Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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}
Question 4 |
The Fourier series to represent x-x^2 for - \pi \leq x\leq \pi is given by
x-x^2=\frac{a_0}{2}+\sum_{n=1}^{\infty }a_n \cos nx +\sum_{n=1}^{\infty }b_n \sin nx
The value of a_0 (round off to two decimal places), is _______.
x-x^2=\frac{a_0}{2}+\sum_{n=1}^{\infty }a_n \cos nx +\sum_{n=1}^{\infty }b_n \sin nx
The value of a_0 (round off to two decimal places), is _______.
-1.52 | |
2.12 | |
-6.58 | |
5.23 |
Question 4 Explanation:
\begin{aligned} a_0&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx\\ &=\frac{1}{\pi}\int_{-\pi}^{\pi} (x-x^2)dx\\ &=\frac{-1}{\pi}\int_{0}^{\pi}2x^2 dx\\ &=\frac{-1}{\pi}\left ( \frac{2x^3}{3} \right )_0^{\pi}\\ &=-\frac{2}{3 \pi}[\pi^3]=\frac{-2 \pi^2}{3}\\ &=-6.58 \end{aligned}
Question 5 |
The Laplace transform of \sinh (at) is
\frac{a}{s^2-a^2} | |
\frac{a}{s^2+a^2} | |
\frac{s}{s^2-a^2} | |
\frac{s}{s^2+a^2} |
Question 5 Explanation:
L(\sinh (at))=\frac{a}{s^2-a^2}
Question 6 |
Consider a two-dimensional flow through isotropic soil along x direction and z direction. If h is the hydraulic head, the Laplace's equation of continuity is expressed as
\frac{\partial h}{\partial x}+\frac{\partial h}{\partial z}=0 | |
\frac{\partial h}{\partial x}+\frac{\partial h}{\partial x}\frac{\partial h}{\partial z}+\frac{\partial h}{\partial z}=0 | |
\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial z^2}=0 | |
\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial x \partial z}+\frac{\partial^2 h}{\partial z^2}=0 |
Question 6 Explanation:
The Laplace's equation of continuity for two dimensional flow in a soil is expressed as:
k_x\frac{\partial^2 h}{\partial x^2}+k_z\frac{\partial^2 h}{\partial z^2}=0... for anisotropic soil [k_x\neq k_z]
:
\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial z^2}=0... for isotropic soil [k_x = k_z]
k_x\frac{\partial^2 h}{\partial x^2}+k_z\frac{\partial^2 h}{\partial z^2}=0... for anisotropic soil [k_x\neq k_z]
:
\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial z^2}=0... for isotropic soil [k_x = k_z]
Question 7 |
The laplace transform F(s) of the exponential function ,f(t)=e^{at}\; when \; t\geq 0, where a is a constant and (s-a)\gt 0, is
\frac{1}{s+a} | |
\frac{1}{s-a} | |
\frac{1}{a-s} | |
\infty |
Question 7 Explanation:
\begin{aligned} L(e^{at}) &=\frac{1}{s-a} \\ L(e^{at}) &=\int_{0}^{\infty }e^{-st}e^{at}dt \\ &= \int_{0}^{\infty }e^{-(s-a)t}dt\\ &= \left.\begin{matrix} \frac{e^{-(s-a)t}}{-(s-a)} \end{matrix}\right|_{0}^{\infty }\\ &=-\frac{1}{s-a}(0-1)\\ &=\frac{1}{s-a} \end{aligned}
Question 8 |
The Fourier series of the function,
\begin{matrix} f(x) & =0 & -\pi \lt x \leq 0 \\ f(x) &=\pi-x & 0 \lt x \lt \pi \end{matrix}
in the interval [-\pi ,\pi ] is
f(x)=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\cdots\: \cdots \ \cdots \right ] + \left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots \: \cdots\: \cdot \right ]
The convergence of the above Fourier series at x = 0 gives
\begin{matrix} f(x) & =0 & -\pi \lt x \leq 0 \\ f(x) &=\pi-x & 0 \lt x \lt \pi \end{matrix}
in the interval [-\pi ,\pi ] is
f(x)=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\cdots\: \cdots \ \cdots \right ] + \left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots \: \cdots\: \cdot \right ]
The convergence of the above Fourier series at x = 0 gives
\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6} | |
\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi ^{2}}{12} | |
\sum_{n=1}^{\infty }\frac{1}{(2n-1)^{2}}=\frac{\pi ^{2}}{8} | |
\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{(2n-1)}=\frac{\pi}{4} |
Question 8 Explanation:
The function is f(x)=0
-p\lt x\leq 0
=p-x,\, 0 \lt x \lt \pi
And Fourier series is,
f\left ( x \right )=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\frac{\cos 5x}{5^{2}}+... \right ]+\left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+... \right ] ...\left ( i \right )
At x=0, (a point of discontinuity), the fourier series converges to \frac{1}{2}\left [ f\left ( 0^{-1} \right )+f\left ( 0^{+} \right ) \right ]
where f\left ( 0^{-} \right )=\lim_{x\rightarrow 0}\left ( \pi -x \right )=\pi
Hence, eq. (i), we get,
\frac{\pi }{2}=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{1}{1^{2}}+\frac{1}{3^{2}}+... \right ]
\Rightarrow \;\; \frac{1}{1}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+...\frac{\pi ^{2}}{8}
-p\lt x\leq 0
=p-x,\, 0 \lt x \lt \pi
And Fourier series is,
f\left ( x \right )=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\frac{\cos 5x}{5^{2}}+... \right ]+\left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+... \right ] ...\left ( i \right )
At x=0, (a point of discontinuity), the fourier series converges to \frac{1}{2}\left [ f\left ( 0^{-1} \right )+f\left ( 0^{+} \right ) \right ]
where f\left ( 0^{-} \right )=\lim_{x\rightarrow 0}\left ( \pi -x \right )=\pi
Hence, eq. (i), we get,
\frac{\pi }{2}=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{1}{1^{2}}+\frac{1}{3^{2}}+... \right ]
\Rightarrow \;\; \frac{1}{1}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+...\frac{\pi ^{2}}{8}
Question 9 |
The infinite series 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+... corresponds to
sec x | |
e^{x} | |
cos x | |
1+sin^{2}x |
Question 9 Explanation:
e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}...
(By McLaurin's series expansion)
(By McLaurin's series expansion)
Question 10 |
Laplace transform for the function f (x) = cosh(ax) is
\frac{a}{s^{2}-a^{2}} | |
\frac{s}{s^{2}-a^{2}} | |
\frac{a}{s^{2}+a^{2}} | |
\frac{s}{s^{2}+a^{2}} |
Question 10 Explanation:
It is a standard result that
L\left ( \cosh at \right )=\frac{s}{s^{2}-a^{2}}
L\left ( \cosh at \right )=\frac{s}{s^{2}-a^{2}}
There are 10 questions to complete.