} b_{n} \sin \left(\frac{n \pi x}{5}\right)
From fourier series,
b_{n}=0 for even value of n
\therefore \mathrm{T}(\mathrm{x}, \mathrm{t})=\sum_{\mathrm{n}=13,5 \ldots .}^{\infty} \mathrm{b}_{\mathrm{n}} \sin \left(\frac{\mathrm{n} \pi \mathrm{x}}{5}\right) \mathrm{e}^{-\frac{\mathrm{n}^{2} \pi^{2} \mathrm{t}}{25}}
Given: T(x, t)=\sum_{n=13,5 \ldots .}^{\infty} C_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\beta n^{2} t}
On comparison;
\beta=\frac{\pi^{2}}{25}=0.395 \mathrm{sec}^{-1}
\frac{\partial T}{\partial t}=\mathrm{D} \frac{\partial^{2} \mathrm{~T}}{\partial \mathrm{x}^{2}}
And also,
D=1
\mathrm{T}(0, \mathrm{t})=0
\mathrm{T}(5,5)=0
T(x, 0)=T_{0}
We know, solution of equation is given by,
T(x, t)=\left(C_{1} \cos \alpha x+C_{2} \sin \alpha x\right) C_{3} e^{-\alpha^{2} t} \;\;\; ...(1)
For \mathrm{T}(0, t)=0
\left(C_{1}+0\right) C_{3} e^{-\alpha^{2} t}=0
\Rightarrow \mathrm{C}_{1}=0
For \mathrm{T}(\mathrm{s}, \mathrm{t})=0
\left(C_{2} \sin \alpha 5\right) C_{3} \mathrm{e}^{-25 t}=0
\Rightarrow \sin 5 \alpha=0
\Rightarrow \alpha=\frac{\mathrm{n} \pi}{5}
Now, from equation (1)
T(x, t)=C_{2} C_{3} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2}}{25} t}
=b_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2} t}{25}}
Generalize solution
T(x, t)=\sum_{n=1}^{\infty} b_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2} t}{25}}
For \mathrm{T}(\mathrm{x}, 0)=\mathrm{T}_{0}
\Rightarrow T_{0}=\sum_{n=1}^{\infty} b_{n} \sin \left(\frac{n \pi x}{5}\right)
From fourier series,
b_{n}=0 for even value of n
\therefore \mathrm{T}(\mathrm{x}, \mathrm{t})=\sum_{\mathrm{n}=13,5 \ldots .}^{\infty} \mathrm{b}_{\mathrm{n}} \sin \left(\frac{\mathrm{n} \pi \mathrm{x}}{5}\right) \mathrm{e}^{-\frac{\mathrm{n}^{2} \pi^{2} \mathrm{t}}{25}}
Given: T(x, t)=\sum_{n=13,5 \ldots .}^{\infty} C_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\beta n^{2} t}
On comparison;
\beta=\frac{\pi^{2}}{25}=0.395 \mathrm{sec}^{-1}
Question 2 |
The steady-state temperature distribution in a square plate A B C D is governed by the 2-dimensional Laplace equation. The side A B is kept at a temperature of 100{ }^{\circ} \mathrm{C} and the other three sides are kept at a temperature of 0{ }^{\circ} \mathrm{C}. Ignoring the effect of discontinuities in the boundary conditions at the corners, the steady-state temperature at the center of the plate is obtained as T_0 ^{\circ} \mathrm{C}. Due to symmetry, the steady-state temperature at the center will be same \left(\mathrm{T_0}{ }^{\circ} \mathrm{C}\right), when any one side of the square is kept at a temperature of 100{ }^{\circ} \mathrm{C} and the remaining three sides are kept at a temperature of 0{ }^{\circ} \mathrm{C}. Using the principle of superposition, the value of T_0 is ___ (rounded off to two decimal places).
25.35 | |
14.25 | |
19.92 | |
28.25 |
Question 2 Explanation:
We know, Laplace equation in two dimensional on a unit square with Dirichlet boundary condition :
\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0 \lt (x, y) \lt 1
This Laplace equation is called harmonic function. Solution of Laplace equation,
u(x, y)=\left(C_{1} \cos \beta x+C_{2} \sin \beta x\right)\left(C_{3} e^{\text {By }}+C_{4} e^{-\beta y}\right)
Let square A B C D is unit square.
Now, given
\begin{aligned} & u(x, 0)=0 \\ & u(0, y)=0 \\ & u(1, y)=0 \\ & u(x, 1)=100 \end{aligned}
For u(x, 0)=0
\quad 0=\left(C_{1} \cos \beta x+0\right)(C_3+C_4)
\Rightarrow \quad \mathrm{C}_{3}=-\mathrm{C}_{4}
For u(0, y)=0;
0=\left(C_{1}+0\right)\left(C_{3} e^{\beta y}+C_{4} e^{-\beta y}\right)
\Rightarrow \quad \mathrm{C}_{1}=0
Now,
\begin{aligned} u(x, y) & =0+C_{2} \sin \beta x\left(C_{3} e^{\beta y}+C_{4} e^{-\beta y}\right) \\ & =a_{x} \sin \beta x\left(e^{\beta y}-e^{-\beta y}\right) \quad ...(1) \end{aligned}
\left[\because \mathrm{C}_{3}=-\mathrm{C}_{4}\right. and Let \left.\mathrm{C}_{2} \mathrm{C}_{3}=\mathrm{a}_{\mathrm{x}}\right]
For u(1, y)=0
0=a_{x} \sin \beta x\left(e^{\beta y}-e^{\beta y}\right)
\Rightarrow \quad \beta=\mathrm{n} \pi
For u(x, 1)=100
a_{x} \sin n\pi x\left(e^{n\pi}-e^{-n \pi}\right)=100
a_{x}=\frac{100}{\sin n\pi x\left(e^{n\pi}-e^{-n \pi}\right)}
From eqn. (1), we get
\begin{gathered} u(x, y)=\sum_{n=1}^{\infty }\frac{100}{\sin n \pi x\left(e^{n\pi}-e^{-n\pi}\right)} \times \sin n\pi x\left(e^{n \pi y}-e^{- n\pi y}\right) \\ =\sum_{n=1}^{\infty }\frac{100}{\left(e^{n\pi}-e^{-n \pi}\right)}\left[e^{n\pi y}-e^{-n\pi y}\right] \end{gathered}
Now, at mid point (x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)
u\left(\frac{1}{2}, \frac{1}{2}\right)=T_{0} \text { [given] }
\begin{aligned} T_0&=\sum_{n=1}^{\infty }\frac{100}{\left(\mathrm{e}^{n\pi}-\mathrm{e}^{-n\pi}\right)}\left[\mathrm{e}^{n\pi / 2}-\mathrm{e}^{-n\pi / 2}\right] \\ &=100\left [ \left ( \frac{\mathrm{e}^{\pi / 2}-\mathrm{e}^{-\pi / 2}}{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}} \right ) +\left ( \frac{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}}{\mathrm{e}^{2\pi}-\mathrm{e}^{-2\pi}} \right )+...\right ]\\ &=100\left [ \frac{1}{(\mathrm{e}^{\pi / 2}-\mathrm{e}^{-\pi / 2})}+\frac{1}{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}}+\frac{1}{\mathrm{e}^{3\pi / 2}-\mathrm{e}^{-3\pi / 2}} +...(neglect) \right ]\\ &=100[0.199+0.043+0.008]\\ &\approx 25^{\circ}C \end{aligned}
\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0 \lt (x, y) \lt 1
This Laplace equation is called harmonic function. Solution of Laplace equation,
u(x, y)=\left(C_{1} \cos \beta x+C_{2} \sin \beta x\right)\left(C_{3} e^{\text {By }}+C_{4} e^{-\beta y}\right)
Let square A B C D is unit square.
Now, given
\begin{aligned} & u(x, 0)=0 \\ & u(0, y)=0 \\ & u(1, y)=0 \\ & u(x, 1)=100 \end{aligned}
For u(x, 0)=0
\quad 0=\left(C_{1} \cos \beta x+0\right)(C_3+C_4)
\Rightarrow \quad \mathrm{C}_{3}=-\mathrm{C}_{4}
For u(0, y)=0;
0=\left(C_{1}+0\right)\left(C_{3} e^{\beta y}+C_{4} e^{-\beta y}\right)
\Rightarrow \quad \mathrm{C}_{1}=0
Now,
\begin{aligned} u(x, y) & =0+C_{2} \sin \beta x\left(C_{3} e^{\beta y}+C_{4} e^{-\beta y}\right) \\ & =a_{x} \sin \beta x\left(e^{\beta y}-e^{-\beta y}\right) \quad ...(1) \end{aligned}
\left[\because \mathrm{C}_{3}=-\mathrm{C}_{4}\right. and Let \left.\mathrm{C}_{2} \mathrm{C}_{3}=\mathrm{a}_{\mathrm{x}}\right]
For u(1, y)=0
0=a_{x} \sin \beta x\left(e^{\beta y}-e^{\beta y}\right)
\Rightarrow \quad \beta=\mathrm{n} \pi
For u(x, 1)=100
a_{x} \sin n\pi x\left(e^{n\pi}-e^{-n \pi}\right)=100
a_{x}=\frac{100}{\sin n\pi x\left(e^{n\pi}-e^{-n \pi}\right)}
From eqn. (1), we get
\begin{gathered} u(x, y)=\sum_{n=1}^{\infty }\frac{100}{\sin n \pi x\left(e^{n\pi}-e^{-n\pi}\right)} \times \sin n\pi x\left(e^{n \pi y}-e^{- n\pi y}\right) \\ =\sum_{n=1}^{\infty }\frac{100}{\left(e^{n\pi}-e^{-n \pi}\right)}\left[e^{n\pi y}-e^{-n\pi y}\right] \end{gathered}
Now, at mid point (x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)
u\left(\frac{1}{2}, \frac{1}{2}\right)=T_{0} \text { [given] }
\begin{aligned} T_0&=\sum_{n=1}^{\infty }\frac{100}{\left(\mathrm{e}^{n\pi}-\mathrm{e}^{-n\pi}\right)}\left[\mathrm{e}^{n\pi / 2}-\mathrm{e}^{-n\pi / 2}\right] \\ &=100\left [ \left ( \frac{\mathrm{e}^{\pi / 2}-\mathrm{e}^{-\pi / 2}}{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}} \right ) +\left ( \frac{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}}{\mathrm{e}^{2\pi}-\mathrm{e}^{-2\pi}} \right )+...\right ]\\ &=100\left [ \frac{1}{(\mathrm{e}^{\pi / 2}-\mathrm{e}^{-\pi / 2})}+\frac{1}{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}}+\frac{1}{\mathrm{e}^{3\pi / 2}-\mathrm{e}^{-3\pi / 2}} +...(neglect) \right ]\\ &=100[0.199+0.043+0.008]\\ &\approx 25^{\circ}C \end{aligned}
Question 3 |
The function f(x, y) satisfies the Laplace equation
\triangledown ^2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of f(0, 0) is:
\triangledown ^2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of f(0, 0) is:
0 | |
2 | |
3 | |
1 |
Question 3 Explanation:
According to given condition given function f(x,y) is nothing but constant function i.e. f(x,y)=3 because this is the only function whose value is 3 at any point on the boundary of unit circle and it is also satisfying Laplace equation, so
f(0,0)=3
f(0,0)=3
Question 4 |
The Fourier cosine series of a function is given by:
f(x)=\sum_{n=0}^{\infty }f_n\cos nx
For f(x)=\cos ^4x, the numerical value of f_4+f_5 is ______ . (round off to three decimal places)
f(x)=\sum_{n=0}^{\infty }f_n\cos nx
For f(x)=\cos ^4x, the numerical value of f_4+f_5 is ______ . (round off to three decimal places)
2.255 | |
0.652 | |
0.125 | |
1.585 |
Question 4 Explanation:
\begin{aligned}
\cos ^4 x&=\cos ^2 x\cos ^2 x\\
&=\left ( \frac{1+\cos 2x}{2} \right )\left ( \frac{1+\cos 2x}{2} \right )\\
&=\frac{1}{4}(1+2 \cos 2x+ \cos ^2 2x)\\
&=\frac{1}{4}(1+2 \cos 2x+ \frac{(1+\cos 4x)}{2})\\
&=\frac{1}{4}+\frac{\cos 2x}{2}+\frac{1}{8}+\frac{\cos 4x}{8}\\
&=\frac{3}{8}+\frac{\cos 2x}{2}+\frac{\cos 4x}{8}\\
f_4&=\frac{1}{8}=0.125\\
f_5&=0\\
\rightarrow f_4+f_5&=0.125
\end{aligned}
Question 5 |
Consider the following expression:
z=\sin(y+it)+\cos(y-it)
where z, y, and t are variables, and i=\sqrt{-1} is a complex number. The partial differential equation derived from the above expression is
z=\sin(y+it)+\cos(y-it)
where z, y, and t are variables, and i=\sqrt{-1} is a complex number. The partial differential equation derived from the above expression is
\frac{\partial^2 z}{\partial t^2}+\frac{\partial^2 z}{\partial y^2}=0 | |
\frac{\partial^2 z}{\partial t^2}-\frac{\partial^2 z}{\partial y^2}=0 | |
\frac{\partial z}{\partial t}-i\frac{\partial z}{\partial y}=0 | |
\frac{\partial z}{\partial t}+i\frac{\partial z}{\partial y}=0 |
Question 5 Explanation:
\begin{aligned}
z&=\sin(y+it)+ \cos (y-it)\\
\frac{\partial z}{\partial y}&=\cos (y+it)-\sin (y-it)\\
\frac{\partial ^2 z}{\partial ^2 y^2}&=-\sin(y+it)- \cos (y-it)\\
\frac{\partial ^2 z}{\partial ^2 y^2}&=-z \;\;...(i)\\
\frac{\partial z}{\partial t}&=i \cos (y+it)+i\sin (y-it)\\
\frac{\partial ^2 z}{\partial ^2 t^2}&=+\sin(y+it)+ \cos (y-it)\\
\frac{\partial ^2 z}{\partial ^2 t^2}&=z\;\;...(ii)\\
&\text{Adding (i) and (ii)}\\
&\frac{\partial ^2 z}{\partial ^2 y^2}+\frac{\partial ^2 z}{\partial ^2 t^2}=0
\end{aligned}
There are 5 questions to complete.