Question 1 |

If y(x) satisfies the differential equation

(\sin x)\frac{dy}{dx}+y \cos x =1

subject to the condition y(\pi /2)=\pi /2, then y(\pi /6) is

(\sin x)\frac{dy}{dx}+y \cos x =1

subject to the condition y(\pi /2)=\pi /2, then y(\pi /6) is

0 | |

\frac{\pi}{6} | |

\frac{\pi}{3} | |

\frac{\pi}{2} |

Question 1 Explanation:

\begin{aligned} \frac{d y}{d x}+y \cot x&=\text{cosec} x\\ 1.F. \qquad&=e^{\int \cot x d x}=e^{\log \sin x}=\sin x\\ \Rightarrow \quad y(\sin x)&=\int \text{cosec} x \sin x d x+c\\ \Rightarrow \qquad y \sin x&=x+c\\ \Rightarrow \qquad \frac{\pi}{2} \sin \frac{\pi}{2} & =\frac{\pi}{2}+c \\ \Rightarrow \qquad \frac{\pi}{2} & =\frac{\pi}{2}+c \quad \Rightarrow c=0 \\ \Rightarrow \qquad y \sin x & =x\\ \Rightarrow \qquad y \sin \frac{\pi}{6}&=\frac{\pi}{6}\\ \Rightarrow \qquad y\left(\frac{1}{2}\right) &=\frac{\pi}{6} \\ \Rightarrow y &=\frac{\pi}{3} \end{aligned}

Question 2 |

The value of \lim_{x \to 0}\left ( \frac{1- \cos x}{x^2} \right ) is

\frac{1}{4} | |

\frac{1}{3} | |

\frac{1}{2} | |

1 |

Question 2 Explanation:

\begin{aligned} \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)&=? \;\;\;\;\;\;\left(\frac{0}{0} \text { form }\right) \\ \text { Applying } L \cdot H \text { rule } & =\lim _{x \rightarrow 0} \frac{\sin x}{2 x}\left(\frac{0}{0}\right)=\lim _{x \rightarrow 0} \frac{\cos x}{2}=\frac{1}{2} \end{aligned}

Question 3 |

The Dirac-delta function (\delta (t-t_0)) \text{ for }t,t_0 \in \mathbb{R}, has the following property

\int_{a}^{b}\varphi (t)\delta (t-t_0)dt=\left\{\begin{matrix} \varphi (t_0) & a \lt t_0 \lt b\\ 0 &\text{otherwise} \end{matrix}\right.

The Laplace transform of the Dirac-delta function \delta (t-a) for a \gt 0;

\mathcal{L} (\delta (t-a))=F(s) is

\int_{a}^{b}\varphi (t)\delta (t-t_0)dt=\left\{\begin{matrix} \varphi (t_0) & a \lt t_0 \lt b\\ 0 &\text{otherwise} \end{matrix}\right.

The Laplace transform of the Dirac-delta function \delta (t-a) for a \gt 0;

\mathcal{L} (\delta (t-a))=F(s) is

0 | |

\infty | |

e^{sa} | |

e^{-sa} |

Question 3 Explanation:

\begin{aligned} \because \qquad \int_{0}^{-} f(t) \delta(t-a) d t&=f(a) \\ \therefore \qquad L\{\delta(t-a)\}&=\int_{0}^{-} e^{-s t} \delta(t-a) d t=e^{-a s} \end{aligned}

Question 4 |

The ordinary differential equation \frac{dy}{dt}=-\pi y subject to an initial condition y(0)=1 is solved numerically using the following scheme:

\frac{y(t_{n+1})-y(t_n)}{h}=-\pi y(t_n)

where h is the time step, t_n=nh, and n=0,1,2,.... This numerical scheme is stable for all values of h in the interval.

\frac{y(t_{n+1})-y(t_n)}{h}=-\pi y(t_n)

where h is the time step, t_n=nh, and n=0,1,2,.... This numerical scheme is stable for all values of h in the interval.

0 \lt h \lt \frac{2}{\pi} | |

0 \lt h \lt 1 | |

0 \lt h \lt \frac{\pi}{2} | |

for all h \gt 0 |

Question 4 Explanation:

\begin{aligned} \frac{y\left(t_{n+1}\right)-y\left(t_{n}\right)}{h} &=-\pi y\left(t_{n}\right) \\ y_{n+1} &=-\pi / y_{n}+y_{n}=(-\pi h+1) y_{n} \end{aligned}

It is recursion relation between y_{n+1} and y_{n}

So solution will be stable if

\begin{aligned} |-\pi h+1| & \lt 1 \\ -1 \lt -\pi h+1 & \lt 1 \\ -2 \lt -\pi h & \lt 0 \\ 0 & \lt \pi h \lt 2 \\ 0 & \lt h \lt \frac{2}{\pi} \end{aligned}

Therefore option (A) is correct.

It is recursion relation between y_{n+1} and y_{n}

So solution will be stable if

\begin{aligned} |-\pi h+1| & \lt 1 \\ -1 \lt -\pi h+1 & \lt 1 \\ -2 \lt -\pi h & \lt 0 \\ 0 & \lt \pi h \lt 2 \\ 0 & \lt h \lt \frac{2}{\pi} \end{aligned}

Therefore option (A) is correct.

Question 5 |

Consider a binomial random variable X. If X_1,X_2,..., X_n are independent and identically distributed samples from the distribution of X with sum Y=\sum_{i=1}^{n}X_i, then the distribution of Y as n\rightarrow \infty can be approximated as

Exponential | |

Bernoulli | |

Binomial | |

Normal |

Question 6 |

The loading and unloading response of a metal is shown in the figure. The elastic and plastic strains corresponding to 200 MPa stress, respectively, are

0.01 and 0.01 | |

0.02 and 0.01 | |

0.01 and 0.02 | |

0.02 and 0.02 |

Question 6 Explanation:

Elastic strain : Which can be recovered = 0.03 - 0.01 = 0.02

Plastic strain : Permanent strain = 0.01

Plastic strain : Permanent strain = 0.01

Question 7 |

In a machining operation, if a cutting tool traces the workpiece such that the directrix is perpendicular to the plane of the generatrix as shown in figure, the surface generated is

plane | |

cylindrical | |

spherical | |

a surface of revolution |

Question 7 Explanation:

Question 8 |

The correct sequence of machining operations to be performed to finish a large diameter through hole is

drilling, boring, reaming | |

boring, drilling, reaming | |

drilling, reaming, boring | |

boring, reaming, drilling |

Question 8 Explanation:

Drilling: to produce a hole, which then may be followed by boring it to improve its dimensional accuracy and surface finish.

Boring: to enlarge a hole or cylindrical cavity made by a previous process or to produce circular internal grooves.

Reaming: is an operation used to (a) make an existing hole dimensionally more accurate than can br achived by drilling alone and (b) improve its surface finish. The most accurate holes in workpieces generally are produced by the following sequence of operation.

Centering -> Drilling -> Boring -> Reaming.

Boring: to enlarge a hole or cylindrical cavity made by a previous process or to produce circular internal grooves.

Reaming: is an operation used to (a) make an existing hole dimensionally more accurate than can br achived by drilling alone and (b) improve its surface finish. The most accurate holes in workpieces generally are produced by the following sequence of operation.

Centering -> Drilling -> Boring -> Reaming.

Question 9 |

In modern CNC machine tools, the backlash has been eliminated by

preloaded ballscrews | |

rack and pinion | |

ratchet and pinion | |

slider crank mechanism |

Question 9 Explanation:

Question 10 |

Consider the surface roughness profile as shown in the figure.

The center line average roughness (R_a \text{ in }\mu m) of the measured length (L) is

The center line average roughness (R_a \text{ in }\mu m) of the measured length (L) is

0 | |

1 | |

2 | |

4 |

Question 10 Explanation:

R_{G}=\frac{\sum_{i=1}^{n} y}{n}=\frac{4}{4}=1

There are 10 questions to complete.

Great job. Can you add the marks for each question so we can differentiate between 2- and 1-mark questions?