GATE ME 2009

Question 1
For a matrix [M]=\begin{bmatrix} \frac{3}{5} & \frac{4}{5}\\ x & \frac{3}{5} \end{bmatrix} the transpose of the matrix is equal to the inverse of the matrix [M]^{T}= [M]^{-1}. The value of x is given by
A
-\frac{4}{5}
B
-\frac{3}{5}
C
\frac{3}{5}
D
\frac{4}{5}
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
\text{If }\quad A^{T}=A^{-1}
then A is orthogonal matrix.
Therefore A . A^{-1}=A^{-1} A=I
\text{and }A^{T} A=A A^{T}=I
since \mathrm{M} is orthogonal matrix
M^{T} M=I
=\left[\begin{array}{cc}\frac{3}{5} & x \\ \frac{4}{5} & \frac{3}{5}\end{array}\right]\left[\begin{array}{cc}\frac{3}{5} & \frac{4}{5} \\ x & \frac{3}{5}\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]
\left[\begin{aligned} \left(\frac{3}{5}\right)^{2}+x^{2} \quad\left(\frac{3}{5} \cdot \frac{4}{5}\right)+\frac{3}{5} x \\ \left(\frac{4}{5} \cdot \frac{3}{5}\right)+\frac{3}{5} \cdot x\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2} \end{aligned} \right]
=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \Rightarrow \text{Compare both sides} a_{12}
a_{12}=\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)+\left(\frac{3}{5}\right) x=0
\Rightarrow \quad \frac{3}{5} x=-\frac{3}{5} \cdot \frac{4}{5}
\Rightarrow x=-\frac{4}{5}
Question 2
The divergence of the vector field 3xz\hat{i}+2xy\hat{j}-yz^{2}\hat{k} at a point (1,1,1 ) is equal to
A
7
B
4
C
3
D
0
Engineering Mathematics   Calculus
Question 2 Explanation: 
\text{Vector field,}
\vec{f}=3 x z \hat{i}+2 x y \hat{j}-y z^{2} \hat{k}
\vec{f}=\left[f_{1} \hat{i}+f_{2} \hat{j}+f_{3} \hat{k}\right]
\begin{aligned}&\text{Divergence of vector field} \vec{f}=\nabla \cdot \vec{f} \\ &\text{Div}(f)=\left(i \frac{\partial}{\partial x}+\hat{j} \frac{\partial}{\partial y}+\hat{k} \frac{\partial}{\partial z}\right) \cdot\left(t_{1} \hat{i}+f_{2} \hat{j}+f_{3} \hat{k}\right) \\ &\text{Div}(f)=\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z} \\ &\text{Div}(f)=\frac{\partial}{\partial x}[3 x z]+\frac{\partial}{\partial y}[2 x y]+\frac{\partial}{\partial z}\left[-2 y z^{2}\right] \\ &\text{Div}(f)=3 z+2 x-2 z y \\ &\text{Div}(f)|_{(1,1,1)}=3(1)+2(1)-2(1)(1)=3 \end{aligned}
Question 3
The inverse Laplace transform of \frac{1}{(s^{2}+s)} is
A
1+e^{t}
B
1-e^{t}
C
1-e^{-t}
D
1+e^{-t}
Engineering Mathematics   Differential Equations
Question 3 Explanation: 
\begin{aligned} L^{-1}\left(\frac{1}{s^{2}+s}\right) &=? \\ \frac{1}{s^{2}+s} &=\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1} \\ L^{-1}\left(\frac{1}{s^{2}+s}\right) &=L^{-1}\left(\frac{1}{s}\right)--L^{-1}\left(\frac{1}{s+1}\right) \\ &=1-e^{-t} \end{aligned}
[Using standard formulae]
Standard formula:
\begin{aligned} L^{-1}\left(\frac{1}{s}\right) &=1 \\ L^{-1}\left(\frac{1}{s+a}\right) &=e^{-a t} \\ L^{-1}\left(\frac{1}{s-a}\right) &=e^{a t} \end{aligned}
Question 4
If three coins are tossed simultaneously, the probability of getting at least one head
A
1/8
B
3/8
C
1/2
D
7/8
Engineering Mathematics   Probability and Statistics
Question 4 Explanation: 
Three coins are tossed simultaneously, the total
number of ways =2^{3}=8
Number of ways to get all tails = 1
Probability to get all tails
=\frac{\text { Number of ways for tails }}{\text { Total number of ways }}=\frac{1}{8}
Therefore probability to get at least one head is
=1-\frac{1}{8}=\frac{7}{8}
or, Total number of ways with at least one head is
\left.\begin{matrix} & &\left.\begin{array}{lll} H & T & T \\ T & H & T \\ T & T & H \end{array} \right] \\ &&\left.\begin{array}{lll} H & T & T \\ T & H & T \\ T & T & H \end{array} \right] \\ &&\left.\begin{array}{lll} H & H & H \end{array} \right] \end{matrix}\right\}
Therefore probability to get at least one head =\frac{7}{8}
Question 5
If a closed system is undergoing an irreversible process, the entropy of the system
A
Must increase
B
Always remains constant
C
Must decrease
D
Can increase, decrease or remain constant
Thermodynamics   Thermodynamic System and Processes
Question 5 Explanation: 
If a closed system is undergoing an irreversible process, the entropy of the system can increase, decrease or remain constant
Question 6
A coolant fluid at 30^{\circ}C flows over a heated flat plate maintained at a constant temperature of 100^{\circ}C. The boundary layer temperature distribution at a given location on the plate may be approximated as T=30+70 exp(-y) where y (in m) is the distance normal to the plate and T is in ^{\circ}C. If thermal conductivity of the fluid is 1.0W/mK, the local convective heat transfer coefficient (in W/m^{2}K ) at that location will be
A
0.2
B
1
C
5
D
10
Heat Transfer   Conduction
Question 6 Explanation: 


\begin{array}{l} T_{s}=100^{\circ} \mathrm{C} \\ \text{At}\; \mathrm{y}=0 ; q_{\mathrm{cond}}=q_{\mathrm{conv}} \\ -\left.k_{f} \frac{\partial T}{\partial y}\right|_{y=0}=h \Delta T=h\left(T_{s}-T_{\infty}\right) \\ -\left.(1) \frac{\partial}{\partial y}\left[30+70 e^{-y}\right]\right|_{y=0}=h(100-30) \\ -\left.\left[0+70(-1) e^{-y}\right]\right|_{y=0}=h(70) \\ 70 e^{-0}=h(70) \\ ^{0} \mathrm{h}=1 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \end{array}
Alternatively
Given data:
\begin{aligned} T_{\infty}&=30^{\circ} \mathrm{C} \\ T_{s}&=100^{\circ} \mathrm{C} \\ k_{f}&=1 \mathrm{W} / \mathrm{mK} \\ T&=30+70 \exp (-y) \end{aligned}
Differentiating w.r.t y, we get
\begin{aligned} \text{At }\quad y &=0\\ \frac{d T}{d y}&=-70 e^{-y}\\ \left(\frac{d T}{d y}\right)^{y=0}&=-70 \end{aligned}
We know that local convective heat transfer coefficient:
h_{x}=\frac{-k_{1}\left(\frac{d T}{d y}\right)_{y=0}}{T_{s}-T_{\infty}}=\frac{-1 \times(-70)}{100-30}=1 \mathrm{W} / \mathrm{m}^{2} \mathrm{K}
Question 7
A frictionless piston-cylinder device contains a gas initially at 0.8MPa and 0.015 m^{3}. It expands quasi-statically at constant temperature to a final volume of 0.030 m^{3}. The work output (in kJ) during this process will be
A
8.32
B
12
C
554.67
D
8320
Thermodynamics   First Law, Heat, Work and Energy
Question 7 Explanation: 
Given data for constant temperature process
\begin{array}{l} p_{1}=0.8 \mathrm{MPa}=800 \mathrm{kPa} \\ V_{1}=0.015 \mathrm{m}^{3} \\ V_{2}=0.030 \mathrm{m}^{3} \end{array}
We know that work done for constant temperature process,
\begin{aligned} W &=p_{1} V_{1} \log _{e} \frac{V_{2}}{V_{1}} \\ &=800 \times 0.015 \log _{e} \frac{0.030}{0.015} \\ &=8.317 \mathrm{kJ} \approx 8.32 \mathrm{kJ} \end{aligned}
Question 8
In an ideal vapour compression refrigeration cycle, the specific enthalpy of refrigerant (in kJ/kg) at the following states is given as:
Inlet of condenser: 283
Exit of condenser: 116
Exit of evaporator: 232
The COP of this cycle is
A
2.27
B
2.75
C
3.27
D
3.75
Refrigeration and Air-conditioning   Vapor and Gas Refrigeration
Question 8 Explanation: 
Given data:
\begin{aligned} h_{2} &=283 \mathrm{kJ} / \mathrm{kg} \\ h_{3} &=116 \mathrm{kJ} / \mathrm{kg} \\ &=h_{4} \\ h_{1} &=232 \mathrm{kJ} / \mathrm{kg} \end{aligned}

Coefficient of performance,
\begin{aligned} \mathrm{COP} &=\frac{\text { Refrigerating effect: Re }}{\text { Work input: } w} \\ \text { where } \mathrm{Re} &=h_{1}-h_{4}=232-116 \\ &=116 \mathrm{kJ} / \mathrm{kg}\\ \text { and } \quad w &=h_{2}-h_{1}=283-232 \\ &=51 \mathrm{kJ} / \mathrm{kg} \\ \therefore \quad \mathrm{COP} &=\frac{116}{51}=2.27 \end{aligned}
Question 9
A compressor undergoes a reversible, steady flow process. The gas at inlet and outlet of the compressor is designated as state 1 and state 2 respectively. Potential and kinetic energy changes are to be ignored. The following notations are used:

v= specific volume and P=pressure of the gas.
The specific work required to be supplied to the compressor for this gas compression process is
A
\int_{1}^{2}Pdv
B
\int_{1}^{2}vdP
C
v_{1}(P_{2}-P_{1})
D
-P_{2}(v_{1}-v_{2})
Thermodynamics   First Law, Heat, Work and Energy
Question 9 Explanation: 
Reversible steady flow process


State 1: Inlet State
2: Outlet
Potential and kinetic energy change ignored.
v= Specific volume of the gas
P= Pressure of the gas
Specific work required to be supplied to the
compressor for this gas compression process is
w=\int_{1}^{2} v d p \quad
(Open system flow in compressor)


Question 10
A block weighing 981N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is \mu=0.2A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right?
A
176.2
B
196
C
481
D
981
Engineering Mechanics   Friction
Question 10 Explanation: 
Free body Diagram

Assume normal reaction =R (Newton)
Balancing force in horizontal direction
(\Sigma F x=0)
\begin{aligned} \mu R &=100 \\ (0.2) R &=100 \\ R &=500 \mathrm{N} \end{aligned}
Balancing the force in vertical direction
\begin{aligned} (\Sigma F y&=0) \\ T+R-W &=0 \\ \Rightarrow T+500-981 &=0 \\ T &=481 \mathrm{N} \end{aligned}
There are 10 questions to complete.

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