Question 1 |
If x=a(\theta +\sin \theta )
and y=a(1-\cos \theta ), then \frac{\mathrm{d} y}{\mathrm{d} x}
will be equal to
\sin \left ( \frac{\theta }{2} \right ) | |
\cos \left ( \frac{\theta }{2} \right ) | |
\tan \left ( \frac{\theta }{2} \right ) | |
\cot \left ( \frac{\theta }{2} \right ) |
Question 1 Explanation:
\begin{array}{l} Given, \quad \begin{array}{l} x=a(\theta+\sin \theta) \\ y=a(1-\cos \theta) \end{array} \\ \begin{aligned} \therefore \quad \frac{d x}{d \theta} &=a(1+\cos \theta), \frac{d y}{d \theta}=a \sin \theta \\ \frac{d y}{d x} &=\frac{d y / d \theta}{d x / d \theta}=\frac{a \sin \theta}{a(1+\cos \theta)} \\ &=\frac{2 \operatorname{asin}\left(\frac{\theta}{2}\right) \cdot \cos \left(\frac{\theta}{2}\right)}{a \times 2 \cos ^{2}\left(\frac{\theta}{2}\right)} \\ &=\tan \theta / 2 \end{aligned} \end{array}
Question 2 |
The angle between two unit-magnitude coplanar vectors P(0.86, 0.500,0) and Q(0.259, 0.956, 0) will be
0^{\circ} | |
30^{\circ} | |
45^{\circ} | |
60^{\circ} |
Question 2 Explanation:
\begin{aligned} \vec{P}=0.866 \hat{i}+0.500 \hat{j}+0 \hat{k} &\\ \vec{Q}=0.259 \hat{i}+0.966 \hat{j}+0 \hat{k} \end{aligned}
\therefore \vec{P} \cdot \vec{Q}=|\vec{P}||\vec{Q}| \cos \theta
Or, (0.866 \hat{i}+0.5 \hat{j}+0 \hat{k}) .(0.259 \hat{i}+0.966 \hat{j}+0 \hat{k})
Or, \begin{aligned}=& \sqrt{(0.866)^{2}+(0.5)^{2}} \\ & \times \sqrt{(0.259)^{2}+(0.966)^{2}} \cdot \cos \theta \\ \therefore \cos \theta=& \frac{0.866 \times 0.259+0.5 \times 0.966}{\sqrt{0.99} \times \sqrt{1.001}} \\ \cos \theta &= 0.707 \\ \theta& = 45\end{aligned}
\therefore \vec{P} \cdot \vec{Q}=|\vec{P}||\vec{Q}| \cos \theta
Or, (0.866 \hat{i}+0.5 \hat{j}+0 \hat{k}) .(0.259 \hat{i}+0.966 \hat{j}+0 \hat{k})
Or, \begin{aligned}=& \sqrt{(0.866)^{2}+(0.5)^{2}} \\ & \times \sqrt{(0.259)^{2}+(0.966)^{2}} \cdot \cos \theta \\ \therefore \cos \theta=& \frac{0.866 \times 0.259+0.5 \times 0.966}{\sqrt{0.99} \times \sqrt{1.001}} \\ \cos \theta &= 0.707 \\ \theta& = 45\end{aligned}
Question 3 |
The sum of the eigen values of the matrix given below is \begin{pmatrix} 1 & 1 & 3\\ 1 & 5 & 1\\ 3 & 1 & 1 \end{pmatrix}
5 | |
7 | |
9 | |
18 |
Question 3 Explanation:
Sum of eigen values of given matrix
= sum of diagonal element of given matrix
=1+5+1=7
= sum of diagonal element of given matrix
=1+5+1=7
Question 4 |
The figure shows a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The member LN of the truss is subjected to a load of
0 Newton | |
490 Newtons in compression | |
981 Newtons in compression | |
981 Newtons in tension |
Question 4 Explanation:
\Sigma F_H=0 and \Sigma F_V=0
At joint "L"
\therefore \;\; F_{LK}-F_{LM}=0\;\;(\Sigma F_H=0)
F_{LN}=0\;\;\;(\Sigma F_V=0)
Hence no force is acting on the truss number LN.
Question 5 |
In terms of Poission's ratio (v) the ratio of Young's Modulus (E) to Shear Modulus (G) of elastic materials is
2(1 + v) | |
2(1 - v) | |
\frac{1}{2}(1+v) | |
\frac{1}{2}(1-v) |
Question 5 Explanation:
We know that
\begin{aligned} E &=2 G(1+v) \\ \text{or}\quad \frac{E}{G} &=2(1+v) \end{aligned}
\begin{aligned} E &=2 G(1+v) \\ \text{or}\quad \frac{E}{G} &=2(1+v) \end{aligned}
Question 6 |
Two mating spur gears have 40 and 120 teeth respectively. The pinion rotates at 1200 rpm and transmits a torque of 20 N.m. The torque transmitted by the gear is
6.6 Nm | |
20 Nm | |
40 Nm | |
60 Nm |
Question 6 Explanation:
Power transmitted,
P=\frac{2 \pi N_{G} T_{G}}{60}=\frac{2 \pi N_{P} T_{P}}{60}
\begin{aligned} \Rightarrow \quad N_{G} T_{G}&=N_{p} T_{P} \;\; \text{(where T is torque)} \\ and, \quad \frac{N_{p}}{N_{G}}&=\frac{Z_{G}}{Z_{p}} \\ &\text{(where Z is number of teeths)}\\ \Rightarrow \quad \frac{1200}{N_{G}}&=\frac{120}{40} \\ N_{G}&=400 \mathrm{r} . \mathrm{p.m} \\ \text{Hence,}\\ 400 \times T_{G}&=1200 \times 20 \\ T_{G}&=60 \mathrm{N} . \mathrm{m} \end{aligned}
P=\frac{2 \pi N_{G} T_{G}}{60}=\frac{2 \pi N_{P} T_{P}}{60}
\begin{aligned} \Rightarrow \quad N_{G} T_{G}&=N_{p} T_{P} \;\; \text{(where T is torque)} \\ and, \quad \frac{N_{p}}{N_{G}}&=\frac{Z_{G}}{Z_{p}} \\ &\text{(where Z is number of teeths)}\\ \Rightarrow \quad \frac{1200}{N_{G}}&=\frac{120}{40} \\ N_{G}&=400 \mathrm{r} . \mathrm{p.m} \\ \text{Hence,}\\ 400 \times T_{G}&=1200 \times 20 \\ T_{G}&=60 \mathrm{N} . \mathrm{m} \end{aligned}
Question 7 |
The figure shows the state of stress at a certain point in a stressed body. The magnitudes of normal stresses in the x and y direction are 100 MPa and 20 MPA respectively. The radius of Mohr's stress circle representing this state of stress is
120 | |
80 | |
60 | |
40 |
Question 7 Explanation:
\begin{aligned} \text { Radius } &=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}} \\ \text{Given,} \quad \sigma_{x} &=100 \mathrm{MPa} \\ \sigma_{y}&=-20 \mathrm{MPa} \\ \tau_{x y} &=0 \\ \therefore \text { Radius } &=\sqrt{\left(\frac{100+20}{2}\right)^{2}+0} \\ &=60 \mathrm{MPa} \end{aligned}
Question 8 |
For a mechanism shown below, the mechanical advantage for the given configuration is
0 | |
0.5 | |
1 | |
\infty |
Question 8 Explanation:
Mechanical advantage
=\frac{T_{\text {output }}}{T_{\text {input }}}=\frac{\omega_{\text {input }}}{\omega_{\text {output }}}
since output link is a slider
Hence,\omega_{\text {output }}=0
\therefore Mechanical advantage =\infty
=\frac{T_{\text {output }}}{T_{\text {input }}}=\frac{\omega_{\text {input }}}{\omega_{\text {output }}}
since output link is a slider
Hence,\omega_{\text {output }}=0
\therefore Mechanical advantage =\infty
Question 9 |
A vibrating machine is isolated from the floor using springs. If the ratio of excitation frequency of vibration of machine to the natural frequency of the isolation system is equal to 0.5, the transmissibility of ratio of isolation is
\frac{1}{2} | |
\frac{3}{4} | |
\frac{4}{3} | |
2 |
Question 9 Explanation:
Given \quad r=0.5
Due to isolation
\xi_{S}=0
Transmissibility ratio
\begin{aligned} T R &=\frac{\sqrt{1+(2 \xi r)^{2}}}{\sqrt{\left(1-r^{2}\right)^{2}+(2 \xi r)^{2}}} \\ &=\frac{1}{\sqrt{\left(1-(0.5)^{2}\right)^{2}}} \\ &=\frac{1}{1-0.25}=\frac{4}{3} \end{aligned}
Due to isolation
\xi_{S}=0
Transmissibility ratio
\begin{aligned} T R &=\frac{\sqrt{1+(2 \xi r)^{2}}}{\sqrt{\left(1-r^{2}\right)^{2}+(2 \xi r)^{2}}} \\ &=\frac{1}{\sqrt{\left(1-(0.5)^{2}\right)^{2}}} \\ &=\frac{1}{1-0.25}=\frac{4}{3} \end{aligned}
Question 10 |
A torque of 10 Nm is transmitted through a stepped shaft as shown in figure. The torsional stiffnesses of individual sections of lengths MN, NO and OP are 20 Nm/rad, 30 Nm/rad and 60 Nm/rad respectively. The angular deflection between the ends M and P of the shaft is
0.5 rad | |
0.1 rad | |
5.0 rad | |
1.0 rad |
Question 10 Explanation:
\begin{aligned} \theta_{M N}&=\frac{10}{20}=\frac{10}{20} \text { radian } \\ \theta_{N O}&=\frac{10}{30}=\frac{10}{30} \text { radian } \\ \theta_{O P}&=\frac{10}{60}= \frac{10}{60} \text { radian } \end{aligned}
The parts MN, NO and OP are connected in series.
\theta_{M P}=\theta_{M N}+\theta_{N O}+\theta_{O P}=1 \text { radian }
The parts MN, NO and OP are connected in series.
\theta_{M P}=\theta_{M N}+\theta_{N O}+\theta_{O P}=1 \text { radian }
There are 10 questions to complete.