Question 1 |
Two vectors \left[\begin{array}{llll}2 & 1 & 0 & 3\end{array}\right]^{\top} and \left[\begin{array}{llll}1 & 0 & 1 & 2\end{array}\right]^{\top} belong to the null space of a 4 \times 4 matrix of rank 2 . Which one of the following vectors also belongs to the null space?
\left[\begin{array}{llll}1 & 1 & -1 & 1\end{array}\right]^{\top} | |
\left[\begin{array}{llll}2 & 0 & 1 & 2\end{array}\right]^{\top} | |
\left[\begin{array}{llll}0 & -2 & 1 & -1\end{array}\right]^{\top} | |
\left[\begin{array}{llll}3 & 1 & 1 & 2\end{array}\right]^{\top} |
Question 1 Explanation:
Given matrix is 4 \times 4 and rank of matrix is 2 .
Therefore, rank of matrix \neq No. of variables Thus, there two linearly dependent vectors \& two linearly independent vectors are present.
\begin{aligned} & X_{1}=\left[\begin{array}{llll} 2 & 1 & 0 & 3 \end{array}\right]^{\top} \\ & X_{2}=\left[\begin{array}{llll} 1 & 0 & 1 & 2 \end{array}\right]^{\top} \end{aligned}
\therefore \quad X=\mathrm{K}_{1}\left[\begin{array}{l}2 \\ 1 \\ 0 \\ 3\end{array}\right]+\mathrm{K}_{2}\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 2\end{array}\right]
For \mathrm{K}_{1}=1 \text{ and } \mathrm{~K}_{2}=-1
X=\left[\begin{array}{c}1 \\ 1 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{llll}1 & 1 & -1 & 1\end{array}\right]^{\top}
Therefore, rank of matrix \neq No. of variables Thus, there two linearly dependent vectors \& two linearly independent vectors are present.
\begin{aligned} & X_{1}=\left[\begin{array}{llll} 2 & 1 & 0 & 3 \end{array}\right]^{\top} \\ & X_{2}=\left[\begin{array}{llll} 1 & 0 & 1 & 2 \end{array}\right]^{\top} \end{aligned}
\therefore \quad X=\mathrm{K}_{1}\left[\begin{array}{l}2 \\ 1 \\ 0 \\ 3\end{array}\right]+\mathrm{K}_{2}\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 2\end{array}\right]
For \mathrm{K}_{1}=1 \text{ and } \mathrm{~K}_{2}=-1
X=\left[\begin{array}{c}1 \\ 1 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{llll}1 & 1 & -1 & 1\end{array}\right]^{\top}
Question 2 |
Let \phi be a scalar field, and \mathbf{u} be a vector field. Which of the following identities is true for div(\phi \mathbf{u}) ?
\operatorname{div}(\phi \mathbf{u})=\phi \operatorname{div}(\mathbf{u})+\mathbf{u} \cdot \operatorname{grad}(\phi) | |
\operatorname{div}(\phi \mathbf{u})=\phi \operatorname{div}(\mathbf{u})+\mathbf{u} \times \operatorname{grad}(\phi) | |
\operatorname{div}(\phi \mathbf{u})=\phi \operatorname{grad}(\mathbf{u})+\mathbf{u} \cdot \operatorname{grad}(\phi) | |
\operatorname{div}(\phi \mathbf{u})=\phi \operatorname{grad}(\mathbf{u})+\mathbf{u} \times \operatorname{grad}(\phi) |
Question 2 Explanation:
div(\phi \mathrm{u})=\phi div(\mu)+ugrad(\phi)
Question 3 |
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 3 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 4 |
The value of the integral \iint_{R} x y d x d y over the region R, given in the figure, is ___
(rounded off to the nearest integer).
(rounded off to the nearest integer).
0 | |
1 | |
2 | |
3 |
Question 4 Explanation:
\begin{aligned} I & =\iint_{R} x y d x d y \\ & =\int_{y=0}^{1} \int_{x=-y}^{y} x y d x d y+\int_{y=1}^{2} \int_{x=y-2}^{2-y} x y d x d y \\ & =\int_{0}^{1} y\left(\frac{x^{2}}{2}\right)_{-y}^{y} d y+\int_{1}^{2} y\left(\frac{x^{2}}{2}\right)_{y-2}^{2-y} d y \\ & =0+0=0 \end{aligned}
Question 5 |
The smallest perimeter that a rectangle with area of 4 square units can have is
______ units.
(Answer in integer)
4 | |
6 | |
8 | |
10 |
Question 5 Explanation:
Given \quad B \times L=4
Perimeter (P)=2 B+2 L
perimeter to be smallest possible
\frac{d P}{d B}=0=\frac{d}{d B}\left(2 B+2 \times \frac{4}{B}\right)=0
2-\frac{8}{B^{2}}=0
B=\pm 2 \Rightarrow B=2 \quad L=2
Smallest perimeter =2(B+L)=2(2+2)=8
There are 5 questions to complete.
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In question 27, the limit is tending to 1 instead of 0. Please update it accordingly.
Kindly seperate topic wise in maths also
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