Question 1 |
The \lim_{x\rightarrow 0}\frac{sin[\frac{2}{3}x]}{x} is
\frac{2}{3} | |
1 | |
\frac{3}{2} | |
\infty |
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
\frac{1}{8} | |
\frac{1}{6} | |
\frac{1}{4} | |
\frac{1}{2} |
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
\frac{d^{3}y}{dx^{3}}+4\sqrt{(\frac{dy}{dx})^{3}}+y^{2}=0 are respectively
3 and 2 | |
2 and 3 | |
3 and 3 | |
3 and 1 |
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.
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
WL/8 | |
WL/16 | |
WL/32 | |
zero |
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}
\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
0 | |
\frac{P}{\sqrt{2}} | |
P | |
\sqrt{2}P |
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}
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
0MPa | |
3MPa | |
6MPa | |
9MPa |
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}
\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
4 | |
3 | |
2 | |
1 |
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
0.5L | |
0.7L | |
1.414L | |
2L |
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
M20 for both | |
M40 and M30 | |
M15 and M20 | |
M30 and M40 |
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
\frac{16TL}{\pi d^{4}G} | |
\frac{32TL}{\pi d^{4}G} | |
\frac{64TL}{\pi d^{4}G} | |
\frac{128TL}{\pi d^{4}G} |
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}
\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.
