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PHXII04:MOVING CHARGES AND MAGNETISM

362993 Best method to increase the sensitivity of the moving coil galvanometer is to decrease.

1 Radius of the coil

2 Number of turns of the coil

3 External magnetic field

4 Couple per unit twist

PHXII04:MOVING CHARGES AND MAGNETISM

362994 The relation between voltage sensitivity $σ_{V}$ and the current sensitivity $σ_{i}$ of a moving coil galvanometer is (Given that $G$ is the resistance of the galvanometer)

1

σ_{V} = G σ_{i}

2

σ_{V} = σ_{i} / G

3

σ_{V} σ_{i} = G

4

σ_{V} σ_{i} = 1 / G

Explanation:

$σ_{V} = \frac{θ}{V} = \frac{θ}{i R} = \frac{σ_{i}}{R} = \frac{σ_{i}}{G} (R = G)$

PHXII04:MOVING CHARGES AND MAGNETISM

362995 A galvanometer of resistance $G$ is connected in a circuit. Now, a resistance $R$ is connected in series of galvanometer. To keep the main current in the circuit unchanged, the resistance to be put in parallel with the series combination of $G$ and $R$ is

1

\frac{G^{2}}{R} + G

2

\frac{R^{2}}{G} + G

3

\frac{G^{2}}{R} - G

4

\frac{R^{2}}{G} - G

Explanation:

$G^{'} = G$
$\frac{(G + R) S}{G + R + S} = G$
$S = \frac{G^{2}}{R} + G$

PHXII04:MOVING CHARGES AND MAGNETISM

362996 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double
Statement B :
Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Current sensitivity, $I_{S} = \frac{Q}{I} = \frac{N B A}{C}$
Voltage sensitivity, $V_{S} = \frac{Q}{V} = \frac{N B A}{C R}$
When $N$ is doubled, current sensitivity also doubled. So, statement A is true.
But as $N$ is doubled resistance $R$ will also change and hence voltage sensitivity cannot be changed as $\frac{N}{R}$ ratio is constant, so statement B is false.

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362997 The restoring couple in the moving coil galvanometer is due to

1 Current in the coil

2 Magnetic field of the magnet

3 Material of the coil

4 Twist produced in the suspension wire

Explanation:

Due to the twist produced in the suspension wire produces the restoring couple in $M C G$

PHXII04:MOVING CHARGES AND MAGNETISM

362993 Best method to increase the sensitivity of the moving coil galvanometer is to decrease.

1 Radius of the coil

2 Number of turns of the coil

3 External magnetic field

4 Couple per unit twist

PHXII04:MOVING CHARGES AND MAGNETISM

362994 The relation between voltage sensitivity $σ_{V}$ and the current sensitivity $σ_{i}$ of a moving coil galvanometer is (Given that $G$ is the resistance of the galvanometer)

1

σ_{V} = G σ_{i}

2

σ_{V} = σ_{i} / G

3

σ_{V} σ_{i} = G

4

σ_{V} σ_{i} = 1 / G

Explanation:

$σ_{V} = \frac{θ}{V} = \frac{θ}{i R} = \frac{σ_{i}}{R} = \frac{σ_{i}}{G} (R = G)$

PHXII04:MOVING CHARGES AND MAGNETISM

362995 A galvanometer of resistance $G$ is connected in a circuit. Now, a resistance $R$ is connected in series of galvanometer. To keep the main current in the circuit unchanged, the resistance to be put in parallel with the series combination of $G$ and $R$ is

1

\frac{G^{2}}{R} + G

2

\frac{R^{2}}{G} + G

3

\frac{G^{2}}{R} - G

4

\frac{R^{2}}{G} - G

Explanation:

$G^{'} = G$
$\frac{(G + R) S}{G + R + S} = G$
$S = \frac{G^{2}}{R} + G$

PHXII04:MOVING CHARGES AND MAGNETISM

362996 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double
Statement B :
Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Current sensitivity, $I_{S} = \frac{Q}{I} = \frac{N B A}{C}$
Voltage sensitivity, $V_{S} = \frac{Q}{V} = \frac{N B A}{C R}$
When $N$ is doubled, current sensitivity also doubled. So, statement A is true.
But as $N$ is doubled resistance $R$ will also change and hence voltage sensitivity cannot be changed as $\frac{N}{R}$ ratio is constant, so statement B is false.

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362997 The restoring couple in the moving coil galvanometer is due to

1 Current in the coil

2 Magnetic field of the magnet

3 Material of the coil

4 Twist produced in the suspension wire

Explanation:

Due to the twist produced in the suspension wire produces the restoring couple in $M C G$

PHXII04:MOVING CHARGES AND MAGNETISM

362993 Best method to increase the sensitivity of the moving coil galvanometer is to decrease.

1 Radius of the coil

2 Number of turns of the coil

3 External magnetic field

4 Couple per unit twist

PHXII04:MOVING CHARGES AND MAGNETISM

362994 The relation between voltage sensitivity $σ_{V}$ and the current sensitivity $σ_{i}$ of a moving coil galvanometer is (Given that $G$ is the resistance of the galvanometer)

1

σ_{V} = G σ_{i}

2

σ_{V} = σ_{i} / G

3

σ_{V} σ_{i} = G

4

σ_{V} σ_{i} = 1 / G

Explanation:

$σ_{V} = \frac{θ}{V} = \frac{θ}{i R} = \frac{σ_{i}}{R} = \frac{σ_{i}}{G} (R = G)$

PHXII04:MOVING CHARGES AND MAGNETISM

362995 A galvanometer of resistance $G$ is connected in a circuit. Now, a resistance $R$ is connected in series of galvanometer. To keep the main current in the circuit unchanged, the resistance to be put in parallel with the series combination of $G$ and $R$ is

1

\frac{G^{2}}{R} + G

2

\frac{R^{2}}{G} + G

3

\frac{G^{2}}{R} - G

4

\frac{R^{2}}{G} - G

Explanation:

$G^{'} = G$
$\frac{(G + R) S}{G + R + S} = G$
$S = \frac{G^{2}}{R} + G$

PHXII04:MOVING CHARGES AND MAGNETISM

362996 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double
Statement B :
Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Current sensitivity, $I_{S} = \frac{Q}{I} = \frac{N B A}{C}$
Voltage sensitivity, $V_{S} = \frac{Q}{V} = \frac{N B A}{C R}$
When $N$ is doubled, current sensitivity also doubled. So, statement A is true.
But as $N$ is doubled resistance $R$ will also change and hence voltage sensitivity cannot be changed as $\frac{N}{R}$ ratio is constant, so statement B is false.

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362997 The restoring couple in the moving coil galvanometer is due to

1 Current in the coil

2 Magnetic field of the magnet

3 Material of the coil

4 Twist produced in the suspension wire

Explanation:

Due to the twist produced in the suspension wire produces the restoring couple in $M C G$

PHXII04:MOVING CHARGES AND MAGNETISM

362993 Best method to increase the sensitivity of the moving coil galvanometer is to decrease.

1 Radius of the coil

2 Number of turns of the coil

3 External magnetic field

4 Couple per unit twist

PHXII04:MOVING CHARGES AND MAGNETISM

362994 The relation between voltage sensitivity $σ_{V}$ and the current sensitivity $σ_{i}$ of a moving coil galvanometer is (Given that $G$ is the resistance of the galvanometer)

1

σ_{V} = G σ_{i}

2

σ_{V} = σ_{i} / G

3

σ_{V} σ_{i} = G

4

σ_{V} σ_{i} = 1 / G

Explanation:

$σ_{V} = \frac{θ}{V} = \frac{θ}{i R} = \frac{σ_{i}}{R} = \frac{σ_{i}}{G} (R = G)$

PHXII04:MOVING CHARGES AND MAGNETISM

362995 A galvanometer of resistance $G$ is connected in a circuit. Now, a resistance $R$ is connected in series of galvanometer. To keep the main current in the circuit unchanged, the resistance to be put in parallel with the series combination of $G$ and $R$ is

1

\frac{G^{2}}{R} + G

2

\frac{R^{2}}{G} + G

3

\frac{G^{2}}{R} - G

4

\frac{R^{2}}{G} - G

Explanation:

$G^{'} = G$
$\frac{(G + R) S}{G + R + S} = G$
$S = \frac{G^{2}}{R} + G$

PHXII04:MOVING CHARGES AND MAGNETISM

362996 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double
Statement B :
Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Current sensitivity, $I_{S} = \frac{Q}{I} = \frac{N B A}{C}$
Voltage sensitivity, $V_{S} = \frac{Q}{V} = \frac{N B A}{C R}$
When $N$ is doubled, current sensitivity also doubled. So, statement A is true.
But as $N$ is doubled resistance $R$ will also change and hence voltage sensitivity cannot be changed as $\frac{N}{R}$ ratio is constant, so statement B is false.

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362997 The restoring couple in the moving coil galvanometer is due to

1 Current in the coil

2 Magnetic field of the magnet

3 Material of the coil

4 Twist produced in the suspension wire

Explanation:

Due to the twist produced in the suspension wire produces the restoring couple in $M C G$

PHXII04:MOVING CHARGES AND MAGNETISM

362993 Best method to increase the sensitivity of the moving coil galvanometer is to decrease.

1 Radius of the coil

2 Number of turns of the coil

3 External magnetic field

4 Couple per unit twist

PHXII04:MOVING CHARGES AND MAGNETISM

362994 The relation between voltage sensitivity $σ_{V}$ and the current sensitivity $σ_{i}$ of a moving coil galvanometer is (Given that $G$ is the resistance of the galvanometer)

1

σ_{V} = G σ_{i}

2

σ_{V} = σ_{i} / G

3

σ_{V} σ_{i} = G

4

σ_{V} σ_{i} = 1 / G

Explanation:

$σ_{V} = \frac{θ}{V} = \frac{θ}{i R} = \frac{σ_{i}}{R} = \frac{σ_{i}}{G} (R = G)$

PHXII04:MOVING CHARGES AND MAGNETISM

362995 A galvanometer of resistance $G$ is connected in a circuit. Now, a resistance $R$ is connected in series of galvanometer. To keep the main current in the circuit unchanged, the resistance to be put in parallel with the series combination of $G$ and $R$ is

1

\frac{G^{2}}{R} + G

2

\frac{R^{2}}{G} + G

3

\frac{G^{2}}{R} - G

4

\frac{R^{2}}{G} - G

Explanation:

$G^{'} = G$
$\frac{(G + R) S}{G + R + S} = G$
$S = \frac{G^{2}}{R} + G$

PHXII04:MOVING CHARGES AND MAGNETISM

362996 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double
Statement B :
Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Current sensitivity, $I_{S} = \frac{Q}{I} = \frac{N B A}{C}$
Voltage sensitivity, $V_{S} = \frac{Q}{V} = \frac{N B A}{C R}$
When $N$ is doubled, current sensitivity also doubled. So, statement A is true.
But as $N$ is doubled resistance $R$ will also change and hence voltage sensitivity cannot be changed as $\frac{N}{R}$ ratio is constant, so statement B is false.

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362997 The restoring couple in the moving coil galvanometer is due to

1 Current in the coil

2 Magnetic field of the magnet

3 Material of the coil

4 Twist produced in the suspension wire

Explanation:

Due to the twist produced in the suspension wire produces the restoring couple in $M C G$

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