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
Removing radicals we get,
\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}

Question 6
The major and minor principal stresses at a point are 3MPa and -3MPa respectively. The maximum shear stress at the point is
A
0MPa
B
3MPa
C
6MPa
D
9MPa
Solid Mechanics   Principal Stress and Principal Strain
Question 6 Explanation: 
Maximum shear stress at the point is given by
\tau_{\max }=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{3-(-3)}{2}=3 \mathrm{MPa}
Question 7
The number of independent elastic constants for a linear elastic isotropic and homogeneous material is
A
4
B
3
C
2
D
1
Solid Mechanics   Properties of Metals, Stress and Strain
Question 7 Explanation: 


Question 8
The effective length of a column of length L fixed against rotation and translation at one end and free at the other end is
A
0.5L
B
0.7L
C
1.414L
D
2L
Solid Mechanics   Theory of Columns and Shear Centre
Question 9
As per India standard code of practice for pre stressed concrete (IS:1343-1980) the minimum grades of concrete to be used for post-tensioned and pre-tensioned structural elements are respectively
A
M20 for both
B
M40 and M30
C
M15 and M20
D
M30 and M40
RCC Structures   Prestressed Concrete Beams
Question 10
A solid circular shaft of diameter d and length L is fixed at one end free at the other end. A torque T is applied at the free end. The shear modulus of the material is G. The angle of twist at the free end is
A
\frac{16TL}{\pi d^{4}G}
B
\frac{32TL}{\pi d^{4}G}
C
\frac{64TL}{\pi d^{4}G}
D
\frac{128TL}{\pi d^{4}G}
Solid Mechanics   Torsion of Shafts and Pressure Vessels
Question 10 Explanation: 
Angle of twist is given by,
\begin{aligned} \theta &=\frac{T L}{G I_{P}} \text { but } I_{P}=\frac{\pi d^{4}}{32} \\ \Rightarrow \quad \theta&=\frac{T L}{G \times \frac{\pi d^{4}}{32}}=\frac{32 T L}{\pi d^{4} G} \end{aligned}
There are 10 questions to complete.

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