Question 1 |
The closed curve shown in the figure is described by
r=1+\cos \theta, where r=\sqrt{x^{2}+y^{2}} x=r \cos \theta, y=r \sin \theta
The magnitude of the line integral of the vector field F=-y \hat{i}+x \hat{j} around the closed curve is ___(Round off to 2 decimal places).
r=1+\cos \theta, where r=\sqrt{x^{2}+y^{2}} x=r \cos \theta, y=r \sin \theta
The magnitude of the line integral of the vector field F=-y \hat{i}+x \hat{j} around the closed curve is ___(Round off to 2 decimal places).
9.42 | |
6.36 | |
2.45 | |
7.54 |
Question 1 Explanation:
\begin{aligned}
I & =\int_{0}^{2 \pi} \vec{F} \cdot \overrightarrow{d l} \\
& =\int_{0}^{2 \pi}(-y \hat{i}+x)(d x \hat{i}+d y) \\
& =\int_{0}^{2 \pi}(-y d x+x d y)
\end{aligned}
Given : \quad x=r \cos \theta and y=r \sin \theta
\begin{aligned} \therefore I&=\int_{0}^{2 \pi}[(-r \sin \theta)(-r \sin \theta) d \theta+(r \cos \theta)(r \cos \theta) d \theta] \\ &=\int_{0}^{2 \pi} r^{2} d \theta \\ &=\int_{0}^{2 \pi}(1+\cos \theta)^{2} d \theta \\ &=\int_{0}^{2 \pi}\left(1+\cos ^{2} \theta+2 \cos \theta\right) d \theta \\ &=\int_{0}^{2 \pi}\left(1+\frac{1+\cos 2 \theta}{2}+2 \cos \theta\right) \mathrm{d} \theta \\ &=3 \pi=9.425 \end{aligned}
Given : \quad x=r \cos \theta and y=r \sin \theta
\begin{aligned} \therefore I&=\int_{0}^{2 \pi}[(-r \sin \theta)(-r \sin \theta) d \theta+(r \cos \theta)(r \cos \theta) d \theta] \\ &=\int_{0}^{2 \pi} r^{2} d \theta \\ &=\int_{0}^{2 \pi}(1+\cos \theta)^{2} d \theta \\ &=\int_{0}^{2 \pi}\left(1+\cos ^{2} \theta+2 \cos \theta\right) d \theta \\ &=\int_{0}^{2 \pi}\left(1+\frac{1+\cos 2 \theta}{2}+2 \cos \theta\right) \mathrm{d} \theta \\ &=3 \pi=9.425 \end{aligned}
Question 2 |
Consider the following equation in a 2-D realspace.
\left|x_{1}\right|^{p}+\left|x_{2}\right|^{p}=1 for p \gt 0
Which of the following statement(s) is/are true.
\left|x_{1}\right|^{p}+\left|x_{2}\right|^{p}=1 for p \gt 0
Which of the following statement(s) is/are true.
When \mathrm{p}=2, the area enclosed by the curve is \pi. | |
When p tends to \infty, the area enclosed by the curve tends to 4. | |
When p tends to 0 , the area enclosed by the curve is 1. | |
When p=1, the area enclosed by the curve is 2. |
Question 2 Explanation:
Check option (A),
put P=2
x_{1}^{2}+x_{2}^{2}=1
Which is equation of circle whose radius is 1 .
\therefore \quad Area =\pi r^{2}=\pi
Check option (B),
If x_{2} \lt 1 Then, P \rightarrow \infty,\left|x_{2}\right|^{P} \rightarrow 0
\therefore For equation \left|x_{1}\right|^{\mathrm{P}}+\left|\mathrm{x}_{2}\right|^{\mathrm{P}}=1 satisfaction,
\mathrm{x}_{1} \rightarrow 1 \text { or } \dot{Y}
Curve :
\therefore Area tends to 4 .
Check option (D),
Put P=1,
\left|x_{1}\right|+\left|\mathrm{x}_{2}\right|=1
Curve:
\therefore Area of curve (square) =(\sqrt{2})^{2}=2
put P=2
x_{1}^{2}+x_{2}^{2}=1
Which is equation of circle whose radius is 1 .
\therefore \quad Area =\pi r^{2}=\pi
Check option (B),
If x_{2} \lt 1 Then, P \rightarrow \infty,\left|x_{2}\right|^{P} \rightarrow 0
\therefore For equation \left|x_{1}\right|^{\mathrm{P}}+\left|\mathrm{x}_{2}\right|^{\mathrm{P}}=1 satisfaction,
\mathrm{x}_{1} \rightarrow 1 \text { or } \dot{Y}
Curve :
\therefore Area tends to 4 .
Check option (D),
Put P=1,
\left|x_{1}\right|+\left|\mathrm{x}_{2}\right|=1
Curve:
\therefore Area of curve (square) =(\sqrt{2})^{2}=2
Question 3 |
One million random numbers are generated from a statistically stationary process with a Gaussian distribution with mean zero and standard deviation \sigma_{0}.
The \sigma_{0} is estimated by randomly drawing out 10,000 numbers of samples \left(x_{n}\right). The estimates \hat{\sigma}_{1}, \hat{\sigma}_{2} are computed in the following two ways.
\hat{\sigma}_{1}^{2}=\frac{1}{10000} \sum_{n=1}^{10000} x_{n}^{2} \quad \hat{\sigma}_{2}^{2}=\frac{1}{9999} \sum_{n=1}^{10000} x_{n}^{2}
Which of the following statements is true?
The \sigma_{0} is estimated by randomly drawing out 10,000 numbers of samples \left(x_{n}\right). The estimates \hat{\sigma}_{1}, \hat{\sigma}_{2} are computed in the following two ways.
\hat{\sigma}_{1}^{2}=\frac{1}{10000} \sum_{n=1}^{10000} x_{n}^{2} \quad \hat{\sigma}_{2}^{2}=\frac{1}{9999} \sum_{n=1}^{10000} x_{n}^{2}
Which of the following statements is true?
E\left(\hat{\sigma}_{2}^{2}\right)=\sigma_{0}^{2} | |
\mathrm{E}\left(\hat{\sigma}_{2}\right)=\sigma_{0} | |
E\left(\hat{\sigma}_{1}^{2}\right)=\sigma_{0}^{2} | |
\mathrm{E}\left(\hat{\sigma}_{1}\right)=\mathrm{E}\left(\hat{\sigma}_{2}\right) |
Question 3 Explanation:
We know,
\mathrm{E}\left[(\mathrm{X}-\overline{\mathrm{X}})^{2}\right]=\operatorname{Var}
Given, \quad \bar{X}=0
\therefore \quad E[X]=\operatorname{Var}.
We have,
Standard deviation, \sigma^{2} =\frac{1}{N-1} \sum_{n=1}^{n} x_{n}^{2} \Rightarrow \text { for sample }
and \quad \sigma^{2}=\frac{1}{N} \sum_{n=1}^{n} x_{n}^{2} \Rightarrow For population
Here, we take samples from 10^{6} numbers.
\therefore \quad \mathrm{E}\left(\sigma_{2}^{2}\right)=\sigma_{0}^{2} will be correct.
\mathrm{E}\left[(\mathrm{X}-\overline{\mathrm{X}})^{2}\right]=\operatorname{Var}
Given, \quad \bar{X}=0
\therefore \quad E[X]=\operatorname{Var}.
We have,
Standard deviation, \sigma^{2} =\frac{1}{N-1} \sum_{n=1}^{n} x_{n}^{2} \Rightarrow \text { for sample }
and \quad \sigma^{2}=\frac{1}{N} \sum_{n=1}^{n} x_{n}^{2} \Rightarrow For population
Here, we take samples from 10^{6} numbers.
\therefore \quad \mathrm{E}\left(\sigma_{2}^{2}\right)=\sigma_{0}^{2} will be correct.
Question 4 |
In the figure, the vectors u and v are related as: A u=v by a transformation matrix A. The correct choice of A is
\left[\begin{array}{cc}\frac{4}{5} & \frac{3}{5} \\ -\frac{3}{5} & \frac{4}{5}\end{array}\right] | |
\left[\begin{array}{cc}\frac{4}{5} & -\frac{3}{5} \\ \frac{3}{5} & \frac{4}{5}\end{array}\right] | |
\left[\begin{array}{ll}\frac{4}{5} & \frac{3}{5} \\ \frac{3}{5} & \frac{4}{5}\end{array}\right] | |
\left[\begin{array}{rr}\frac{4}{5} & -\frac{3}{5} \\ \frac{3}{5} & -\frac{4}{5}\end{array}\right] |
Question 4 Explanation:
Given :
\begin{aligned} & |u|=\sqrt{4^{2}+3^{2}}=5 \\ & |V|=5 \end{aligned}
and \quad \theta=\tan ^{-1} \frac{3}{4} \Rightarrow \tan \theta=\frac{3}{4}
\therefore \quad \sin \theta=\frac{3}{5} and \cos \theta=\frac{4}{5}
For clockwise relation by \theta, transformation matrix,
\begin{aligned} A & =\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ & =\left[\begin{array}{cc} \frac{4}{5} & \frac{3}{5} \\ -\frac{3}{5} & \frac{4}{5} \end{array}\right] \end{aligned}
[\because F or clockwise \theta \rightarrow N egative ]
\begin{aligned} & |u|=\sqrt{4^{2}+3^{2}}=5 \\ & |V|=5 \end{aligned}
and \quad \theta=\tan ^{-1} \frac{3}{4} \Rightarrow \tan \theta=\frac{3}{4}
\therefore \quad \sin \theta=\frac{3}{5} and \cos \theta=\frac{4}{5}
For clockwise relation by \theta, transformation matrix,
\begin{aligned} A & =\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ & =\left[\begin{array}{cc} \frac{4}{5} & \frac{3}{5} \\ -\frac{3}{5} & \frac{4}{5} \end{array}\right] \end{aligned}
[\because F or clockwise \theta \rightarrow N egative ]
Question 5 |
For a given vector w=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]^{\top}, the vector normal to the plane defined by \mathbf{w}^{\top} x=1 is
\left[\begin{array}{lll}-2 & -2 & 2\end{array}\right]^{T} | |
\left[\begin{array}{lll}3 & 0 & -1\end{array}\right]^{T} | |
\left[\begin{array}{lll}3 & 2 & 1\end{array}\right]^{T} | |
\left[\begin{array}{llll}1 & 2 & 3\end{array}\right]^{T} |
Question 5 Explanation:
Given, W^{T}=1
\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=1
We have, vector normal to the plane =\nabla F
\begin{aligned} & =i \frac{\partial F}{\partial x}+\hat{j} \frac{\partial F}{\partial y}+\hat{k} \frac{\partial F}{\partial z} \\ & =\hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}
\therefore Normal vector =\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]^{T}
\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=1
We have, vector normal to the plane =\nabla F
\begin{aligned} & =i \frac{\partial F}{\partial x}+\hat{j} \frac{\partial F}{\partial y}+\hat{k} \frac{\partial F}{\partial z} \\ & =\hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}
\therefore Normal vector =\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]^{T}
There are 5 questions to complete.