Question 1 |
In the following partial differential equation, \theta
is a function of t and z, and D and K are
functions of \theta
D(\theta )\frac{\partial^2 \theta }{\partial z^2}+\frac{\partial K(\theta )}{\partial z}-\frac{\partial \theta }{\partial t}=0
The above equation is
D(\theta )\frac{\partial^2 \theta }{\partial z^2}+\frac{\partial K(\theta )}{\partial z}-\frac{\partial \theta }{\partial t}=0
The above equation is
a second order linear equation | |
a second degree linear equation | |
a second order non-linear equation | |
a second degree non-linear equation |
Question 1 Explanation:
\because \;\;1^{st}
term of given D. Equation contains product of dependent variable with it's derivative, so it is non-linear and also we have 2nd order derivative so it's order is two
i.e., 2nd order non linear equation.
i.e., 2nd order non linear equation.
Question 2 |
The value of \lim_{x \to \infty }\frac{x^2-5x+4}{4x^2+2x}
0 | |
\frac{1}{4} | |
\frac{1}{2} | |
1 |
Question 2 Explanation:
It is in \left (\frac{\infty }{\infty } \right )
from so by L-Hospital Rule
\begin{aligned} =&\lim_{x \to \infty }\left ( \frac{2x-5}{8x+2} \right )=\frac{\infty }{\infty }\\ =&\lim_{x \to \infty }\left ( \frac{2}{8} \right )=\frac{1}{4} \end{aligned}
\begin{aligned} =&\lim_{x \to \infty }\left ( \frac{2x-5}{8x+2} \right )=\frac{\infty }{\infty }\\ =&\lim_{x \to \infty }\left ( \frac{2}{8} \right )=\frac{1}{4} \end{aligned}
Question 3 |
The true value of ln(2) is 0.69. If the value of ln(2) is obtained by linear interpolation
between ln(1) and ln(6), the percentage of absolute error (round off to the nearest integer),
is
35 | |
48 | |
69 | |
84 |
Question 3 Explanation:
True value of \ln 2=0.69=T
\begin{aligned} &x &&y=\ln x \\ &x_0=1 & &0 \\ &x_1=6& &1.79 \end{aligned}
Divided differentiation
\begin{aligned} \frac{1.79-0}{6-1}&=0.358=f[x_0,x_1] \\ \text{Approx:}\;\;\ln 2 &=f[x_0]+(x-x_0)f[x_0,x_1] \\ &= 0+(2-1)0.358\\ &= 0.358=A\\ \% \; error &= \frac{T-A}{T}\times 100=48.11\% \end{aligned}
\begin{aligned} &x &&y=\ln x \\ &x_0=1 & &0 \\ &x_1=6& &1.79 \end{aligned}
Divided differentiation
\begin{aligned} \frac{1.79-0}{6-1}&=0.358=f[x_0,x_1] \\ \text{Approx:}\;\;\ln 2 &=f[x_0]+(x-x_0)f[x_0,x_1] \\ &= 0+(2-1)0.358\\ &= 0.358=A\\ \% \; error &= \frac{T-A}{T}\times 100=48.11\% \end{aligned}
Question 4 |
The area of an ellipse represented by an equation \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is
\frac{\pi a b}{4} | |
\frac{\pi a b}{2} | |
\pi a b | |
\frac{4 \pi a b}{3} |
Question 4 Explanation:
\begin{aligned} \text{Area} &=\int \int (1)dydx \\ &=\int_{x=-a}^{a}\int_{y=-\frac{b}{a}}^{+\frac{b}{a}}(1)dydx \\ &=4\int_{x=0}^{a} \int_{y=0}^{\frac{b}{a}\sqrt{a^2-x^2}}(1)dydx\\ &= 4 \int_{x=0}^{a}\int_{y=0}^{\frac{b}{a}\sqrt{a^2-x^2}} dx\\ &= \pi ab \end{aligned}
Question 5 |
Consider the planar truss shown in the figure (not drawn to the scale)
Neglecting self-weight of the members, the number of zero-force members in the truss under the action of the load P, is
Neglecting self-weight of the members, the number of zero-force members in the truss under the action of the load P, is
6 | |
7 | |
8 | |
9 |
Question 5 Explanation:
As \Delta _{AB}=0, hence F _{AB}=0
Total number of zero force member = 8
There are 5 questions to complete.