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 |
Cholesky decomposition is carried out on the following square matrix [\mathrm{A}].
[A]=\left[\begin{array}{cc} 8 & -5 \\ -5 & a_{22} \end{array}\right]
Let \mathrm{I}_{\mathrm{ij}} and \mathrm{a}_{\mathrm{ij}} be the (i, j)^{\text {th }} elements of matrices [L] and [A], respectively. If the element I_{22} of the decomposed lower triangular matrix [\mathrm{L}] is 1.968 , what is the value (rounded off to the nearest integer) of the element a_{22} ?
[A]=\left[\begin{array}{cc} 8 & -5 \\ -5 & a_{22} \end{array}\right]
Let \mathrm{I}_{\mathrm{ij}} and \mathrm{a}_{\mathrm{ij}} be the (i, j)^{\text {th }} elements of matrices [L] and [A], respectively. If the element I_{22} of the decomposed lower triangular matrix [\mathrm{L}] is 1.968 , what is the value (rounded off to the nearest integer) of the element a_{22} ?
5 | |
7 | |
9 | |
11 |
Question 2 Explanation:
We know, cholesky decomposition,
A=LL^{T}
Where, L= lower tringular matrix
\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{cc}L_{11} & 0 \\ L_{21} & L_{22}\end{array}\right]\left[\begin{array}{cc}\mathrm{L}_{11} & L_{21} \\ 0 & L_{22}\end{array}\right]
\left[\begin{array}{cc}8 & -5 \\ -5 & \mathrm{a}_{22}\end{array}\right]=\left[\begin{array}{cc}\mathrm{L}_{11}^{2} & \mathrm{~L}_{11} \mathrm{~L}_{21} \\ \mathrm{~L}_{11} \mathrm{~L}_{21} & \mathrm{~L}_{21}^{2}+\mathrm{L}_{22}^{2}\end{array}\right]
On comparison on both sides,
\mathrm{L}_{11}=\sqrt{8}=2 \sqrt{2}
and,\;\; L_{11} L_{21}=-5
\mathrm{L}_{21}=\frac{-5}{2 \sqrt{2}}
and,\;\; a_{22}=L_{21}^{2}+L_{22}^{2}
=\left(\frac{-5}{2 \sqrt{2}}\right)^{2}+1.968^{2}
=6.998 \approx 7
A=LL^{T}
Where, L= lower tringular matrix
\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{cc}L_{11} & 0 \\ L_{21} & L_{22}\end{array}\right]\left[\begin{array}{cc}\mathrm{L}_{11} & L_{21} \\ 0 & L_{22}\end{array}\right]
\left[\begin{array}{cc}8 & -5 \\ -5 & \mathrm{a}_{22}\end{array}\right]=\left[\begin{array}{cc}\mathrm{L}_{11}^{2} & \mathrm{~L}_{11} \mathrm{~L}_{21} \\ \mathrm{~L}_{11} \mathrm{~L}_{21} & \mathrm{~L}_{21}^{2}+\mathrm{L}_{22}^{2}\end{array}\right]
On comparison on both sides,
\mathrm{L}_{11}=\sqrt{8}=2 \sqrt{2}
and,\;\; L_{11} L_{21}=-5
\mathrm{L}_{21}=\frac{-5}{2 \sqrt{2}}
and,\;\; a_{22}=L_{21}^{2}+L_{22}^{2}
=\left(\frac{-5}{2 \sqrt{2}}\right)^{2}+1.968^{2}
=6.998 \approx 7
Question 3 |
Two vectors \left[\begin{array}{llll}2 & 1 & 0 & 3\end{array}\right]^{\top} and \left[\begin{array}{llll}1 & 0 & 1 & 2\end{array}\right]^{\top} belong to the null space of a 4 \times 4 matrix of rank 2 . Which one of the following vectors also belongs to the null space?
\left[\begin{array}{llll}1 & 1 & -1 & 1\end{array}\right]^{\top} | |
\left[\begin{array}{llll}2 & 0 & 1 & 2\end{array}\right]^{\top} | |
\left[\begin{array}{llll}0 & -2 & 1 & -1\end{array}\right]^{\top} | |
\left[\begin{array}{llll}3 & 1 & 1 & 2\end{array}\right]^{\top} |
Question 3 Explanation:
Given matrix is 4 \times 4 and rank of matrix is 2 .
Therefore, rank of matrix \neq No. of variables Thus, there two linearly dependent vectors \& two linearly independent vectors are present.
\begin{aligned} & X_{1}=\left[\begin{array}{llll} 2 & 1 & 0 & 3 \end{array}\right]^{\top} \\ & X_{2}=\left[\begin{array}{llll} 1 & 0 & 1 & 2 \end{array}\right]^{\top} \end{aligned}
\therefore \quad X=\mathrm{K}_{1}\left[\begin{array}{l}2 \\ 1 \\ 0 \\ 3\end{array}\right]+\mathrm{K}_{2}\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 2\end{array}\right]
For \mathrm{K}_{1}=1 \text{ and } \mathrm{~K}_{2}=-1
X=\left[\begin{array}{c}1 \\ 1 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{llll}1 & 1 & -1 & 1\end{array}\right]^{\top}
Therefore, rank of matrix \neq No. of variables Thus, there two linearly dependent vectors \& two linearly independent vectors are present.
\begin{aligned} & X_{1}=\left[\begin{array}{llll} 2 & 1 & 0 & 3 \end{array}\right]^{\top} \\ & X_{2}=\left[\begin{array}{llll} 1 & 0 & 1 & 2 \end{array}\right]^{\top} \end{aligned}
\therefore \quad X=\mathrm{K}_{1}\left[\begin{array}{l}2 \\ 1 \\ 0 \\ 3\end{array}\right]+\mathrm{K}_{2}\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 2\end{array}\right]
For \mathrm{K}_{1}=1 \text{ and } \mathrm{~K}_{2}=-1
X=\left[\begin{array}{c}1 \\ 1 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{llll}1 & 1 & -1 & 1\end{array}\right]^{\top}
Question 4 |
The solution of the differential equation.
\frac{d^{3} y}{d x^{3}}-5.5 \frac{d^{2} y}{d x^{2}}+9.5 \frac{d y}{d x}-5 y=0
is expressed as y=C_{1} e^{2.5 x}+C_{2} e^{\alpha . x}+C_{3} e^{\beta x}, where \mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3} and \alpha and \beta are constants, with \alpha and \beta being distinct and not equal to 2.5. which of the following options is correct for the values of \alpha and \beta ?
\frac{d^{3} y}{d x^{3}}-5.5 \frac{d^{2} y}{d x^{2}}+9.5 \frac{d y}{d x}-5 y=0
is expressed as y=C_{1} e^{2.5 x}+C_{2} e^{\alpha . x}+C_{3} e^{\beta x}, where \mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3} and \alpha and \beta are constants, with \alpha and \beta being distinct and not equal to 2.5. which of the following options is correct for the values of \alpha and \beta ?
1 and 2 | |
-1 and -2 | |
2 and 3 | |
-2 and -3 |
Question 4 Explanation:
The differential eqn. can be written as :
D^{3}-5.5 D^{2}+9.5 D-5=0
Solving we get,
D=1,2,2.5
Hence, the solution is given as
y=C_{1} e^{2.5 x}+C_{2} e^{1 x}+C_{3} e^{2 x}
Comparing we get,
\begin{aligned} & \alpha=1 \\ & \beta=2 \end{aligned}
Hence, option (A) is correct.
D^{3}-5.5 D^{2}+9.5 D-5=0
Solving we get,
D=1,2,2.5
Hence, the solution is given as
y=C_{1} e^{2.5 x}+C_{2} e^{1 x}+C_{3} e^{2 x}
Comparing we get,
\begin{aligned} & \alpha=1 \\ & \beta=2 \end{aligned}
Hence, option (A) is correct.
Question 5 |
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 5 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}
There are 5 questions to complete.
Good question answer very nice