# Electronics Enginnering

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Ankit Pandey ♦♦

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#### Posts by Users

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... Transfer etc. The application will be easy and multilingual to be configured in Local Language. Provide suitable Analysis and Interface design for the same.
**Mumbai University > Computer Engineering > Sem 8 > Human Machine Interation**
**Marks:** 10 M
**Difficulty:** Medium ...

written 2 days ago by
Ankit Pandey ♦♦

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... The portal consists of online meter logging facility, Bill Payments, VDS i.e Voluntary Deposit Scheme for Bill. Complaints and other facilities. Being a Subject Matter Experts (SME) provide the detailed analysis and for the same provide the Interface that will used by people in all Districts of Maha ...

written 2 days ago by
Ankit Pandey ♦♦

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... **Mumbai University > Computer Engineering > Sem 8 > Human Machine Interation**
**Marks:** 10 M
**Difficulty:** Medium ...

written 2 days ago by
Ankit Pandey ♦♦

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... Solution:
Let $v=\iiint x^{2} \ d x \ d y \ d z$
Region of integration is volume bounded by the planes $x=0, y=0, z=0$
And $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Put $x=a u, y=b v, z=c w$
$\therefore \quad \ d x \ d y \ d z=a \ b \ c \ d u . d v$
![enter image description here][1]
The inter ...

written 3 days ago by
Ankit Pandey ♦♦

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... Solution:
let $\quad I=\int_{0}^{\frac{a}{\sqrt{2}}} \int_{y}^{\sqrt{a^{2}-y^{2}}} \log \left(x^{2}+y^{2}\right) d x \ d y$
$\begin{aligned} \text { Region of integration: } & y \leq x \leq \sqrt{a^{2}-y^{2}} \\ & 0 \leq y \leq \frac{a}{\sqrt{2}} \end{aligned}$
The line $x=y$ is inclined ...

written 3 days ago by
Ankit Pandey ♦♦

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... Solution:
let $\quad I=\int_{0}^{\pi / 6} \cos ^{6} 3 \theta \cdot \sin ^{2} 6 \theta d \theta$
Put $\quad 3 \theta=t$
Diff. w.r.t $\theta$
$\quad d \theta=\frac{d t}{3} \quad$ limits: $\left[0, \frac{\pi}{2}\right]$
$\begin{aligned} \therefore \mathrm{I} &=\frac{1}{3} \int_{0}^{\pi / 2} \c ...

written 3 days ago by
Ankit Pandey ♦♦

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Answer:
A: Compute the value of

... Solution:
let $I=\int_{0.2}^{1.4}\left(\sin x-\ln x+e^{x}\right) d x$
Dividing limits in six subintervals.
$\quad \therefore \mathrm{n}=6 \quad \therefore \mathrm{h}=\frac{b-a}{n}=\frac{1.4-0.2}{6}=\frac{1}{5}$
$\begin{array}{|c|c|c|c|c|c|c|}\hline x_{0=0.2} & {x_{1}=0.4} & {x_{2}=0.6} & ...

written 3 days ago by
Ankit Pandey ♦♦

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... **(1)**
$\begin{aligned} \frac{d y}{d x} &=x y+1 \quad, x_{0}=0, y_{0}=0, \mathrm{h}=0.1 \\ f(x, y) &=1+x y \end{aligned}$
$\begin{array}{ll}{y^{\prime}=1+x y} & {y_{0}^{\prime}=1} \\ {y^{\prime \prime}=x y^{\prime}+y} & {y_{0}^{\prime \prime}=0} \\ {y^{\prime \prime \prime}=x y^{\ ...

written 3 days ago by
Ankit Pandey ♦♦

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... Solution:
**(1)**
$\frac{d y}{d x}=\frac{y-x}{y+x} \quad x_{0}=0, y_{0}=1, h=0.2$
$f(x, y)=\frac{y-x}{y+x}$
$k_{1}=h . f\left(x_{0}, y_{0}\right)=0.2 f(0,1)=0.2$
$k_{2}=h . f\left(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{1}}{2}\right)=0.2 . f(0.1,1.1)=0.1666$
$k_{3}=h . f\left(x_{0}+\frac{h}{2}, y_{ ...

written 3 days ago by
Ankit Pandey ♦♦

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... Solution:
Given curve: astroid $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$
The line $\theta=\pi / 6$ cuts the asroid in 1 st quadrant.
![enter image description here][1]
C is the point on the curve which cuts the arc.
Length of astroid in first quadrant:
Put $\quad x=a \cos ^{3} t \quad$ and $\quad y=a \s ...

written 3 days ago by
Ankit Pandey ♦♦

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