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Current Electricity

268340 A potentiometer having a wire of $4 m$ length is connected to the terminals of a battery with a steady voltage. A leclanche cell has a null point at $1 m$ . If thelength of the potentiometer wire is increased by $1 m$ , The position of the null point is

1

1.5 m

2

1.25 m

3

10.05 m

4

1.31 m

Explanation:

$I α L \Rightarrow \frac{I_{1}}{I_{2}} = \frac{L_{1}}{L_{2}}$

Current Electricity

268341 Theemf of a battery $A$ is balanced by a length of $80 cm$ on a potentio meter wire. The emf of a standard cell $1 v$ is balanced by $50 cm$ . The emf of $A$ is

1

2 V

2

1.4 V

3

1.5 V

4

1.6 V

Explanation:

$v = i ρ l$

Current Electricity

268342 When 6 identical cells of no internal resistance are connected in series in the secondary circuit of a potentio meter, the balancing length is ' $l$ ', balancing length becomes $I / 3$ when some cells are connected wrongly, the number of cells conected wrongly are

1 1

2 3

3 2

4 4

Explanation:

$E^{'} = (n - 2 m) E$

Current Electricity

268343 In a potentiometer experiment, the balancing length with a cell is $560 cm$ . When an external resistance of $100 hms$ is connected in parallel to the cell the balancing length changs by $60 cm$ . Theinternal resistance of thecell in ohm is

1 3.6

2 2.4

3 1.2

4 0.6

Explanation:

$r = R [\frac{ℓ_{1} - I_{2}}{ℓ_{2}}]$

Current Electricity

268344 The resistivity of a potentio meter wire is, if the area of cross section of the wire is $4 {cm}^{2}$ . The current flowing in the circuit is 1A, the poetntial gradient is $7.5 v / m$

1

3 \times 10^{- 3} Ω - m

2

2 \times 10^{- 6} Ω - m

3

4 \times 10^{- 6} Ω - m

4

5 \times 10^{- 4} Ω - m

Explanation:

$P . G = \frac{i R}{I}$ ; Resistivity $ρ = \frac{R A}{L}$

Current Electricity

268340 A potentiometer having a wire of $4 m$ length is connected to the terminals of a battery with a steady voltage. A leclanche cell has a null point at $1 m$ . If thelength of the potentiometer wire is increased by $1 m$ , The position of the null point is

1

1.5 m

2

1.25 m

3

10.05 m

4

1.31 m

Explanation:

$I α L \Rightarrow \frac{I_{1}}{I_{2}} = \frac{L_{1}}{L_{2}}$

Current Electricity

268341 Theemf of a battery $A$ is balanced by a length of $80 cm$ on a potentio meter wire. The emf of a standard cell $1 v$ is balanced by $50 cm$ . The emf of $A$ is

1

2 V

2

1.4 V

3

1.5 V

4

1.6 V

Explanation:

$v = i ρ l$

Current Electricity

268342 When 6 identical cells of no internal resistance are connected in series in the secondary circuit of a potentio meter, the balancing length is ' $l$ ', balancing length becomes $I / 3$ when some cells are connected wrongly, the number of cells conected wrongly are

1 1

2 3

3 2

4 4

Explanation:

$E^{'} = (n - 2 m) E$

Current Electricity

268343 In a potentiometer experiment, the balancing length with a cell is $560 cm$ . When an external resistance of $100 hms$ is connected in parallel to the cell the balancing length changs by $60 cm$ . Theinternal resistance of thecell in ohm is

1 3.6

2 2.4

3 1.2

4 0.6

Explanation:

$r = R [\frac{ℓ_{1} - I_{2}}{ℓ_{2}}]$

Current Electricity

268344 The resistivity of a potentio meter wire is, if the area of cross section of the wire is $4 {cm}^{2}$ . The current flowing in the circuit is 1A, the poetntial gradient is $7.5 v / m$

1

3 \times 10^{- 3} Ω - m

2

2 \times 10^{- 6} Ω - m

3

4 \times 10^{- 6} Ω - m

4

5 \times 10^{- 4} Ω - m

Explanation:

$P . G = \frac{i R}{I}$ ; Resistivity $ρ = \frac{R A}{L}$

Current Electricity

268340 A potentiometer having a wire of $4 m$ length is connected to the terminals of a battery with a steady voltage. A leclanche cell has a null point at $1 m$ . If thelength of the potentiometer wire is increased by $1 m$ , The position of the null point is

1

1.5 m

2

1.25 m

3

10.05 m

4

1.31 m

Explanation:

$I α L \Rightarrow \frac{I_{1}}{I_{2}} = \frac{L_{1}}{L_{2}}$

Current Electricity

268341 Theemf of a battery $A$ is balanced by a length of $80 cm$ on a potentio meter wire. The emf of a standard cell $1 v$ is balanced by $50 cm$ . The emf of $A$ is

1

2 V

2

1.4 V

3

1.5 V

4

1.6 V

Explanation:

$v = i ρ l$

Current Electricity

268342 When 6 identical cells of no internal resistance are connected in series in the secondary circuit of a potentio meter, the balancing length is ' $l$ ', balancing length becomes $I / 3$ when some cells are connected wrongly, the number of cells conected wrongly are

1 1

2 3

3 2

4 4

Explanation:

$E^{'} = (n - 2 m) E$

Current Electricity

268343 In a potentiometer experiment, the balancing length with a cell is $560 cm$ . When an external resistance of $100 hms$ is connected in parallel to the cell the balancing length changs by $60 cm$ . Theinternal resistance of thecell in ohm is

1 3.6

2 2.4

3 1.2

4 0.6

Explanation:

$r = R [\frac{ℓ_{1} - I_{2}}{ℓ_{2}}]$

Current Electricity

268344 The resistivity of a potentio meter wire is, if the area of cross section of the wire is $4 {cm}^{2}$ . The current flowing in the circuit is 1A, the poetntial gradient is $7.5 v / m$

1

3 \times 10^{- 3} Ω - m

2

2 \times 10^{- 6} Ω - m

3

4 \times 10^{- 6} Ω - m

4

5 \times 10^{- 4} Ω - m

Explanation:

$P . G = \frac{i R}{I}$ ; Resistivity $ρ = \frac{R A}{L}$

Current Electricity

268340 A potentiometer having a wire of $4 m$ length is connected to the terminals of a battery with a steady voltage. A leclanche cell has a null point at $1 m$ . If thelength of the potentiometer wire is increased by $1 m$ , The position of the null point is

1

1.5 m

2

1.25 m

3

10.05 m

4

1.31 m

Explanation:

$I α L \Rightarrow \frac{I_{1}}{I_{2}} = \frac{L_{1}}{L_{2}}$

Current Electricity

268341 Theemf of a battery $A$ is balanced by a length of $80 cm$ on a potentio meter wire. The emf of a standard cell $1 v$ is balanced by $50 cm$ . The emf of $A$ is

1

2 V

2

1.4 V

3

1.5 V

4

1.6 V

Explanation:

$v = i ρ l$

Current Electricity

268342 When 6 identical cells of no internal resistance are connected in series in the secondary circuit of a potentio meter, the balancing length is ' $l$ ', balancing length becomes $I / 3$ when some cells are connected wrongly, the number of cells conected wrongly are

1 1

2 3

3 2

4 4

Explanation:

$E^{'} = (n - 2 m) E$

Current Electricity

268343 In a potentiometer experiment, the balancing length with a cell is $560 cm$ . When an external resistance of $100 hms$ is connected in parallel to the cell the balancing length changs by $60 cm$ . Theinternal resistance of thecell in ohm is

1 3.6

2 2.4

3 1.2

4 0.6

Explanation:

$r = R [\frac{ℓ_{1} - I_{2}}{ℓ_{2}}]$

Current Electricity

268344 The resistivity of a potentio meter wire is, if the area of cross section of the wire is $4 {cm}^{2}$ . The current flowing in the circuit is 1A, the poetntial gradient is $7.5 v / m$

1

3 \times 10^{- 3} Ω - m

2

2 \times 10^{- 6} Ω - m

3

4 \times 10^{- 6} Ω - m

4

5 \times 10^{- 4} Ω - m

Explanation:

$P . G = \frac{i R}{I}$ ; Resistivity $ρ = \frac{R A}{L}$

Current Electricity

268340 A potentiometer having a wire of $4 m$ length is connected to the terminals of a battery with a steady voltage. A leclanche cell has a null point at $1 m$ . If thelength of the potentiometer wire is increased by $1 m$ , The position of the null point is

1

1.5 m

2

1.25 m

3

10.05 m

4

1.31 m

Explanation:

$I α L \Rightarrow \frac{I_{1}}{I_{2}} = \frac{L_{1}}{L_{2}}$

Current Electricity

268341 Theemf of a battery $A$ is balanced by a length of $80 cm$ on a potentio meter wire. The emf of a standard cell $1 v$ is balanced by $50 cm$ . The emf of $A$ is

1

2 V

2

1.4 V

3

1.5 V

4

1.6 V

Explanation:

$v = i ρ l$

Current Electricity

268342 When 6 identical cells of no internal resistance are connected in series in the secondary circuit of a potentio meter, the balancing length is ' $l$ ', balancing length becomes $I / 3$ when some cells are connected wrongly, the number of cells conected wrongly are

1 1

2 3

3 2

4 4

Explanation:

$E^{'} = (n - 2 m) E$

Current Electricity

268343 In a potentiometer experiment, the balancing length with a cell is $560 cm$ . When an external resistance of $100 hms$ is connected in parallel to the cell the balancing length changs by $60 cm$ . Theinternal resistance of thecell in ohm is

1 3.6

2 2.4

3 1.2

4 0.6

Explanation:

$r = R [\frac{ℓ_{1} - I_{2}}{ℓ_{2}}]$

Current Electricity

268344 The resistivity of a potentio meter wire is, if the area of cross section of the wire is $4 {cm}^{2}$ . The current flowing in the circuit is 1A, the poetntial gradient is $7.5 v / m$

1

3 \times 10^{- 3} Ω - m

2

2 \times 10^{- 6} Ω - m

3

4 \times 10^{- 6} Ω - m

4

5 \times 10^{- 4} Ω - m

Explanation:

$P . G = \frac{i R}{I}$ ; Resistivity $ρ = \frac{R A}{L}$