# GATE CE 2001

 Question 1
The number of boundary conditions required to solve the differential equation $\frac{\partial^2 \phi }{\partial x^2}+\frac{\partial^2 \phi }{\partial y^2}=0$ is
 A 2 B 0 C 1 D 4
Engineering Mathematics   Ordinary Differential Equation
 Question 2
Value of the integral $I=\int_{0}^{\pi /4}cos^{2}xdx$ is
 A $\frac{\pi }{8}+\frac{1}{4}$ B $\frac{\pi }{8}-\frac{1}{4}$ C $-\frac{\pi }{8}-\frac{1}{4}$ D $-\frac{\pi }{8}+\frac{1}{4}$
Engineering Mathematics   Ordinary Differential Equation
Question 2 Explanation:
\begin{aligned} I&=\int_{0}^{\pi /4}\cos ^{2}xdx \\ &=\int_{0}^{\pi /4}\frac{1+\cos 2x}{2}dx \\ &=\int_{0}^{\pi /4}\frac{dX}{2}+\int_{0}^{\pi /4}\frac{\cos 2x}{2}dx \\ &=\frac{1}{2}\left [ x \right ]_{0}^{\pi /4}+\frac{1}{2}\left [ \frac{\sin 2x}{2} \right ]_{0}^{\pi /4} \\ &=\frac{\pi }{8}+\frac{1}{4}\left [ \sin \frac{\pi }{2}-\sin x \right ] \\ &=\frac{\pi }{8}+\frac{1}{4}\times 1 =\frac{\pi }{8}+\frac{1}{4} \end{aligned}

 Question 3
Limit of the following series as x approaches $\frac{\pi }{2}$ is $f(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}$
 A $\frac{2\pi }{3}$ B $\frac{\pi }{2}$ C $\frac{\pi }{3}$ D 1
Engineering Mathematics   Partial Differential Equation
Question 3 Explanation:
$\lim_{x\rightarrow \frac{\pi }{2}}f\left ( x \right )=\lim_{x\rightarrow \frac{\pi }{2}}\sin x=1$
 Question 4
The degree of static indeterminacy, $N_s$, and the degree of kinematic indeterminacy, $N_k$ , for the plane frame shown below, assuming axial deformations to be negligible, are given by
 A $N_s$ = 6 and $N_k$ = 11 B $N_s$ = 6 and $N_k$ = 6 C $N_s$ = 4 and $N_k$ = 6 D $N_s$ = 4 and $N_k$ = 4
Structural Analysis   Determinacy and Indeterminacy
Question 4 Explanation:
Degree of static indeterminacy,
\begin{aligned} N_{s} &=3 m+r_{e}-3 j-r_{r} \\ &=3 \times 5+(3+2+2)-3 \times 6-0 \\ &=4 \end{aligned}
Degree of kinematic indeterminacy,
\begin{aligned} N_{k} &=3 j-r_{e}-m \\ &=3 \times 6-(3+2+2)-5=6 \\ N_{s} &=4 \text { and } N_{k}=6 \end{aligned}
 Question 5
The bending moment (in kNm units) at the mid span location X in the beam with overhangs shown below is equal to
 A 0 B -10 C -15 D -20
Solid Mechanics   Shear Force and Bending Moment
Question 5 Explanation:

\begin{aligned} \mathrm{R}+\mathrm{V}&=30 &\ldots(i)\\ \text { Now, } \mathrm{V} &\times 2-20 \times 3+10 \times 1=0 \\ \mathrm{V}&=\frac{60-10}{2}=25 \mathrm{kN}\\ \therefore \quad \mathrm{R}&=5 \mathrm{kN}\\ \mathrm{BM} \text { at }\quad X&=25 \times 1-20 \times 2=-15 \mathrm{kNm} \end{aligned}

There are 5 questions to complete.