Question 1 |
Which one of the following is correct?
\lim_{x\rightarrow 0}\left ( \frac{sin4x}{sin 2x} \right )=2 and \lim_{x\rightarrow 0}\left ( \frac{tanx}{x} \right )=1
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\lim_{x\rightarrow 0}\left ( \frac{sin4x}{sin 2x} \right )=1 and \lim_{x\rightarrow 0}\left ( \frac{tanx}{x} \right )=1
| |
\lim_{x\rightarrow 0}\left ( \frac{sin4x}{sin 2x} \right )= \infty and \lim_{x\rightarrow 0}\left ( \frac{tanx}{x} \right )=1
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\lim_{x\rightarrow 0}\left ( \frac{sin4x}{sin 2x} \right )=2 and \lim_{x\rightarrow 0}\left ( \frac{tanx}{x} \right )= \infty
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Question 1 Explanation:
\lim_{x \to 0}\left ( \frac{\sin 4x}{\sin 2x} \right )=\lim_{x \to 0}\left ( \frac{\frac{\sin 4x}{x}}{\frac{\sin 2x}{x}} \right )=\frac{4}{2}=2 and \lim_{x \to 0}\left ( \frac{\tan x}{x} \right )=1
Question 2 |
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 2 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 3 |
A simple mass-spring oscillatory system consists of a mass m, suspended from a spring of stiffness k. Considering z as the displacementof the system at any time t, the equation of motion for the free vibration of the system is m\ddot{z}+kz=0. The natural frequency of the system is
\frac{k}{m} | |
\sqrt{\frac{k}{m}} | |
\frac{m}{k} | |
\sqrt{\frac{m}{k}} |
Question 3 Explanation:
\begin{aligned} m\ddot{z}+kz&=0 \\ \ddot{z}+\frac{k}{m}z&=0 \\ \text{Comparing with} \\ \ddot{z}+\omega _n^2 z&=0 \\ \text{We get}\;\; \omega _n&=\sqrt{\frac{k}{m}} \end{aligned}
Question 4 |
For a small value of h, the Taylor series expansion for f(x+h) is
f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...\infty | |
f(x)-hf'(x)+\frac{h^2}{2!}f''(x)-\frac{h^3}{3!}f'''(x)+...\infty | |
f(x)+hf'(x)+\frac{h^2}{2}f''(x)+\frac{h^3}{3}f'''(x)+...\infty | |
f(x)-hf'(x)+\frac{h^2}{2}f''(x)-\frac{h^3}{3}f'''(x)+...\infty |
Question 4 Explanation:
We know that Taylor series for small h of f(x + h) is,
f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...
f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...
Question 5 |
A plane truss is shown in the figure
Which one of the options contains ONLY zero force members in the truss?
Which one of the options contains ONLY zero force members in the truss?
FG, FI, HI, RS | |
FG, FH, HI, RS | |
FI, HI, PR, RS | |
FI, FG, RS, PR |
Question 5 Explanation:
So zero force members are FI, FG, RS, PR
Question 6 |
An element is subjected to biaxial normal tensile strains of 0.0030 and 0.0020. The normal strain in the plane of maximum shear strain is
Zero | |
0.001 | |
0.0025 | |
0.005 |
Question 6 Explanation:
\varepsilon _x=0.0030
\varepsilon _y=0.0020
Normal strain in the plane of maximum shear strain
\varepsilon _{avg}=\frac{\varepsilon _x+\varepsilon _y}{2}=\frac{0.0030+0.0020}{2}=0.0025
\varepsilon _y=0.0020
Normal strain in the plane of maximum shear strain
\varepsilon _{avg}=\frac{\varepsilon _x+\varepsilon _y}{2}=\frac{0.0030+0.0020}{2}=0.0025
Question 7 |
Consider the pin-jointed plane truss shown in the figure. Let R_P,R_Q and R_R denote the vertical reactions (upward positive) applied by the supports at P, Q, and R, respectively, on the truss. The correct combination of (R_P,R_Q,R_R)is represented by
(30, -30, 30) kN | |
(20, 0, 10) kN | |
(10, 30, -10) kN | |
(0, 60, -30) kN |
Question 7 Explanation:
Adopting method of sections and taking LHS of the section
\begin{aligned} &\Sigma F_y= 0\\ &R_P=30kN \\ &\text{For complete truss,} \\ &\Sigma M_R =0 \\ &9R_P-30 \times 6 -R_Q \times 3=0 \\ &R_Q=30kN(\downarrow ) \\ &\text{Taking RHS of section,} \\ &\Sigma F_y=0\Rightarrow R_R=-R_Q \\ &\text{Thus},\;\; R_Q= 30kN(\downarrow )\\ &R_R=30kN(\uparrow ) \end{aligned}
Question 8 |
Assuming that there is no possibility of shear buckling in the web, the maximum reduction permitted by IS 800-2007 in the (low-shear) design bending strength of a semi-compact steel section due to high shear is
Zero | |
25% | |
50% | |
governed by the area of the flange |
Question 8 Explanation:
As per IS 800 : 2007
For semi compact section
(i) In low shear case (V \leq 0.6 V_d)
M_d = Z_ef_y/\gamma _{mo}
(ii) In high shear case (V \gt 0.6 V_d)
M_d = Z_ef_y/\gamma _{mo}
So reduction is zero.
For semi compact section
(i) In low shear case (V \leq 0.6 V_d)
M_d = Z_ef_y/\gamma _{mo}
(ii) In high shear case (V \gt 0.6 V_d)
M_d = Z_ef_y/\gamma _{mo}
So reduction is zero.
Question 9 |
In the reinforced beam section shown in the figure, the nominal cover provided at the bottom of the beam as per IS 456-2000, is
30 mm | |
36 mm | |
42 mm | |
50 mm |
Question 9 Explanation:
Nominal cover = Effective cover -\frac{\phi _m}{2}-\phi _{st}
=50-\frac{16}{2}-12=30mm
Nominal cover is the distance from extreme concrete fbre to the surface of stirrup.
=50-\frac{16}{2}-12=30mm
Nominal cover is the distance from extreme concrete fbre to the surface of stirrup.
Question 10 |
The interior angles of four triangles are given below:
Which of the triangles are ill-conditioned and should be avoided in Triangulation surveys?
Which of the triangles are ill-conditioned and should be avoided in Triangulation surveys?
Both P and R | |
Both Q and R | |
Both P and S | |
Both Q and S |
Question 10 Explanation:
For an ill conditioned traingle in triangulation survey, any angle can be less than 38^{\circ}, and can be greater than 38^{\circ}.
For traingles Q and S, the above condition is valid.
For traingles Q and S, the above condition is valid.
There are 10 questions to complete.
