Question 1 |

If any two columns of a determinant P= \begin{vmatrix} 4 & 7 & 8\\ 3 & 1 & 5\\ 9& 6 & 2 \end{vmatrix} are interchanged, which one of the following statements regarding the value of the determinant is CORRECT?

Absolute value remains unchanged but sign will change. | |

Both absolute value and sign will change. | |

Absolute value will change but sign will not change | |

Both absolute value and sign will remain unchanged |

Question 1 Explanation:

Property of determinant: If any two row or column are interchanged, then mangnitude of determinant remains same but sign changes.

Question 2 |

Among the four normal distributions with probability density functions as shown below, which one has the lowest variance?

I | |

II | |

III | |

IV |

Question 2 Explanation:

We know that probability density function of normal distribution is given by

F(x)=\frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}\left ( \frac{x-\mu }{\sigma } \right )^2}

For having lowest variance (\sigma ^2), standard deviation (\sigma ) will be lowest. From the density function f(x), we can say that as σ decreases F(x) will increase, so curve having highest peak has lowest standard deviation and variance.

F(x)=\frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}\left ( \frac{x-\mu }{\sigma } \right )^2}

For having lowest variance (\sigma ^2), standard deviation (\sigma ) will be lowest. From the density function f(x), we can say that as σ decreases F(x) will increase, so curve having highest peak has lowest standard deviation and variance.

Question 3 |

Simpson's 1/3 rule is used to integrate the function \frac{3}{5}x^{2}+\frac{9}{5} between x = 0 and x = 1 using the least number of equal sub -intervals. The value of the integral is _____________

1 | |

2 | |

3 | |

4 |

Question 3 Explanation:

f(x)=\frac{3}{5} x^{2}+\frac{9}{5}

\begin{array}{|c|c|c|c|}\hline x & 0 & 0.5 & 1 \\\hline f(x) & 1.8 & 1.95 & 2.4 \\\hline\end{array}\\ \Rightarrow\int_{0}^{1} f(x)=\frac{h}{3}\left[y_{0}+4 y_{1}+y_{2}\right]

=\frac{0.5}{3}[1.8+4(1.95)+2.4]=2

\begin{array}{|c|c|c|c|}\hline x & 0 & 0.5 & 1 \\\hline f(x) & 1.8 & 1.95 & 2.4 \\\hline\end{array}\\ \Rightarrow\int_{0}^{1} f(x)=\frac{h}{3}\left[y_{0}+4 y_{1}+y_{2}\right]

=\frac{0.5}{3}[1.8+4(1.95)+2.4]=2

Question 4 |

The value of \lim_{x\rightarrow 0}\frac{1-cos(x^{2})}{2x^{4}} is

0 | |

1/2 | |

1/4 | |

undefined |

Question 4 Explanation:

\lim _{x \rightarrow 0} \frac{1-\cos \left(x^{2}\right)}{2 x^{4}} \quad putting the x \rightarrow 0

we get \frac{0}{0} form

Applying L' Hospital rule

\Rightarrow \lim _{x \rightarrow 0} \frac{2 x \sin \left(x^{2}\right)}{8 x^{3}} \quad

\Rightarrow \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{4 x^{2}}

\Rightarrow \frac{1}{4} \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}

\Rightarrow \frac{1}{4} \lim _{x^{2} \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}=\frac{1}{4} \times 1=\frac{1}{4}

we get \frac{0}{0} form

Applying L' Hospital rule

\Rightarrow \lim _{x \rightarrow 0} \frac{2 x \sin \left(x^{2}\right)}{8 x^{3}} \quad

\Rightarrow \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{4 x^{2}}

\Rightarrow \frac{1}{4} \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}

\Rightarrow \frac{1}{4} \lim _{x^{2} \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}=\frac{1}{4} \times 1=\frac{1}{4}

Question 5 |

Given two complex numbers z_{1}=5+(5\sqrt{3})i and z_{2}=2/\sqrt{3}+2i, the argument of z_{1}/z_{2} in degrees is

0 | |

30 | |

60 | |

90 |

Question 5 Explanation:

\begin{aligned} z_{1} &=5+(5 \sqrt{3}) i \\ z_{2} &=\frac{2}{\sqrt{3}}+2 i \\ \arg \left(z_{1}\right) &=\theta_{1}=\tan ^{-1}\left(\frac{5 \sqrt{3}}{5}\right) \\ \theta_{1} &=60^{\circ} \\ \arg \left(z_{2}\right) &=\theta_{2}=\tan ^{-1}\left(\frac{2}{2 \sqrt{3}}\right) \\ \theta_{2} &=60^{\circ} \\ \arg \left(\frac{z_{1}}{z_{2}}\right) &=\arg \left(z_{1}\right)-\arg \left(z_{2}\right) \\ &=60-60=0^{\circ} \end{aligned}

Question 6 |

Consider fully developed flow in a circular pipe with negligible entrance length effects. Assuming the mass flow rate, density and friction factor to be constant, if the length of the pipe is doubled and the diameter is halved, the head loss due to friction will increase by a factor of

4 | |

16 | |

32 | |

64 |

Question 6 Explanation:

Head loss due to friction,

\begin{aligned} h_{f}&=\frac{f L V^{2}}{2 g d}\\ &\text{Mass flow rate,}\\ m&=\rho A V=\rho \times \frac{\pi}{4} d^{2} V\\ or\quad V&=\frac{4 m}{\rho \pi d^{2}} \\ \therefore \quad h_{f}&=\frac{f L}{2 g d} \times \frac{16 m^{2}}{\rho^{2} \pi^{2} d^{4}}\\ &=\frac{8 f m^{2}}{\rho^{2} \pi^{2}} \times \frac{L}{d^{5}} \\ &=C \times \frac{L}{d^{5}} \text { where } C=\frac{8 f m^{2}}{\rho^{2} \pi^{2}} \\ h_{f_{1}}&=C \times \frac{L}{d^{5}}\\ and\quad h_{2} &=C \times \frac{2 L}{(d / 2)^{5}}=\frac{C \times 2 L}{\frac{d^{5}}{2^{5}}} \\ &=\frac{C \times 2 \times 2^{5} L}{d^{5}}=\frac{C \times 64 L}{d^{5}} \\ h_{1 / 2} &=64 h_{11} \end{aligned}

The head loss due to friction will increase by a factor of 64

\begin{aligned} h_{f}&=\frac{f L V^{2}}{2 g d}\\ &\text{Mass flow rate,}\\ m&=\rho A V=\rho \times \frac{\pi}{4} d^{2} V\\ or\quad V&=\frac{4 m}{\rho \pi d^{2}} \\ \therefore \quad h_{f}&=\frac{f L}{2 g d} \times \frac{16 m^{2}}{\rho^{2} \pi^{2} d^{4}}\\ &=\frac{8 f m^{2}}{\rho^{2} \pi^{2}} \times \frac{L}{d^{5}} \\ &=C \times \frac{L}{d^{5}} \text { where } C=\frac{8 f m^{2}}{\rho^{2} \pi^{2}} \\ h_{f_{1}}&=C \times \frac{L}{d^{5}}\\ and\quad h_{2} &=C \times \frac{2 L}{(d / 2)^{5}}=\frac{C \times 2 L}{\frac{d^{5}}{2^{5}}} \\ &=\frac{C \times 2 \times 2^{5} L}{d^{5}}=\frac{C \times 64 L}{d^{5}} \\ h_{1 / 2} &=64 h_{11} \end{aligned}

The head loss due to friction will increase by a factor of 64

Question 7 |

The Blasius equation related to boundary layer theory is a

third-order linear partial differential equation | |

third-order nonlinear partial differential equation | |

second-order nonlinear ordinary differential equation | |

third-order nonlinear ordinary differential equation |

Question 7 Explanation:

2 \frac{d^{3} f}{d \eta^{3}}+f \frac{d^{2} f}{d \eta^{2}}=0

third-order non-linear differential equation.

third-order non-linear differential equation.

Question 8 |

For flow of viscous fluid over a flat plate, if the fluid temperature is the same as the plate temperature, the thermal boundary layer is

thinner than the velocity boundary layer | |

thicker than the velocity boundary layer | |

of the same thickness as the velocity boundary layer | |

not formed at all |

Question 8 Explanation:

For the flow of viscous fluid over a flat plate if the fluid temperature is the same as the plate temperature the thermal boundary layer is not formed at all because boundary layer formation took place if there is some difference in fluid property of no slip layer and remaining fluid.

As here given that fluid is viscous and flowing over a flat plate. Here, definitely a kinematic boundary layer will be formed but as there is no temperature difference so no formation of thermal boundary layer.

As here given that fluid is viscous and flowing over a flat plate. Here, definitely a kinematic boundary layer will be formed but as there is no temperature difference so no formation of thermal boundary layer.

Question 9 |

For an ideal gas with constant values of specific heats, for calculation of the specific enthalpy,

it is sufficient to know only the temperature | |

both temperature and pressure are required to be known | |

both temperature and volume are required to be known | |

both temperature and mass are required to be known |

Question 9 Explanation:

For constant valves of specific heats

dh = F(T) only

So for calculation of specific enthalpy it is sufficient to know only the temperature.

dh = F(T) only

So for calculation of specific enthalpy it is sufficient to know only the temperature.

Question 10 |

A Carnot engine (CE-1) works between two temperature reservoirs A and B, where T_{A} = 900 K and T_{B} = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs B and C, where T_{C} = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 to reservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 from reservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is ______________

20 | |

30 | |

50 | |

60 |

Question 10 Explanation:

For Carnot engine (CE-2)

\begin{aligned} \eta_{2}&=1-\frac{T_{C}}{T_{B}}\\ \text{also}\qquad \eta_{2} &=1-\frac{Q_{C}}{Q_{B}} \\ \therefore \quad 1-\frac{T_{C}}{T_{B}} &=1-\frac{Q_{C}}{Q_{B}}\\ \text{or}\qquad\frac{T_{C}}{T_{B}} &=\frac{Q_{C}}{Q_{B}} \\ \frac{300}{500} &=\frac{Q_{C}}{\frac{5 \times 150}{9}}\\ \text{or}\qquad Q_{c}&=50 \mathrm{MJ} \end{aligned}

For Carnot engine (CE-2)

\begin{aligned} \eta_{2}&=1-\frac{T_{C}}{T_{B}}\\ \text{also}\qquad \eta_{2} &=1-\frac{Q_{C}}{Q_{B}} \\ \therefore \quad 1-\frac{T_{C}}{T_{B}} &=1-\frac{Q_{C}}{Q_{B}}\\ \text{or}\qquad\frac{T_{C}}{T_{B}} &=\frac{Q_{C}}{Q_{B}} \\ \frac{300}{500} &=\frac{Q_{C}}{\frac{5 \times 150}{9}}\\ \text{or}\qquad Q_{c}&=50 \mathrm{MJ} \end{aligned}

There are 10 questions to complete.