Question 1 |
Consider the matrix
P=\begin{bmatrix} 1 & 1 &0 \\ 0&1 &1 \\ 0& 0 & 1 \end{bmatrix}
The number of distinct eigenvalues of P is
P=\begin{bmatrix} 1 & 1 &0 \\ 0&1 &1 \\ 0& 0 & 1 \end{bmatrix}
The number of distinct eigenvalues of P is
0 | |
1 | |
2 | |
3 |
Question 1 Explanation:
\text { Given: } A=\left[\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]
It is an upper triangular matrix. It's diagonal elements are eigen values.
The eigen values of the matrix are 1, 1, 1.
\therefore Number of distinct eigen values = 1
Hence, option (B) is correct.
It is an upper triangular matrix. It's diagonal elements are eigen values.
The eigen values of the matrix are 1, 1, 1.
\therefore Number of distinct eigen values = 1
Hence, option (B) is correct.
Question 2 |
A parabola x=y^2 \; with \; 0\leq x\leq 1
is shown in the figure. The volume of the solid of rotation obtained by rotating the shaded area by 360^{\circ} around the x-axis is
\frac{\pi}{4} | |
\frac{\pi}{2} | |
\pi | |
2 \pi |
Question 2 Explanation:
\text { Given: } y^{2}=x, 0 \leq x \leq 1
The value of solid obtained by rotating the area bounded by the curve
\begin{aligned} \mathrm{y}^{2}&=\mathrm{x}, 0 \leq \mathrm{x} \leq 1 \text{ about x-axis is}\\ V &=\int_{a}^{b} \pi y^{2} d x \\ V &=\int_{0}^{1} \pi x d x \\ &=\left(\frac{\pi \mathrm{x}^{2}}{2}\right)_{0}^{1} \\ &=\frac{\pi}{2} \end{aligned}
The value of solid obtained by rotating the area bounded by the curve
\begin{aligned} \mathrm{y}^{2}&=\mathrm{x}, 0 \leq \mathrm{x} \leq 1 \text{ about x-axis is}\\ V &=\int_{a}^{b} \pi y^{2} d x \\ V &=\int_{0}^{1} \pi x d x \\ &=\left(\frac{\pi \mathrm{x}^{2}}{2}\right)_{0}^{1} \\ &=\frac{\pi}{2} \end{aligned}
Question 3 |
For the equation \frac{dy}{dx} + 7x^2 y=0, if y(0)=3/7, then the value of y(1)is
\frac{7}{3}e^{-7/3} | |
\frac{7}{3}e^{-3/7} | |
\frac{3}{7}e^{-7/3}
| |
\frac{3}{7}e^{-3/7}
|
Question 3 Explanation:
Given \frac{d y}{d x}+7 x^{2} y=0 \ldots(1)
With y(0)=\frac{3}{7} \ldots(2)
Now,(1) is written as
\Rightarrow \int \frac{1}{y} d y+\int 7 x^{2} d x=C
\Rightarrow \log y+\frac{7 x^{3}}{3}=C
\Rightarrow y=e^{\frac{7 x^{3}}{3}+c} \; \; ...(3)
Using (2),(3) becomes \frac{3}{7}=\mathrm{e}^{0} \cdot \mathrm{e}^{\mathrm{C}}(\mathrm{or}) \mathrm{e}^{\mathrm{C}}=\frac{3}{7} \; \; ...(4)
\therefore The solution of (1) with (3) &(4) is given by
y=y(x)=e^{\frac{-7 x^{3}}{3}+c}=e^{\frac{-7 x^{3}}{3}} \cdot e^{c}=\frac{3}{7} \cdot e^{\frac{-7 x^{3}}{3}}
Hence, y(1)=y=\frac{3}{7} \cdot e^{-\frac{7}{3}}
With y(0)=\frac{3}{7} \ldots(2)
Now,(1) is written as
\Rightarrow \int \frac{1}{y} d y+\int 7 x^{2} d x=C
\Rightarrow \log y+\frac{7 x^{3}}{3}=C
\Rightarrow y=e^{\frac{7 x^{3}}{3}+c} \; \; ...(3)
Using (2),(3) becomes \frac{3}{7}=\mathrm{e}^{0} \cdot \mathrm{e}^{\mathrm{C}}(\mathrm{or}) \mathrm{e}^{\mathrm{C}}=\frac{3}{7} \; \; ...(4)
\therefore The solution of (1) with (3) &(4) is given by
y=y(x)=e^{\frac{-7 x^{3}}{3}+c}=e^{\frac{-7 x^{3}}{3}} \cdot e^{c}=\frac{3}{7} \cdot e^{\frac{-7 x^{3}}{3}}
Hence, y(1)=y=\frac{3}{7} \cdot e^{-\frac{7}{3}}
Question 4 |
The lengths of a large stock of titanium rods follow a normal distribution with a mean (\mu) of 440 mm and a standard deviation (\sigma) of 1mm. What is the percentage of rods whose lengths lie between 438 mm and 441 mm?
81.85% | |
68.40% | |
99.75% | |
86.64% |
Question 4 Explanation:
\begin{array}{l} \mathrm{Z}=\frac{x-\mu}{\sigma} \\ \mathrm{Z}(\mathrm{x}=438)=\frac{438-440}{1}=-2 \\ \mathrm{P}(\mathrm{Z}=-2)=2.28 \% \\ \mathrm{Z}(\mathrm{x}=441)=\frac{441-440}{1}=1 \\ \mathrm{P}(\mathrm{Z}=1)=84.13 \% \\ \end{array}
The percentage of rods whose lengths lie between 438 mm and 441 mm =
\begin{array}{c} =\mathrm{P}(\mathrm{Z}=1)-\mathrm{P}(\mathrm{Z}=-2) \\ =84.13 \%-2.28 \%=81.85 \% \end{array}
Question 5 |
A flat-faced follower is driven using a circular eccentric cam rotating at a constant angular velocity \omega. At time t=0, the vertical position of the followeris y(0)=0, and the system is in the configuration shown below.
The vertical position of the follower face, y(t) is given by
The vertical position of the follower face, y(t) is given by
e \sin \omega t | |
e(1+ \cos 2\omega t) | |
e(1- \cos \omega t) | |
e \sin 2 \omega t |
Question 5 Explanation:
\begin{aligned} \mathrm{AA}_{1}=\mathrm{y} &=\mathrm{e}(1-\cos \theta) \\ &=\mathrm{e}(1-\cos \omega \mathrm{t}) \end{aligned}
Question 6 |
The natural frequencies corresponding to the spring-mass systems I and II are \omega _I \; and \; \omega _{II}, respectively. The ratio \frac{\omega _I}{\omega _{II}} is
\frac{1}{4} | |
\frac{1}{2} | |
2 | |
4 |
Question 6 Explanation:
System: I
k_{e q}=\frac{k \cdot k}{k+k}=\frac{k}{2}
\omega_{\mathrm{n}_{1}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}}
System - II
\mathrm{k}_{\mathrm{eq}}=2 \mathrm{k} \quad \omega_{\mathrm{n}_{2}}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}
\frac{\omega_{\mathrm{n}_{1}}}{\omega_{\mathrm{n}_{2}}}=\frac{\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}}}{\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}} \times \frac{\mathrm{m}}{2 \mathrm{k}}}=\frac{1}{2}
k_{e q}=\frac{k \cdot k}{k+k}=\frac{k}{2}
\omega_{\mathrm{n}_{1}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}}
System - II
\mathrm{k}_{\mathrm{eq}}=2 \mathrm{k} \quad \omega_{\mathrm{n}_{2}}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}
\frac{\omega_{\mathrm{n}_{1}}}{\omega_{\mathrm{n}_{2}}}=\frac{\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}}}{\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}} \times \frac{\mathrm{m}}{2 \mathrm{k}}}=\frac{1}{2}
Question 7 |
A spur gear with 20^{\circ} full depth teeth is transmitting 20 kW at 200 rad/s. The pitch circle diameter of the gear is 100 mm. The magnitude of the force applied on the gear in the radial direction is
0.36 kN | |
0.73 kN | |
1.39 kN | |
2.78kN |
Question 7 Explanation:
\begin{array}{l} \phi=20^{\circ}, P=20 k W, \omega=200 \mathrm{rad} / \mathrm{s}, d=100 \mathrm{mm}=0.1 \mathrm{m} \\ \text { Torque }=\text { Power } / \omega \\ \mathrm{T}=\frac{20000}{200}=100 \mathrm{Nm} \\ \text { Now, } \mathrm{T}=\mathrm{F}_{\mathrm{T}} \times \frac{\mathrm{d}}{2} \\ \Rightarrow 100=\mathrm{F}_{\mathrm{T}} \times \frac{0.1}{2} \\ \Rightarrow \mathrm{F}_{\mathrm{T}}=2000 \mathrm{N}\\ \frac{F_{R}}{F_{T}}=\tan \phi \\ \Rightarrow \mathrm{F}_{\mathrm{R}}=2000 \times \tan 20^{\circ} \\ \Rightarrow \mathrm{F}_{\mathrm{R}}=727.94 \mathrm{N}=0.73 \mathrm{kN} \end{array}
Question 8 |
During a non-flow thermodynamic process(1-2) executed by a perfect gas, the heat interaction is equal to the work interaction (Q_{1-2}=W_{1-2}) when the process is
Isentropic | |
Polytropic | |
Isothermal | |
Adiabatic |
Question 8 Explanation:
Given, \mathrm{Q}_{1-2}=\mathrm{W}_{1-2}
\therefore \Delta \mathrm{U}_{1-2}=0
\Rightarrow \mathrm{c}_{\mathrm{v}}\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]=0
\Rightarrow \mathrm{T}_{1}=\mathrm{T}_{2}
So, the process is isothermal.
\therefore \Delta \mathrm{U}_{1-2}=0
\Rightarrow \mathrm{c}_{\mathrm{v}}\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]=0
\Rightarrow \mathrm{T}_{1}=\mathrm{T}_{2}
So, the process is isothermal.
Question 9 |
For a hydrodynamically and thermally fully developed laminar flow through a circular pipe of constant cross-section, the Nusselt number at constant wall heat flux (Nu_q) and that at constant wall temperature (Nu_T) are related as
Nu_q \gt Nu_T | |
Nu_q \lt Nu_T | |
Nu_q = Nu_T | |
Nu_q = ( Nu_T)^2 |
Question 9 Explanation:
For laminar flow through circular tube:
\mathrm{Nu}_{\mathrm{q}}=4.36 (For constant heat flux)
\mathrm{Nu}_{\mathrm{T}}=3.66 (For constant wall temperature)
\mathrm{Nu}_{\mathrm{q}} \gt \mathrm{Nu}_{\mathrm{T}}
\mathrm{Nu}_{\mathrm{q}}=4.36 (For constant heat flux)
\mathrm{Nu}_{\mathrm{T}}=3.66 (For constant wall temperature)
\mathrm{Nu}_{\mathrm{q}} \gt \mathrm{Nu}_{\mathrm{T}}
Question 10 |
As per common design practice, the three types of hydraulic turbines, in descending order of flow rate, are
Kaplan, Francis, Pelton | |
Pelton, Francis, Kaplan | |
Francis, Kaplan, Pelton | |
Pelton, Kaplan, Francis |
Question 10 Explanation:
Kaplan turbine has highest flow area hence it can handle highest discharge. On the other hand, Pelton turbine has lowest flow area hence it works on low discharge.
\therefore \mathrm{Q}_{\mathrm{Kaplan}} \gt \mathrm{Q}_{\mathrm{Francis}}>\mathrm{Q}_{\mathrm{Pclton}}
\therefore \mathrm{Q}_{\mathrm{Kaplan}} \gt \mathrm{Q}_{\mathrm{Francis}}>\mathrm{Q}_{\mathrm{Pclton}}
There are 10 questions to complete.