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Current Electricity

152885 The meter bridge shown in the balance position with $\frac{P}{Q} = \frac{l_{1}}{l_{2}}$ .If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, that will be balanced condition?

1 yes,

\frac{P}{Q} = \frac{l_{2} - l_{1}}{l_{2} + l_{1}}

2 no, no null point

3 yes,

\frac{P}{Q} = \frac{l_{2}}{l_{1}}

4 yes,

\frac{P}{Q} = \frac{l_{1}}{l_{2}}

Explanation:

D The meter bridge is balanced against length $l_{1}$ and $l_{2}$ are-

When galvanometer and cells are interchanged, the balance condition remains unchanged.

For balancing condition,
$\frac{P}{Q} = \frac{l_{1}}{l_{2}}$
Yes, the bridge will work for balanced condition.

NEET Odisha-2019

Current Electricity

152887 You are given an ammeter, a galvanometer and a voltmeter. From these, the device having maximum resistance is:

1 ammeter

2 galvanometer

3 voltmeter

4 all will have the same resistance

Explanation:

C You are given an ammeter, a galvanometer and voltmeter. From these the device having maximum resistance is voltmeter.
(i) Voltmeter

Equivalent resistance of voltmeter $(R_{eq}) = R_{G} + R_{V}$
(ii) Galvanometer

Equivalent resistance of galvanometer, $R_{e g} (G) = R_{G}$ (iii) Ammeter

Equivalent resistance of ammeter, $R_{e q} (A) = \frac{R_{S} R_{G}}{R_{G} + R_{S}}$
Hence, the order of resistance
$R_{eq (V)} > R_{eq (G)} > R_{eq (A)}$
So, that maximum resistance device is voltmeter.

UPSEE 2019

Current Electricity

152888 In a potentiometer experiment, the balancing point with a cell is at a length $240 cm$ . On shunting the cell with a resistance of $2 Ω$ . the balancing length becomes $120 cm$ . The internal resistance of the cell is:

1

4 Ω

2

2 Ω

3

1 Ω

4

0.5 Ω

Explanation:

B Given that,
$l_{1} = 240$
$l_{2} = 120$
$R = 2 Ω$
$r = ?$
$r = R [\frac{l_{1}}{l_{2}} - 1] = 2 [\frac{240}{120} - 1]$
$= 2 [2 - 1] = 2 \times 1$
$= 2 Ω$

Karnataka CET-2019

Current Electricity

152889 The number of turns in a coil of galvanometer is tripled, then :

1 voltage sensitivity increases 3 times and current sensitivity remains constant

2 voltage sensitivity remains constant and current sensitivity increases 3 times

3 both voltage and current sensitivity remains constant

4 both voltage and current sensitivity decreases by

33 %

Explanation:

B Current sensitivity $(I_{s}) = \frac{N B A}{K}$
Where, N- number of turns
B- magnetic field
A- area of coils
$K$ - figure of miret Now, $I_{s} \propto N$
Therefore, current sensitivity will increase 3 times
And, voltage sensitivity $(V_{s}) = \frac{I_{s}}{R}$
$V_{s} = \frac{NBA}{K} \times \frac{1}{R}$
From above expression we can conclude
$N \propto R and I_{s} \propto N$
Thus, $V_{s}$ will remain constant.
Voltage sensitivity remains constant and current sensitivity increases 3 times.

Karnataka CET-2019

Current Electricity

152885 The meter bridge shown in the balance position with $\frac{P}{Q} = \frac{l_{1}}{l_{2}}$ .If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, that will be balanced condition?

1 yes,

\frac{P}{Q} = \frac{l_{2} - l_{1}}{l_{2} + l_{1}}

2 no, no null point

3 yes,

\frac{P}{Q} = \frac{l_{2}}{l_{1}}

4 yes,

\frac{P}{Q} = \frac{l_{1}}{l_{2}}

Explanation:

D The meter bridge is balanced against length $l_{1}$ and $l_{2}$ are-

When galvanometer and cells are interchanged, the balance condition remains unchanged.

For balancing condition,
$\frac{P}{Q} = \frac{l_{1}}{l_{2}}$
Yes, the bridge will work for balanced condition.

NEET Odisha-2019

Current Electricity

152887 You are given an ammeter, a galvanometer and a voltmeter. From these, the device having maximum resistance is:

1 ammeter

2 galvanometer

3 voltmeter

4 all will have the same resistance

Explanation:

C You are given an ammeter, a galvanometer and voltmeter. From these the device having maximum resistance is voltmeter.
(i) Voltmeter

Equivalent resistance of voltmeter $(R_{eq}) = R_{G} + R_{V}$
(ii) Galvanometer

Equivalent resistance of galvanometer, $R_{e g} (G) = R_{G}$ (iii) Ammeter

Equivalent resistance of ammeter, $R_{e q} (A) = \frac{R_{S} R_{G}}{R_{G} + R_{S}}$
Hence, the order of resistance
$R_{eq (V)} > R_{eq (G)} > R_{eq (A)}$
So, that maximum resistance device is voltmeter.

UPSEE 2019

Current Electricity

152888 In a potentiometer experiment, the balancing point with a cell is at a length $240 cm$ . On shunting the cell with a resistance of $2 Ω$ . the balancing length becomes $120 cm$ . The internal resistance of the cell is:

1

4 Ω

2

2 Ω

3

1 Ω

4

0.5 Ω

Explanation:

B Given that,
$l_{1} = 240$
$l_{2} = 120$
$R = 2 Ω$
$r = ?$
$r = R [\frac{l_{1}}{l_{2}} - 1] = 2 [\frac{240}{120} - 1]$
$= 2 [2 - 1] = 2 \times 1$
$= 2 Ω$

Karnataka CET-2019

Current Electricity

152889 The number of turns in a coil of galvanometer is tripled, then :

1 voltage sensitivity increases 3 times and current sensitivity remains constant

2 voltage sensitivity remains constant and current sensitivity increases 3 times

3 both voltage and current sensitivity remains constant

4 both voltage and current sensitivity decreases by

33 %

Explanation:

B Current sensitivity $(I_{s}) = \frac{N B A}{K}$
Where, N- number of turns
B- magnetic field
A- area of coils
$K$ - figure of miret Now, $I_{s} \propto N$
Therefore, current sensitivity will increase 3 times
And, voltage sensitivity $(V_{s}) = \frac{I_{s}}{R}$
$V_{s} = \frac{NBA}{K} \times \frac{1}{R}$
From above expression we can conclude
$N \propto R and I_{s} \propto N$
Thus, $V_{s}$ will remain constant.
Voltage sensitivity remains constant and current sensitivity increases 3 times.

Karnataka CET-2019

Current Electricity

152885 The meter bridge shown in the balance position with $\frac{P}{Q} = \frac{l_{1}}{l_{2}}$ .If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, that will be balanced condition?

1 yes,

\frac{P}{Q} = \frac{l_{2} - l_{1}}{l_{2} + l_{1}}

2 no, no null point

3 yes,

\frac{P}{Q} = \frac{l_{2}}{l_{1}}

4 yes,

\frac{P}{Q} = \frac{l_{1}}{l_{2}}

Explanation:

D The meter bridge is balanced against length $l_{1}$ and $l_{2}$ are-

When galvanometer and cells are interchanged, the balance condition remains unchanged.

For balancing condition,
$\frac{P}{Q} = \frac{l_{1}}{l_{2}}$
Yes, the bridge will work for balanced condition.

NEET Odisha-2019

Current Electricity

152887 You are given an ammeter, a galvanometer and a voltmeter. From these, the device having maximum resistance is:

1 ammeter

2 galvanometer

3 voltmeter

4 all will have the same resistance

Explanation:

C You are given an ammeter, a galvanometer and voltmeter. From these the device having maximum resistance is voltmeter.
(i) Voltmeter

Equivalent resistance of voltmeter $(R_{eq}) = R_{G} + R_{V}$
(ii) Galvanometer

Equivalent resistance of galvanometer, $R_{e g} (G) = R_{G}$ (iii) Ammeter

Equivalent resistance of ammeter, $R_{e q} (A) = \frac{R_{S} R_{G}}{R_{G} + R_{S}}$
Hence, the order of resistance
$R_{eq (V)} > R_{eq (G)} > R_{eq (A)}$
So, that maximum resistance device is voltmeter.

UPSEE 2019

Current Electricity

152888 In a potentiometer experiment, the balancing point with a cell is at a length $240 cm$ . On shunting the cell with a resistance of $2 Ω$ . the balancing length becomes $120 cm$ . The internal resistance of the cell is:

1

4 Ω

2

2 Ω

3

1 Ω

4

0.5 Ω

Explanation:

B Given that,
$l_{1} = 240$
$l_{2} = 120$
$R = 2 Ω$
$r = ?$
$r = R [\frac{l_{1}}{l_{2}} - 1] = 2 [\frac{240}{120} - 1]$
$= 2 [2 - 1] = 2 \times 1$
$= 2 Ω$

Karnataka CET-2019

Current Electricity

152889 The number of turns in a coil of galvanometer is tripled, then :

1 voltage sensitivity increases 3 times and current sensitivity remains constant

2 voltage sensitivity remains constant and current sensitivity increases 3 times

3 both voltage and current sensitivity remains constant

4 both voltage and current sensitivity decreases by

33 %

Explanation:

B Current sensitivity $(I_{s}) = \frac{N B A}{K}$
Where, N- number of turns
B- magnetic field
A- area of coils
$K$ - figure of miret Now, $I_{s} \propto N$
Therefore, current sensitivity will increase 3 times
And, voltage sensitivity $(V_{s}) = \frac{I_{s}}{R}$
$V_{s} = \frac{NBA}{K} \times \frac{1}{R}$
From above expression we can conclude
$N \propto R and I_{s} \propto N$
Thus, $V_{s}$ will remain constant.
Voltage sensitivity remains constant and current sensitivity increases 3 times.

Karnataka CET-2019

Current Electricity

152885 The meter bridge shown in the balance position with $\frac{P}{Q} = \frac{l_{1}}{l_{2}}$ .If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, that will be balanced condition?

1 yes,

\frac{P}{Q} = \frac{l_{2} - l_{1}}{l_{2} + l_{1}}

2 no, no null point

3 yes,

\frac{P}{Q} = \frac{l_{2}}{l_{1}}

4 yes,

\frac{P}{Q} = \frac{l_{1}}{l_{2}}

Explanation:

D The meter bridge is balanced against length $l_{1}$ and $l_{2}$ are-

When galvanometer and cells are interchanged, the balance condition remains unchanged.

For balancing condition,
$\frac{P}{Q} = \frac{l_{1}}{l_{2}}$
Yes, the bridge will work for balanced condition.

NEET Odisha-2019

Current Electricity

152887 You are given an ammeter, a galvanometer and a voltmeter. From these, the device having maximum resistance is:

1 ammeter

2 galvanometer

3 voltmeter

4 all will have the same resistance

Explanation:

C You are given an ammeter, a galvanometer and voltmeter. From these the device having maximum resistance is voltmeter.
(i) Voltmeter

Equivalent resistance of voltmeter $(R_{eq}) = R_{G} + R_{V}$
(ii) Galvanometer

Equivalent resistance of galvanometer, $R_{e g} (G) = R_{G}$ (iii) Ammeter

Equivalent resistance of ammeter, $R_{e q} (A) = \frac{R_{S} R_{G}}{R_{G} + R_{S}}$
Hence, the order of resistance
$R_{eq (V)} > R_{eq (G)} > R_{eq (A)}$
So, that maximum resistance device is voltmeter.

UPSEE 2019

Current Electricity

152888 In a potentiometer experiment, the balancing point with a cell is at a length $240 cm$ . On shunting the cell with a resistance of $2 Ω$ . the balancing length becomes $120 cm$ . The internal resistance of the cell is:

1

4 Ω

2

2 Ω

3

1 Ω

4

0.5 Ω

Explanation:

B Given that,
$l_{1} = 240$
$l_{2} = 120$
$R = 2 Ω$
$r = ?$
$r = R [\frac{l_{1}}{l_{2}} - 1] = 2 [\frac{240}{120} - 1]$
$= 2 [2 - 1] = 2 \times 1$
$= 2 Ω$

Karnataka CET-2019

Current Electricity

152889 The number of turns in a coil of galvanometer is tripled, then :

1 voltage sensitivity increases 3 times and current sensitivity remains constant

2 voltage sensitivity remains constant and current sensitivity increases 3 times

3 both voltage and current sensitivity remains constant

4 both voltage and current sensitivity decreases by

33 %

Explanation:

B Current sensitivity $(I_{s}) = \frac{N B A}{K}$
Where, N- number of turns
B- magnetic field
A- area of coils
$K$ - figure of miret Now, $I_{s} \propto N$
Therefore, current sensitivity will increase 3 times
And, voltage sensitivity $(V_{s}) = \frac{I_{s}}{R}$
$V_{s} = \frac{NBA}{K} \times \frac{1}{R}$
From above expression we can conclude
$N \propto R and I_{s} \propto N$
Thus, $V_{s}$ will remain constant.
Voltage sensitivity remains constant and current sensitivity increases 3 times.

Karnataka CET-2019

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