QBManager

Motion in Plane

143616 A body is rotating with angular velocity \(\omega=(3 \hat{i}-4 \hat{j}+\hat{k})\). The linear velocity of a point having position vector \(r=(5 \hat{i}-6 \hat{j}+6 \hat{k})\) is

1 \(6 \hat{i}+2 \hat{j}-3 \hat{k}\)

2 \(18 \hat{i}+13 \hat{j}-2 \hat{k}\)

3 \(-18 \hat{i}-13 \hat{j}+2 \hat{k}\)

4 \(6 \hat{i}-2 \hat{j}+8 \hat{k}\)

UP CPMT-2001

Motion in Plane

143617 The vectors \(\vec{A}, \vec{B}\) and \(\vec{C}\) are such that \(|\overrightarrow{\mathbf{A}}|=|\overrightarrow{\mathbf{B}}|,|\overrightarrow{\mathbf{C}}|=\sqrt{2}|\overrightarrow{\mathbf{A}}|\) and \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}=\mathbf{0}\). The angles between \(\vec{A}\) and \(\vec{B}, \vec{B}\) and \(\vec{C}\) respectively are

1 \(45^{\circ}, 90^{\circ}\)

2 \(90^{\circ}, 135^{\circ}\)

3 \(90^{\circ}, 45^{\circ}\)

4 \(45^{\circ}, 135^{\circ}\)

Explanation:

B According to figure.

\(\vec{A}\)
Given that, \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{C}}\)
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|-\overrightarrow{\mathrm{C}}|=|\overrightarrow{\mathrm{C}}|\)
Squaring both side, we get-
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|^{2}=|\overrightarrow{\mathrm{C}}|^{2}\)
\((\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})=|\overrightarrow{\mathrm{C}}|^{2} \quad\left(\because \overrightarrow{\mathrm{C}} \cdot \overrightarrow{\mathrm{C}}=|\overrightarrow{\mathrm{C}}|^{2}\right)\)
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{C}}|^{2}\)
\(|\overrightarrow{\mathrm{A}}|^{2}+|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta+|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta+|\overrightarrow{\mathrm{B}}|^{2}=|\overrightarrow{\mathrm{C}}|^{2}\)
\((\because B=A, C=\sqrt{2} A)\)
\(A^{2}+A^{2} \cos \theta+A^{2} \cos \theta+A^{2}=2 A^{2}\)
\(\Rightarrow \quad(2+2 \cos \theta)=2\)
\(\Rightarrow \quad \cos \theta=0\)
\(\Rightarrow \quad \theta=90^{\circ}\)
\(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{C}}\)
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}\)
\(|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos 90^{\circ}+\mathrm{B}^{2}=-|\overrightarrow{\mathrm{B}}||\overrightarrow{\mathrm{C}}| \cos \phi\)
\(\because \quad \mathrm{A}=\mathrm{B}, \mathrm{C}=\sqrt{2} \mathrm{~A}=\sqrt{2} \mathrm{~B}\)
\(\mathrm{B}^{2}=-\mathrm{B} \cdot \sqrt{2} \mathrm{~B} \cdot \cos \phi\)
\(\cos \phi=\frac{-1}{\sqrt{2}}\)
\(\phi=135^{\circ}\)
So, \(\theta=90^{\circ}, \phi=135^{\circ}\)

TS EAMCET(Medical)-2015

Motion in Plane

143618 An ant starts from the origin and crawls \(10 \mathrm{~cm}\) along the \(x-\) axis and then \(20 \mathrm{~cm}\) along the \(y-\) axis. The dot product of the ant's displacement vector with the position vector of a point that makes \(45^{\circ}\) with the \(x\) - axis and has a magnitude of \(\sqrt{2} \mathrm{~cm}\) is

1 \(30 \mathrm{~cm}\)

2 \(30 \sqrt{2} \mathrm{~cm}\)

3 \(\frac{30}{\sqrt{2}} \mathrm{~cm}\)

4 \(15 \mathrm{~cm}\)

TS EAMCET 18.07.2022

Motion in Plane

143622 Find the angle between the two vectors: \(\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}, \vec{b}=5 \hat{i}+3 \hat{j}+\hat{k}\)

1 \(\cos ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

2 \(\sin ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

3 \(\cos ^{-1}\left(\frac{26}{\sqrt{1335}}\right)\)

4 \(\tan ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

TS EAMCET 04.08.2021

Motion in Plane

143616 A body is rotating with angular velocity \(\omega=(3 \hat{i}-4 \hat{j}+\hat{k})\). The linear velocity of a point having position vector \(r=(5 \hat{i}-6 \hat{j}+6 \hat{k})\) is

1 \(6 \hat{i}+2 \hat{j}-3 \hat{k}\)

2 \(18 \hat{i}+13 \hat{j}-2 \hat{k}\)

3 \(-18 \hat{i}-13 \hat{j}+2 \hat{k}\)

4 \(6 \hat{i}-2 \hat{j}+8 \hat{k}\)

UP CPMT-2001

Motion in Plane

143617 The vectors \(\vec{A}, \vec{B}\) and \(\vec{C}\) are such that \(|\overrightarrow{\mathbf{A}}|=|\overrightarrow{\mathbf{B}}|,|\overrightarrow{\mathbf{C}}|=\sqrt{2}|\overrightarrow{\mathbf{A}}|\) and \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}=\mathbf{0}\). The angles between \(\vec{A}\) and \(\vec{B}, \vec{B}\) and \(\vec{C}\) respectively are

1 \(45^{\circ}, 90^{\circ}\)

2 \(90^{\circ}, 135^{\circ}\)

3 \(90^{\circ}, 45^{\circ}\)

4 \(45^{\circ}, 135^{\circ}\)

Explanation:

B According to figure.

\(\vec{A}\)
Given that, \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{C}}\)
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|-\overrightarrow{\mathrm{C}}|=|\overrightarrow{\mathrm{C}}|\)
Squaring both side, we get-
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|^{2}=|\overrightarrow{\mathrm{C}}|^{2}\)
\((\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})=|\overrightarrow{\mathrm{C}}|^{2} \quad\left(\because \overrightarrow{\mathrm{C}} \cdot \overrightarrow{\mathrm{C}}=|\overrightarrow{\mathrm{C}}|^{2}\right)\)
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{C}}|^{2}\)
\(|\overrightarrow{\mathrm{A}}|^{2}+|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta+|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta+|\overrightarrow{\mathrm{B}}|^{2}=|\overrightarrow{\mathrm{C}}|^{2}\)
\((\because B=A, C=\sqrt{2} A)\)
\(A^{2}+A^{2} \cos \theta+A^{2} \cos \theta+A^{2}=2 A^{2}\)
\(\Rightarrow \quad(2+2 \cos \theta)=2\)
\(\Rightarrow \quad \cos \theta=0\)
\(\Rightarrow \quad \theta=90^{\circ}\)
\(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{C}}\)
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}\)
\(|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos 90^{\circ}+\mathrm{B}^{2}=-|\overrightarrow{\mathrm{B}}||\overrightarrow{\mathrm{C}}| \cos \phi\)
\(\because \quad \mathrm{A}=\mathrm{B}, \mathrm{C}=\sqrt{2} \mathrm{~A}=\sqrt{2} \mathrm{~B}\)
\(\mathrm{B}^{2}=-\mathrm{B} \cdot \sqrt{2} \mathrm{~B} \cdot \cos \phi\)
\(\cos \phi=\frac{-1}{\sqrt{2}}\)
\(\phi=135^{\circ}\)
So, \(\theta=90^{\circ}, \phi=135^{\circ}\)

TS EAMCET(Medical)-2015

Motion in Plane

143618 An ant starts from the origin and crawls \(10 \mathrm{~cm}\) along the \(x-\) axis and then \(20 \mathrm{~cm}\) along the \(y-\) axis. The dot product of the ant's displacement vector with the position vector of a point that makes \(45^{\circ}\) with the \(x\) - axis and has a magnitude of \(\sqrt{2} \mathrm{~cm}\) is

1 \(30 \mathrm{~cm}\)

2 \(30 \sqrt{2} \mathrm{~cm}\)

3 \(\frac{30}{\sqrt{2}} \mathrm{~cm}\)

4 \(15 \mathrm{~cm}\)

TS EAMCET 18.07.2022

Motion in Plane

143622 Find the angle between the two vectors: \(\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}, \vec{b}=5 \hat{i}+3 \hat{j}+\hat{k}\)

1 \(\cos ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

2 \(\sin ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

3 \(\cos ^{-1}\left(\frac{26}{\sqrt{1335}}\right)\)

4 \(\tan ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

TS EAMCET 04.08.2021

Motion in Plane

143616 A body is rotating with angular velocity \(\omega=(3 \hat{i}-4 \hat{j}+\hat{k})\). The linear velocity of a point having position vector \(r=(5 \hat{i}-6 \hat{j}+6 \hat{k})\) is

1 \(6 \hat{i}+2 \hat{j}-3 \hat{k}\)

2 \(18 \hat{i}+13 \hat{j}-2 \hat{k}\)

3 \(-18 \hat{i}-13 \hat{j}+2 \hat{k}\)

4 \(6 \hat{i}-2 \hat{j}+8 \hat{k}\)

UP CPMT-2001

Motion in Plane

143617 The vectors \(\vec{A}, \vec{B}\) and \(\vec{C}\) are such that \(|\overrightarrow{\mathbf{A}}|=|\overrightarrow{\mathbf{B}}|,|\overrightarrow{\mathbf{C}}|=\sqrt{2}|\overrightarrow{\mathbf{A}}|\) and \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}=\mathbf{0}\). The angles between \(\vec{A}\) and \(\vec{B}, \vec{B}\) and \(\vec{C}\) respectively are

1 \(45^{\circ}, 90^{\circ}\)

2 \(90^{\circ}, 135^{\circ}\)

3 \(90^{\circ}, 45^{\circ}\)

4 \(45^{\circ}, 135^{\circ}\)

Explanation:

B According to figure.

\(\vec{A}\)
Given that, \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{C}}\)
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|-\overrightarrow{\mathrm{C}}|=|\overrightarrow{\mathrm{C}}|\)
Squaring both side, we get-
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|^{2}=|\overrightarrow{\mathrm{C}}|^{2}\)
\((\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})=|\overrightarrow{\mathrm{C}}|^{2} \quad\left(\because \overrightarrow{\mathrm{C}} \cdot \overrightarrow{\mathrm{C}}=|\overrightarrow{\mathrm{C}}|^{2}\right)\)
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{C}}|^{2}\)
\(|\overrightarrow{\mathrm{A}}|^{2}+|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta+|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta+|\overrightarrow{\mathrm{B}}|^{2}=|\overrightarrow{\mathrm{C}}|^{2}\)
\((\because B=A, C=\sqrt{2} A)\)
\(A^{2}+A^{2} \cos \theta+A^{2} \cos \theta+A^{2}=2 A^{2}\)
\(\Rightarrow \quad(2+2 \cos \theta)=2\)
\(\Rightarrow \quad \cos \theta=0\)
\(\Rightarrow \quad \theta=90^{\circ}\)
\(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{C}}\)
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}\)
\(|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos 90^{\circ}+\mathrm{B}^{2}=-|\overrightarrow{\mathrm{B}}||\overrightarrow{\mathrm{C}}| \cos \phi\)
\(\because \quad \mathrm{A}=\mathrm{B}, \mathrm{C}=\sqrt{2} \mathrm{~A}=\sqrt{2} \mathrm{~B}\)
\(\mathrm{B}^{2}=-\mathrm{B} \cdot \sqrt{2} \mathrm{~B} \cdot \cos \phi\)
\(\cos \phi=\frac{-1}{\sqrt{2}}\)
\(\phi=135^{\circ}\)
So, \(\theta=90^{\circ}, \phi=135^{\circ}\)

TS EAMCET(Medical)-2015

Motion in Plane

143618 An ant starts from the origin and crawls \(10 \mathrm{~cm}\) along the \(x-\) axis and then \(20 \mathrm{~cm}\) along the \(y-\) axis. The dot product of the ant's displacement vector with the position vector of a point that makes \(45^{\circ}\) with the \(x\) - axis and has a magnitude of \(\sqrt{2} \mathrm{~cm}\) is

1 \(30 \mathrm{~cm}\)

2 \(30 \sqrt{2} \mathrm{~cm}\)

3 \(\frac{30}{\sqrt{2}} \mathrm{~cm}\)

4 \(15 \mathrm{~cm}\)

TS EAMCET 18.07.2022

Motion in Plane

143622 Find the angle between the two vectors: \(\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}, \vec{b}=5 \hat{i}+3 \hat{j}+\hat{k}\)

1 \(\cos ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

2 \(\sin ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

3 \(\cos ^{-1}\left(\frac{26}{\sqrt{1335}}\right)\)

4 \(\tan ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

TS EAMCET 04.08.2021

Motion in Plane

143616 A body is rotating with angular velocity \(\omega=(3 \hat{i}-4 \hat{j}+\hat{k})\). The linear velocity of a point having position vector \(r=(5 \hat{i}-6 \hat{j}+6 \hat{k})\) is

1 \(6 \hat{i}+2 \hat{j}-3 \hat{k}\)

2 \(18 \hat{i}+13 \hat{j}-2 \hat{k}\)

3 \(-18 \hat{i}-13 \hat{j}+2 \hat{k}\)

4 \(6 \hat{i}-2 \hat{j}+8 \hat{k}\)

UP CPMT-2001

Motion in Plane

143617 The vectors \(\vec{A}, \vec{B}\) and \(\vec{C}\) are such that \(|\overrightarrow{\mathbf{A}}|=|\overrightarrow{\mathbf{B}}|,|\overrightarrow{\mathbf{C}}|=\sqrt{2}|\overrightarrow{\mathbf{A}}|\) and \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}=\mathbf{0}\). The angles between \(\vec{A}\) and \(\vec{B}, \vec{B}\) and \(\vec{C}\) respectively are

1 \(45^{\circ}, 90^{\circ}\)

2 \(90^{\circ}, 135^{\circ}\)

3 \(90^{\circ}, 45^{\circ}\)

4 \(45^{\circ}, 135^{\circ}\)

Explanation:

B According to figure.

\(\vec{A}\)
Given that, \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{C}}\)
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|-\overrightarrow{\mathrm{C}}|=|\overrightarrow{\mathrm{C}}|\)
Squaring both side, we get-
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|^{2}=|\overrightarrow{\mathrm{C}}|^{2}\)
\((\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})=|\overrightarrow{\mathrm{C}}|^{2} \quad\left(\because \overrightarrow{\mathrm{C}} \cdot \overrightarrow{\mathrm{C}}=|\overrightarrow{\mathrm{C}}|^{2}\right)\)
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{C}}|^{2}\)
\(|\overrightarrow{\mathrm{A}}|^{2}+|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta+|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta+|\overrightarrow{\mathrm{B}}|^{2}=|\overrightarrow{\mathrm{C}}|^{2}\)
\((\because B=A, C=\sqrt{2} A)\)
\(A^{2}+A^{2} \cos \theta+A^{2} \cos \theta+A^{2}=2 A^{2}\)
\(\Rightarrow \quad(2+2 \cos \theta)=2\)
\(\Rightarrow \quad \cos \theta=0\)
\(\Rightarrow \quad \theta=90^{\circ}\)
\(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{C}}\)
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}\)
\(|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos 90^{\circ}+\mathrm{B}^{2}=-|\overrightarrow{\mathrm{B}}||\overrightarrow{\mathrm{C}}| \cos \phi\)
\(\because \quad \mathrm{A}=\mathrm{B}, \mathrm{C}=\sqrt{2} \mathrm{~A}=\sqrt{2} \mathrm{~B}\)
\(\mathrm{B}^{2}=-\mathrm{B} \cdot \sqrt{2} \mathrm{~B} \cdot \cos \phi\)
\(\cos \phi=\frac{-1}{\sqrt{2}}\)
\(\phi=135^{\circ}\)
So, \(\theta=90^{\circ}, \phi=135^{\circ}\)

TS EAMCET(Medical)-2015

Motion in Plane

143618 An ant starts from the origin and crawls \(10 \mathrm{~cm}\) along the \(x-\) axis and then \(20 \mathrm{~cm}\) along the \(y-\) axis. The dot product of the ant's displacement vector with the position vector of a point that makes \(45^{\circ}\) with the \(x\) - axis and has a magnitude of \(\sqrt{2} \mathrm{~cm}\) is

1 \(30 \mathrm{~cm}\)

2 \(30 \sqrt{2} \mathrm{~cm}\)

3 \(\frac{30}{\sqrt{2}} \mathrm{~cm}\)

4 \(15 \mathrm{~cm}\)

TS EAMCET 18.07.2022

Motion in Plane

143622 Find the angle between the two vectors: \(\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}, \vec{b}=5 \hat{i}+3 \hat{j}+\hat{k}\)

1 \(\cos ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

2 \(\sin ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

3 \(\cos ^{-1}\left(\frac{26}{\sqrt{1335}}\right)\)

4 \(\tan ^{-1}\left(\frac{26}{\sqrt{1330}}\right)\)

TS EAMCET 04.08.2021

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