QBManager

Magnetism and Matter

154277 The strength of earth's magnetic field at a point is $0.4 \times 10^{- 5} T$ . If this field is to be annulled by the magnetic induction produced at the centre of a circular conducting loop of radius $π$ $cm$ , the current to be sent through the loop is

1

2 A

2

0.15 A

3

1.5 A

4

0.2 A

5

1 A

Explanation:

D Given that, $B = 0.4 \times 10^{- 5} T$
Radius ( $r$ ) $= π cm = π \times 10^{- 2} m$
Then, magnetic field at the centre of circular loop.
$B = \frac{μ_{0} I}{2 r}$
$0.4 \times 10^{- 5} = \frac{4 π \times 10^{- 7} \times I}{2 \times π \times 10^{- 2}}$
$I = 0.2 A$

Kerala CEE 04.07.2022

Magnetism and Matter

154276 B-H curve of two different material $P$ and $Q$ are shown in fig. These materials are used to make magnets for electric generator, transformer and electromagnetic core. Then it is proper to use-
original image

1

P

for electric generators and transformers

2 P for electromagnets and

Q

for electric generators

3

P

for transformers and

Q

for electric generators or electromagnets

4 Q for electromagnets and transformers

CG PET-22.05.2022

Magnetism and Matter

154280 A paramagnetic sample shows a net magnetization of $0.8 A / m$ , when placed in an external magnetic field of strength $0.8 T$ at a temperature $5 K$ . When the same sample is placed in an external magnetic field of $0.4 T$ at a temperature of $20 K$ , the magnetization is

1

0.1 {Am}^{- 1}

2

0.8 {Am}^{- 1}

3

0.8 {Am}^{- 2}

4

0.1 Am

Explanation:

A Given, $I_{1} = 0.8 A / m, B = 0.8 T$
And $T_{1} = 5 K, B_{2} = 0.4 T, T_{2} = 20 K$
We know that,
For paramagnetic sample (Curie law) $I \propto \frac{B}{T}$
$∴ \frac{I_{1}}{I_{2}} = \frac{B_{1} / T_{1}}{B_{2} / T_{2}} = \frac{B_{1} \times T_{2}}{B_{2} \times T_{1}}$
$\frac{0.8}{I_{2}} = \frac{0.8 \times 20}{0.4 \times 5}$
$I_{2} = \frac{0.4 \times 5}{2} = 0.1 {Am}^{- 1}$

AP EAMCET-20.08.2021

Magnetism and Matter

154281 An electron having kinetic energy ' $T$ ' is moving in a circular orbit. If radius ' $R$ ' perpendicular to the uniform magnetic field induction $\vec{B}$ . If kinetic energy is doubled and magnetic field induction is tripled, the radius will become

1

R \sqrt{\frac{9}{4}}

2

R \sqrt{\frac{3}{2}}

3

R \sqrt{\frac{2}{9}}

4

R \sqrt{\frac{4}{3}}

Explanation:

C Given, kinetic energy $= T$ , radius $= R$ , magnetic field $= B$
We know that,
Kinetic energy $(T) = \frac{1}{2} {mv}^{2}$
Or $mv = \sqrt{2 mT}$
Radius of the circular orbit $(R) = \frac{mv}{eB}$
$R = \frac{\sqrt{2 mT}}{eB}$
When, kinetic energy double and magnetic field is triple then new radius will be
$R^{'} = \frac{\sqrt{2 m 2 T}}{3 eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times \frac{\sqrt{2 mT}}{eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times R$
$[from equation (i)]$
$So, R^{'} = R \sqrt{\frac{2}{9}}$

AP EAMCET-05.10.2021

Magnetism and Matter

154284 A small cube of side of $1 {mm}^{3}$ is placed at the centre of a circular loop of radius $20 cm$ . If the current in the loop is $2 A$ then the magnetic energy stored inside the cube is
(Assume $μ_{0} = 4 π \times 10^{- 7} TM / A$ )

1

1.57 \times 10^{- 18} J

2

2.57 \times 10^{- 14} J

3

1.57 \times 10^{- 14} J

4

4.57 \times 10^{- 13} J

Explanation:

C Given,
Current in the loop (i) $= 2 A$
Radius of the loop $(r) = 20 cm = 0.2 m$
Volume of the cube $(V) = 1 {mm}^{3} = 1 \times 10^{- 9} m^{3}$
We know that, magnetic field intensity at the centre of circular loop
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{(4 π \times 10^{- 7}) \times 2}{2 \times 0.2} = 2 π \times 10^{- 6} T$
Magnetic energy density $= \frac{B^{2}}{2 μ}$
Total energy stored in volume -
$U = \frac{B^{2} V}{2 μ_{0}}$
$U = \frac{{(2 π \times 10^{- 6})}^{2} \times (1 \times 10^{- 9})}{2 \times (4 π \times 10^{- 7})}$
$U = 1.57 \times 10^{- 14} J$

TS EAMCET 04.08.2021

Magnetism and Matter

154277 The strength of earth's magnetic field at a point is $0.4 \times 10^{- 5} T$ . If this field is to be annulled by the magnetic induction produced at the centre of a circular conducting loop of radius $π$ $cm$ , the current to be sent through the loop is

1

2 A

2

0.15 A

3

1.5 A

4

0.2 A

5

1 A

Explanation:

D Given that, $B = 0.4 \times 10^{- 5} T$
Radius ( $r$ ) $= π cm = π \times 10^{- 2} m$
Then, magnetic field at the centre of circular loop.
$B = \frac{μ_{0} I}{2 r}$
$0.4 \times 10^{- 5} = \frac{4 π \times 10^{- 7} \times I}{2 \times π \times 10^{- 2}}$
$I = 0.2 A$

Kerala CEE 04.07.2022

Magnetism and Matter

154276 B-H curve of two different material $P$ and $Q$ are shown in fig. These materials are used to make magnets for electric generator, transformer and electromagnetic core. Then it is proper to use-
original image

1

P

for electric generators and transformers

2 P for electromagnets and

Q

for electric generators

3

P

for transformers and

Q

for electric generators or electromagnets

4 Q for electromagnets and transformers

CG PET-22.05.2022

Magnetism and Matter

154280 A paramagnetic sample shows a net magnetization of $0.8 A / m$ , when placed in an external magnetic field of strength $0.8 T$ at a temperature $5 K$ . When the same sample is placed in an external magnetic field of $0.4 T$ at a temperature of $20 K$ , the magnetization is

1

0.1 {Am}^{- 1}

2

0.8 {Am}^{- 1}

3

0.8 {Am}^{- 2}

4

0.1 Am

Explanation:

A Given, $I_{1} = 0.8 A / m, B = 0.8 T$
And $T_{1} = 5 K, B_{2} = 0.4 T, T_{2} = 20 K$
We know that,
For paramagnetic sample (Curie law) $I \propto \frac{B}{T}$
$∴ \frac{I_{1}}{I_{2}} = \frac{B_{1} / T_{1}}{B_{2} / T_{2}} = \frac{B_{1} \times T_{2}}{B_{2} \times T_{1}}$
$\frac{0.8}{I_{2}} = \frac{0.8 \times 20}{0.4 \times 5}$
$I_{2} = \frac{0.4 \times 5}{2} = 0.1 {Am}^{- 1}$

AP EAMCET-20.08.2021

Magnetism and Matter

154281 An electron having kinetic energy ' $T$ ' is moving in a circular orbit. If radius ' $R$ ' perpendicular to the uniform magnetic field induction $\vec{B}$ . If kinetic energy is doubled and magnetic field induction is tripled, the radius will become

1

R \sqrt{\frac{9}{4}}

2

R \sqrt{\frac{3}{2}}

3

R \sqrt{\frac{2}{9}}

4

R \sqrt{\frac{4}{3}}

Explanation:

C Given, kinetic energy $= T$ , radius $= R$ , magnetic field $= B$
We know that,
Kinetic energy $(T) = \frac{1}{2} {mv}^{2}$
Or $mv = \sqrt{2 mT}$
Radius of the circular orbit $(R) = \frac{mv}{eB}$
$R = \frac{\sqrt{2 mT}}{eB}$
When, kinetic energy double and magnetic field is triple then new radius will be
$R^{'} = \frac{\sqrt{2 m 2 T}}{3 eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times \frac{\sqrt{2 mT}}{eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times R$
$[from equation (i)]$
$So, R^{'} = R \sqrt{\frac{2}{9}}$

AP EAMCET-05.10.2021

Magnetism and Matter

154284 A small cube of side of $1 {mm}^{3}$ is placed at the centre of a circular loop of radius $20 cm$ . If the current in the loop is $2 A$ then the magnetic energy stored inside the cube is
(Assume $μ_{0} = 4 π \times 10^{- 7} TM / A$ )

1

1.57 \times 10^{- 18} J

2

2.57 \times 10^{- 14} J

3

1.57 \times 10^{- 14} J

4

4.57 \times 10^{- 13} J

Explanation:

C Given,
Current in the loop (i) $= 2 A$
Radius of the loop $(r) = 20 cm = 0.2 m$
Volume of the cube $(V) = 1 {mm}^{3} = 1 \times 10^{- 9} m^{3}$
We know that, magnetic field intensity at the centre of circular loop
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{(4 π \times 10^{- 7}) \times 2}{2 \times 0.2} = 2 π \times 10^{- 6} T$
Magnetic energy density $= \frac{B^{2}}{2 μ}$
Total energy stored in volume -
$U = \frac{B^{2} V}{2 μ_{0}}$
$U = \frac{{(2 π \times 10^{- 6})}^{2} \times (1 \times 10^{- 9})}{2 \times (4 π \times 10^{- 7})}$
$U = 1.57 \times 10^{- 14} J$

TS EAMCET 04.08.2021

Magnetism and Matter

154277 The strength of earth's magnetic field at a point is $0.4 \times 10^{- 5} T$ . If this field is to be annulled by the magnetic induction produced at the centre of a circular conducting loop of radius $π$ $cm$ , the current to be sent through the loop is

1

2 A

2

0.15 A

3

1.5 A

4

0.2 A

5

1 A

Explanation:

D Given that, $B = 0.4 \times 10^{- 5} T$
Radius ( $r$ ) $= π cm = π \times 10^{- 2} m$
Then, magnetic field at the centre of circular loop.
$B = \frac{μ_{0} I}{2 r}$
$0.4 \times 10^{- 5} = \frac{4 π \times 10^{- 7} \times I}{2 \times π \times 10^{- 2}}$
$I = 0.2 A$

Kerala CEE 04.07.2022

Magnetism and Matter

154276 B-H curve of two different material $P$ and $Q$ are shown in fig. These materials are used to make magnets for electric generator, transformer and electromagnetic core. Then it is proper to use-
original image

1

P

for electric generators and transformers

2 P for electromagnets and

Q

for electric generators

3

P

for transformers and

Q

for electric generators or electromagnets

4 Q for electromagnets and transformers

CG PET-22.05.2022

Magnetism and Matter

154280 A paramagnetic sample shows a net magnetization of $0.8 A / m$ , when placed in an external magnetic field of strength $0.8 T$ at a temperature $5 K$ . When the same sample is placed in an external magnetic field of $0.4 T$ at a temperature of $20 K$ , the magnetization is

1

0.1 {Am}^{- 1}

2

0.8 {Am}^{- 1}

3

0.8 {Am}^{- 2}

4

0.1 Am

Explanation:

A Given, $I_{1} = 0.8 A / m, B = 0.8 T$
And $T_{1} = 5 K, B_{2} = 0.4 T, T_{2} = 20 K$
We know that,
For paramagnetic sample (Curie law) $I \propto \frac{B}{T}$
$∴ \frac{I_{1}}{I_{2}} = \frac{B_{1} / T_{1}}{B_{2} / T_{2}} = \frac{B_{1} \times T_{2}}{B_{2} \times T_{1}}$
$\frac{0.8}{I_{2}} = \frac{0.8 \times 20}{0.4 \times 5}$
$I_{2} = \frac{0.4 \times 5}{2} = 0.1 {Am}^{- 1}$

AP EAMCET-20.08.2021

Magnetism and Matter

154281 An electron having kinetic energy ' $T$ ' is moving in a circular orbit. If radius ' $R$ ' perpendicular to the uniform magnetic field induction $\vec{B}$ . If kinetic energy is doubled and magnetic field induction is tripled, the radius will become

1

R \sqrt{\frac{9}{4}}

2

R \sqrt{\frac{3}{2}}

3

R \sqrt{\frac{2}{9}}

4

R \sqrt{\frac{4}{3}}

Explanation:

C Given, kinetic energy $= T$ , radius $= R$ , magnetic field $= B$
We know that,
Kinetic energy $(T) = \frac{1}{2} {mv}^{2}$
Or $mv = \sqrt{2 mT}$
Radius of the circular orbit $(R) = \frac{mv}{eB}$
$R = \frac{\sqrt{2 mT}}{eB}$
When, kinetic energy double and magnetic field is triple then new radius will be
$R^{'} = \frac{\sqrt{2 m 2 T}}{3 eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times \frac{\sqrt{2 mT}}{eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times R$
$[from equation (i)]$
$So, R^{'} = R \sqrt{\frac{2}{9}}$

AP EAMCET-05.10.2021

Magnetism and Matter

154284 A small cube of side of $1 {mm}^{3}$ is placed at the centre of a circular loop of radius $20 cm$ . If the current in the loop is $2 A$ then the magnetic energy stored inside the cube is
(Assume $μ_{0} = 4 π \times 10^{- 7} TM / A$ )

1

1.57 \times 10^{- 18} J

2

2.57 \times 10^{- 14} J

3

1.57 \times 10^{- 14} J

4

4.57 \times 10^{- 13} J

Explanation:

C Given,
Current in the loop (i) $= 2 A$
Radius of the loop $(r) = 20 cm = 0.2 m$
Volume of the cube $(V) = 1 {mm}^{3} = 1 \times 10^{- 9} m^{3}$
We know that, magnetic field intensity at the centre of circular loop
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{(4 π \times 10^{- 7}) \times 2}{2 \times 0.2} = 2 π \times 10^{- 6} T$
Magnetic energy density $= \frac{B^{2}}{2 μ}$
Total energy stored in volume -
$U = \frac{B^{2} V}{2 μ_{0}}$
$U = \frac{{(2 π \times 10^{- 6})}^{2} \times (1 \times 10^{- 9})}{2 \times (4 π \times 10^{- 7})}$
$U = 1.57 \times 10^{- 14} J$

TS EAMCET 04.08.2021

Magnetism and Matter

154277 The strength of earth's magnetic field at a point is $0.4 \times 10^{- 5} T$ . If this field is to be annulled by the magnetic induction produced at the centre of a circular conducting loop of radius $π$ $cm$ , the current to be sent through the loop is

1

2 A

2

0.15 A

3

1.5 A

4

0.2 A

5

1 A

Explanation:

D Given that, $B = 0.4 \times 10^{- 5} T$
Radius ( $r$ ) $= π cm = π \times 10^{- 2} m$
Then, magnetic field at the centre of circular loop.
$B = \frac{μ_{0} I}{2 r}$
$0.4 \times 10^{- 5} = \frac{4 π \times 10^{- 7} \times I}{2 \times π \times 10^{- 2}}$
$I = 0.2 A$

Kerala CEE 04.07.2022

Magnetism and Matter

154276 B-H curve of two different material $P$ and $Q$ are shown in fig. These materials are used to make magnets for electric generator, transformer and electromagnetic core. Then it is proper to use-
original image

1

P

for electric generators and transformers

2 P for electromagnets and

Q

for electric generators

3

P

for transformers and

Q

for electric generators or electromagnets

4 Q for electromagnets and transformers

CG PET-22.05.2022

Magnetism and Matter

154280 A paramagnetic sample shows a net magnetization of $0.8 A / m$ , when placed in an external magnetic field of strength $0.8 T$ at a temperature $5 K$ . When the same sample is placed in an external magnetic field of $0.4 T$ at a temperature of $20 K$ , the magnetization is

1

0.1 {Am}^{- 1}

2

0.8 {Am}^{- 1}

3

0.8 {Am}^{- 2}

4

0.1 Am

Explanation:

A Given, $I_{1} = 0.8 A / m, B = 0.8 T$
And $T_{1} = 5 K, B_{2} = 0.4 T, T_{2} = 20 K$
We know that,
For paramagnetic sample (Curie law) $I \propto \frac{B}{T}$
$∴ \frac{I_{1}}{I_{2}} = \frac{B_{1} / T_{1}}{B_{2} / T_{2}} = \frac{B_{1} \times T_{2}}{B_{2} \times T_{1}}$
$\frac{0.8}{I_{2}} = \frac{0.8 \times 20}{0.4 \times 5}$
$I_{2} = \frac{0.4 \times 5}{2} = 0.1 {Am}^{- 1}$

AP EAMCET-20.08.2021

Magnetism and Matter

154281 An electron having kinetic energy ' $T$ ' is moving in a circular orbit. If radius ' $R$ ' perpendicular to the uniform magnetic field induction $\vec{B}$ . If kinetic energy is doubled and magnetic field induction is tripled, the radius will become

1

R \sqrt{\frac{9}{4}}

2

R \sqrt{\frac{3}{2}}

3

R \sqrt{\frac{2}{9}}

4

R \sqrt{\frac{4}{3}}

Explanation:

C Given, kinetic energy $= T$ , radius $= R$ , magnetic field $= B$
We know that,
Kinetic energy $(T) = \frac{1}{2} {mv}^{2}$
Or $mv = \sqrt{2 mT}$
Radius of the circular orbit $(R) = \frac{mv}{eB}$
$R = \frac{\sqrt{2 mT}}{eB}$
When, kinetic energy double and magnetic field is triple then new radius will be
$R^{'} = \frac{\sqrt{2 m 2 T}}{3 eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times \frac{\sqrt{2 mT}}{eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times R$
$[from equation (i)]$
$So, R^{'} = R \sqrt{\frac{2}{9}}$

AP EAMCET-05.10.2021

Magnetism and Matter

154284 A small cube of side of $1 {mm}^{3}$ is placed at the centre of a circular loop of radius $20 cm$ . If the current in the loop is $2 A$ then the magnetic energy stored inside the cube is
(Assume $μ_{0} = 4 π \times 10^{- 7} TM / A$ )

1

1.57 \times 10^{- 18} J

2

2.57 \times 10^{- 14} J

3

1.57 \times 10^{- 14} J

4

4.57 \times 10^{- 13} J

Explanation:

C Given,
Current in the loop (i) $= 2 A$
Radius of the loop $(r) = 20 cm = 0.2 m$
Volume of the cube $(V) = 1 {mm}^{3} = 1 \times 10^{- 9} m^{3}$
We know that, magnetic field intensity at the centre of circular loop
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{(4 π \times 10^{- 7}) \times 2}{2 \times 0.2} = 2 π \times 10^{- 6} T$
Magnetic energy density $= \frac{B^{2}}{2 μ}$
Total energy stored in volume -
$U = \frac{B^{2} V}{2 μ_{0}}$
$U = \frac{{(2 π \times 10^{- 6})}^{2} \times (1 \times 10^{- 9})}{2 \times (4 π \times 10^{- 7})}$
$U = 1.57 \times 10^{- 14} J$

TS EAMCET 04.08.2021

Magnetism and Matter

154277 The strength of earth's magnetic field at a point is $0.4 \times 10^{- 5} T$ . If this field is to be annulled by the magnetic induction produced at the centre of a circular conducting loop of radius $π$ $cm$ , the current to be sent through the loop is

1

2 A

2

0.15 A

3

1.5 A

4

0.2 A

5

1 A

Explanation:

D Given that, $B = 0.4 \times 10^{- 5} T$
Radius ( $r$ ) $= π cm = π \times 10^{- 2} m$
Then, magnetic field at the centre of circular loop.
$B = \frac{μ_{0} I}{2 r}$
$0.4 \times 10^{- 5} = \frac{4 π \times 10^{- 7} \times I}{2 \times π \times 10^{- 2}}$
$I = 0.2 A$

Kerala CEE 04.07.2022

Magnetism and Matter

154276 B-H curve of two different material $P$ and $Q$ are shown in fig. These materials are used to make magnets for electric generator, transformer and electromagnetic core. Then it is proper to use-
original image

1

P

for electric generators and transformers

2 P for electromagnets and

Q

for electric generators

3

P

for transformers and

Q

for electric generators or electromagnets

4 Q for electromagnets and transformers

CG PET-22.05.2022

Magnetism and Matter

154280 A paramagnetic sample shows a net magnetization of $0.8 A / m$ , when placed in an external magnetic field of strength $0.8 T$ at a temperature $5 K$ . When the same sample is placed in an external magnetic field of $0.4 T$ at a temperature of $20 K$ , the magnetization is

1

0.1 {Am}^{- 1}

2

0.8 {Am}^{- 1}

3

0.8 {Am}^{- 2}

4

0.1 Am

Explanation:

A Given, $I_{1} = 0.8 A / m, B = 0.8 T$
And $T_{1} = 5 K, B_{2} = 0.4 T, T_{2} = 20 K$
We know that,
For paramagnetic sample (Curie law) $I \propto \frac{B}{T}$
$∴ \frac{I_{1}}{I_{2}} = \frac{B_{1} / T_{1}}{B_{2} / T_{2}} = \frac{B_{1} \times T_{2}}{B_{2} \times T_{1}}$
$\frac{0.8}{I_{2}} = \frac{0.8 \times 20}{0.4 \times 5}$
$I_{2} = \frac{0.4 \times 5}{2} = 0.1 {Am}^{- 1}$

AP EAMCET-20.08.2021

Magnetism and Matter

154281 An electron having kinetic energy ' $T$ ' is moving in a circular orbit. If radius ' $R$ ' perpendicular to the uniform magnetic field induction $\vec{B}$ . If kinetic energy is doubled and magnetic field induction is tripled, the radius will become

1

R \sqrt{\frac{9}{4}}

2

R \sqrt{\frac{3}{2}}

3

R \sqrt{\frac{2}{9}}

4

R \sqrt{\frac{4}{3}}

Explanation:

C Given, kinetic energy $= T$ , radius $= R$ , magnetic field $= B$
We know that,
Kinetic energy $(T) = \frac{1}{2} {mv}^{2}$
Or $mv = \sqrt{2 mT}$
Radius of the circular orbit $(R) = \frac{mv}{eB}$
$R = \frac{\sqrt{2 mT}}{eB}$
When, kinetic energy double and magnetic field is triple then new radius will be
$R^{'} = \frac{\sqrt{2 m 2 T}}{3 eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times \frac{\sqrt{2 mT}}{eB}$
$R^{'} = \frac{\sqrt{2}}{3} \times R$
$[from equation (i)]$
$So, R^{'} = R \sqrt{\frac{2}{9}}$

AP EAMCET-05.10.2021

Magnetism and Matter

154284 A small cube of side of $1 {mm}^{3}$ is placed at the centre of a circular loop of radius $20 cm$ . If the current in the loop is $2 A$ then the magnetic energy stored inside the cube is
(Assume $μ_{0} = 4 π \times 10^{- 7} TM / A$ )

1

1.57 \times 10^{- 18} J

2

2.57 \times 10^{- 14} J

3

1.57 \times 10^{- 14} J

4

4.57 \times 10^{- 13} J

Explanation:

C Given,
Current in the loop (i) $= 2 A$
Radius of the loop $(r) = 20 cm = 0.2 m$
Volume of the cube $(V) = 1 {mm}^{3} = 1 \times 10^{- 9} m^{3}$
We know that, magnetic field intensity at the centre of circular loop
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{(4 π \times 10^{- 7}) \times 2}{2 \times 0.2} = 2 π \times 10^{- 6} T$
Magnetic energy density $= \frac{B^{2}}{2 μ}$
Total energy stored in volume -
$U = \frac{B^{2} V}{2 μ_{0}}$
$U = \frac{{(2 π \times 10^{- 6})}^{2} \times (1 \times 10^{- 9})}{2 \times (4 π \times 10^{- 7})}$
$U = 1.57 \times 10^{- 14} J$

TS EAMCET 04.08.2021

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: