# GATE CE 2010

 Question 1
The $\lim_{x\rightarrow 0}\frac{sin[\frac{2}{3}x]}{x}$ is
 A $\frac{2}{3}$ B 1 C $\frac{3}{2}$ D $\infty$
Engineering Mathematics   Calculus
Question 1 Explanation:
\begin{aligned} \lim_{x\rightarrow 0}\frac{\sin \left [ \frac{2}{3}x \right ]}{x} &= \lim_{\frac{2}{3}x\rightarrow 0}\frac{\frac{2}{3}\sin \left [ \frac{2}{3}x \right ]}{\frac{2}{3}x}\\ \text{Let } \theta &=\frac{2}{3}x\\ \text{Required limit }&= \frac{2}{3}\lim_{\theta \rightarrow 0}\frac{\sin \theta }{\theta } \\ &=\frac{2}{3}\times 1=\frac{2}{3}\end{aligned}
 Question 2
Two coins are simultaneously tossed. The probability of two heads simultaneously appearing is
 A $\frac{1}{8}$ B $\frac{1}{6}$ C $\frac{1}{4}$ D $\frac{1}{2}$
Engineering Mathematics   Probability and Statistics
Question 2 Explanation:
p(2 heads) $= \frac{1}{2}\times \frac{1}{2} =\frac{1}{4}$

 Question 3
The order and degree of the differential equation
$\frac{d^{3}y}{dx^{3}}+4\sqrt{(\frac{dy}{dx})^{3}}+y^{2}=0$ are respectively
 A 3 and 2 B 2 and 3 C 3 and 3 D 3 and 1
Engineering Mathematics   Ordinary Differential Equation
Question 3 Explanation:
$\frac{d^{3}y}{dx^{3}}+4\sqrt{\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}+y^{2}}=0$
$\left ( \frac{d^{3}y}{dx^{3}} \right )^{2}=16\left [ \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}+y^{2} \right ]$
$\therefore$ The order is 3 since hghest differential is $\frac{d^{3}y}{dx^{3}}$.
The degree is 2 since power of highest differential is 2.
 Question 4
Two people weighing W each are sitting on a plank of length L floating on water at L/4 from either end. Neglecting the weight of he plank, the bending moment at the centre of the plank is
 A WL/8 B WL/16 C WL/32 D zero
Solid Mechanics   Shear Force and Bending Moment
Question 4 Explanation:
The plank will be balanced by the buoyant force acting under its bottom. Let the intensity of buoyant force be w. \begin{aligned} \text{For equilibrium, }w \times L&=W+W \\ \Rightarrow \quad w&=\frac{2 W}{L}\text{ upwards}\\ \end{aligned}
Thus, the bending moment at the centre of the plank will be
\begin{aligned} M &=\frac{2 W}{L} \times \frac{L}{2} \times \frac{L}{4}-W \times \frac{L}{4} \\ \Rightarrow \quad M&=\frac{W L}{4}-\frac{W L}{4}=0 \end{aligned}
 Question 5
For the truss shown in the figure, the force in the member QR is A 0 B $\frac{P}{\sqrt{2}}$ C P D $\sqrt{2}P$
Structural Analysis   Trusses
Question 5 Explanation:
Using method of joints and considering joint S,
we get,
\begin{aligned} \Sigma F_{x}&=0 \\ \Rightarrow \quad F_{T S}&=0\\ \text{and }\quad \Sigma F_{y}&=0 \\ \Rightarrow F_{R S}-P&=0 \\ \Rightarrow F_{R S}&=P(\text { tensile }) \end{aligned} Considering joint R, we get
\begin{aligned} \Sigma F_{v}&=0 \\ \Rightarrow \quad-F_{I R} \sin 45^{\circ}-P&=0 \\ \Rightarrow \quad F_{I R}&=-P \sqrt{2} \text{(compressive)}\\ \Sigma F_{x}&=0 \\ \Rightarrow \quad-F_{T R} \cos 45^{\circ}-F_{Q R}&=0 \\ \Rightarrow \quad P \sqrt{2} \times \frac{1}{\sqrt{2}}-F_{\alpha R}&=0\\ \Rightarrow \quad F_{Q R}&=P\text{(tension)} \end{aligned} There are 5 questions to complete.