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PHXII03:CURRENT ELECTRICITY

357465 For a metallic wire, the ratio $\frac{V}{i} (V =$ applied potential difference and $i =$ current flowing) is

1 independent of temperature

2 increases as the temperature rises

3 decreases as the temperature rises

4 increases or decreases as temperature rises depending upon the metal

Explanation:

The resistance of a metallic wire at temperature $t^{\circ} C$ is given by $R_{t} = R_{0} (1 + α t)$
where, $α$ is the coefficient of expansion.
The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of $V / i$ is equal to $R$ . Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio $V / i$ also increases.

PHXII03:CURRENT ELECTRICITY

357466 Statement A :
Ohm’s law is not valid if current depends on voltage non-linearly.
Statement B :
Ohm’s law only holds if the temperature is maintained as constant.

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.

PHXII03:CURRENT ELECTRICITY

357467 For a wire $\frac{R}{l} = \frac{1}{2}$ and length of wire is $l = 5 c m$ . If potential difference of $1 V$ is applied across it, then current through wire will be

1

40 A

2

4 A

3

25 A

4

2.5 A

Explanation:

Given, for a wire, $\frac{R}{l} = \frac{1}{2}$
Length of wire, $l = 5 c m = 5 \times 10^{- 2} m$ $∴ R = \frac{l}{2} = 2.5 \times 10^{- 2} Ω$
Potential difference, $V = 1 V$ or IR $= 1$
$I = \frac{1}{R} = \frac{1}{2.5 \times 10^{- 2}} = \frac{100}{2.5} = 40 A$
So, correct option is (1)

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357468 $I$ - $V$ characteristic of a copper wire of length $L$ and area of cross-section $A$ is shown in figure. The slope of the curve becomes
supporting img

1 Less if the area of the wire is increased

2 Less if the length of the wire is increased

3 More if experiment is performed at higher temperature

4 More if a wire of steel of same dimension is used

Explanation:

Here, slope of the curve, $\frac{Δ I}{Δ V} = \frac{1}{R} = \frac{A}{ρ l}$
$∴$ With increase in length ( $l$ ), slope decreases.
$∴$ (2) is the only correct option.

KCET - 2020

PHXII03:CURRENT ELECTRICITY

357469 $\vec{E}$ is the electric field inside a conductor whose material has conductivity $σ$ and resistivity $ρ$ . The current density inside the conductor is $\vec{j}$ . The correct form of $O h m^{'} s$ law is

1

\vec{E} = σ \vec{j}

2

\vec{j} = ρ \vec{E}

3

\vec{E} = ρ \vec{j}

4

\vec{E} \cdot \vec{j} = ρ

Explanation:

Ohm's law
$V = i R$ , and $R = ρ \frac{l}{A}$
$∴ V = i ρ \frac{l}{A}$
$\Rightarrow \frac{V}{l} = \frac{i}{A} ρ$
$∴ \vec{E} = ρ \vec{j}$ .

KCET - 2024

PHXII03:CURRENT ELECTRICITY

357465 For a metallic wire, the ratio $\frac{V}{i} (V =$ applied potential difference and $i =$ current flowing) is

1 independent of temperature

2 increases as the temperature rises

3 decreases as the temperature rises

4 increases or decreases as temperature rises depending upon the metal

Explanation:

The resistance of a metallic wire at temperature $t^{\circ} C$ is given by $R_{t} = R_{0} (1 + α t)$
where, $α$ is the coefficient of expansion.
The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of $V / i$ is equal to $R$ . Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio $V / i$ also increases.

PHXII03:CURRENT ELECTRICITY

357466 Statement A :
Ohm’s law is not valid if current depends on voltage non-linearly.
Statement B :
Ohm’s law only holds if the temperature is maintained as constant.

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.

PHXII03:CURRENT ELECTRICITY

357467 For a wire $\frac{R}{l} = \frac{1}{2}$ and length of wire is $l = 5 c m$ . If potential difference of $1 V$ is applied across it, then current through wire will be

1

40 A

2

4 A

3

25 A

4

2.5 A

Explanation:

Given, for a wire, $\frac{R}{l} = \frac{1}{2}$
Length of wire, $l = 5 c m = 5 \times 10^{- 2} m$ $∴ R = \frac{l}{2} = 2.5 \times 10^{- 2} Ω$
Potential difference, $V = 1 V$ or IR $= 1$
$I = \frac{1}{R} = \frac{1}{2.5 \times 10^{- 2}} = \frac{100}{2.5} = 40 A$
So, correct option is (1)

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357468 $I$ - $V$ characteristic of a copper wire of length $L$ and area of cross-section $A$ is shown in figure. The slope of the curve becomes
supporting img

1 Less if the area of the wire is increased

2 Less if the length of the wire is increased

3 More if experiment is performed at higher temperature

4 More if a wire of steel of same dimension is used

Explanation:

Here, slope of the curve, $\frac{Δ I}{Δ V} = \frac{1}{R} = \frac{A}{ρ l}$
$∴$ With increase in length ( $l$ ), slope decreases.
$∴$ (2) is the only correct option.

KCET - 2020

PHXII03:CURRENT ELECTRICITY

357469 $\vec{E}$ is the electric field inside a conductor whose material has conductivity $σ$ and resistivity $ρ$ . The current density inside the conductor is $\vec{j}$ . The correct form of $O h m^{'} s$ law is

1

\vec{E} = σ \vec{j}

2

\vec{j} = ρ \vec{E}

3

\vec{E} = ρ \vec{j}

4

\vec{E} \cdot \vec{j} = ρ

Explanation:

Ohm's law
$V = i R$ , and $R = ρ \frac{l}{A}$
$∴ V = i ρ \frac{l}{A}$
$\Rightarrow \frac{V}{l} = \frac{i}{A} ρ$
$∴ \vec{E} = ρ \vec{j}$ .

KCET - 2024

PHXII03:CURRENT ELECTRICITY

357465 For a metallic wire, the ratio $\frac{V}{i} (V =$ applied potential difference and $i =$ current flowing) is

1 independent of temperature

2 increases as the temperature rises

3 decreases as the temperature rises

4 increases or decreases as temperature rises depending upon the metal

Explanation:

The resistance of a metallic wire at temperature $t^{\circ} C$ is given by $R_{t} = R_{0} (1 + α t)$
where, $α$ is the coefficient of expansion.
The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of $V / i$ is equal to $R$ . Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio $V / i$ also increases.

PHXII03:CURRENT ELECTRICITY

357466 Statement A :
Ohm’s law is not valid if current depends on voltage non-linearly.
Statement B :
Ohm’s law only holds if the temperature is maintained as constant.

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.

PHXII03:CURRENT ELECTRICITY

357467 For a wire $\frac{R}{l} = \frac{1}{2}$ and length of wire is $l = 5 c m$ . If potential difference of $1 V$ is applied across it, then current through wire will be

1

40 A

2

4 A

3

25 A

4

2.5 A

Explanation:

Given, for a wire, $\frac{R}{l} = \frac{1}{2}$
Length of wire, $l = 5 c m = 5 \times 10^{- 2} m$ $∴ R = \frac{l}{2} = 2.5 \times 10^{- 2} Ω$
Potential difference, $V = 1 V$ or IR $= 1$
$I = \frac{1}{R} = \frac{1}{2.5 \times 10^{- 2}} = \frac{100}{2.5} = 40 A$
So, correct option is (1)

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357468 $I$ - $V$ characteristic of a copper wire of length $L$ and area of cross-section $A$ is shown in figure. The slope of the curve becomes
supporting img

1 Less if the area of the wire is increased

2 Less if the length of the wire is increased

3 More if experiment is performed at higher temperature

4 More if a wire of steel of same dimension is used

Explanation:

Here, slope of the curve, $\frac{Δ I}{Δ V} = \frac{1}{R} = \frac{A}{ρ l}$
$∴$ With increase in length ( $l$ ), slope decreases.
$∴$ (2) is the only correct option.

KCET - 2020

PHXII03:CURRENT ELECTRICITY

357469 $\vec{E}$ is the electric field inside a conductor whose material has conductivity $σ$ and resistivity $ρ$ . The current density inside the conductor is $\vec{j}$ . The correct form of $O h m^{'} s$ law is

1

\vec{E} = σ \vec{j}

2

\vec{j} = ρ \vec{E}

3

\vec{E} = ρ \vec{j}

4

\vec{E} \cdot \vec{j} = ρ

Explanation:

Ohm's law
$V = i R$ , and $R = ρ \frac{l}{A}$
$∴ V = i ρ \frac{l}{A}$
$\Rightarrow \frac{V}{l} = \frac{i}{A} ρ$
$∴ \vec{E} = ρ \vec{j}$ .

KCET - 2024

PHXII03:CURRENT ELECTRICITY

357465 For a metallic wire, the ratio $\frac{V}{i} (V =$ applied potential difference and $i =$ current flowing) is

1 independent of temperature

2 increases as the temperature rises

3 decreases as the temperature rises

4 increases or decreases as temperature rises depending upon the metal

Explanation:

The resistance of a metallic wire at temperature $t^{\circ} C$ is given by $R_{t} = R_{0} (1 + α t)$
where, $α$ is the coefficient of expansion.
The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of $V / i$ is equal to $R$ . Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio $V / i$ also increases.

PHXII03:CURRENT ELECTRICITY

357466 Statement A :
Ohm’s law is not valid if current depends on voltage non-linearly.
Statement B :
Ohm’s law only holds if the temperature is maintained as constant.

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.

PHXII03:CURRENT ELECTRICITY

357467 For a wire $\frac{R}{l} = \frac{1}{2}$ and length of wire is $l = 5 c m$ . If potential difference of $1 V$ is applied across it, then current through wire will be

1

40 A

2

4 A

3

25 A

4

2.5 A

Explanation:

Given, for a wire, $\frac{R}{l} = \frac{1}{2}$
Length of wire, $l = 5 c m = 5 \times 10^{- 2} m$ $∴ R = \frac{l}{2} = 2.5 \times 10^{- 2} Ω$
Potential difference, $V = 1 V$ or IR $= 1$
$I = \frac{1}{R} = \frac{1}{2.5 \times 10^{- 2}} = \frac{100}{2.5} = 40 A$
So, correct option is (1)

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357468 $I$ - $V$ characteristic of a copper wire of length $L$ and area of cross-section $A$ is shown in figure. The slope of the curve becomes
supporting img

1 Less if the area of the wire is increased

2 Less if the length of the wire is increased

3 More if experiment is performed at higher temperature

4 More if a wire of steel of same dimension is used

Explanation:

Here, slope of the curve, $\frac{Δ I}{Δ V} = \frac{1}{R} = \frac{A}{ρ l}$
$∴$ With increase in length ( $l$ ), slope decreases.
$∴$ (2) is the only correct option.

KCET - 2020

PHXII03:CURRENT ELECTRICITY

357469 $\vec{E}$ is the electric field inside a conductor whose material has conductivity $σ$ and resistivity $ρ$ . The current density inside the conductor is $\vec{j}$ . The correct form of $O h m^{'} s$ law is

1

\vec{E} = σ \vec{j}

2

\vec{j} = ρ \vec{E}

3

\vec{E} = ρ \vec{j}

4

\vec{E} \cdot \vec{j} = ρ

Explanation:

Ohm's law
$V = i R$ , and $R = ρ \frac{l}{A}$
$∴ V = i ρ \frac{l}{A}$
$\Rightarrow \frac{V}{l} = \frac{i}{A} ρ$
$∴ \vec{E} = ρ \vec{j}$ .

KCET - 2024

PHXII03:CURRENT ELECTRICITY

357465 For a metallic wire, the ratio $\frac{V}{i} (V =$ applied potential difference and $i =$ current flowing) is

1 independent of temperature

2 increases as the temperature rises

3 decreases as the temperature rises

4 increases or decreases as temperature rises depending upon the metal

Explanation:

The resistance of a metallic wire at temperature $t^{\circ} C$ is given by $R_{t} = R_{0} (1 + α t)$
where, $α$ is the coefficient of expansion.
The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of $V / i$ is equal to $R$ . Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio $V / i$ also increases.

PHXII03:CURRENT ELECTRICITY

357466 Statement A :
Ohm’s law is not valid if current depends on voltage non-linearly.
Statement B :
Ohm’s law only holds if the temperature is maintained as constant.

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.

PHXII03:CURRENT ELECTRICITY

357467 For a wire $\frac{R}{l} = \frac{1}{2}$ and length of wire is $l = 5 c m$ . If potential difference of $1 V$ is applied across it, then current through wire will be

1

40 A

2

4 A

3

25 A

4

2.5 A

Explanation:

Given, for a wire, $\frac{R}{l} = \frac{1}{2}$
Length of wire, $l = 5 c m = 5 \times 10^{- 2} m$ $∴ R = \frac{l}{2} = 2.5 \times 10^{- 2} Ω$
Potential difference, $V = 1 V$ or IR $= 1$
$I = \frac{1}{R} = \frac{1}{2.5 \times 10^{- 2}} = \frac{100}{2.5} = 40 A$
So, correct option is (1)

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357468 $I$ - $V$ characteristic of a copper wire of length $L$ and area of cross-section $A$ is shown in figure. The slope of the curve becomes
supporting img

1 Less if the area of the wire is increased

2 Less if the length of the wire is increased

3 More if experiment is performed at higher temperature

4 More if a wire of steel of same dimension is used

Explanation:

Here, slope of the curve, $\frac{Δ I}{Δ V} = \frac{1}{R} = \frac{A}{ρ l}$
$∴$ With increase in length ( $l$ ), slope decreases.
$∴$ (2) is the only correct option.

KCET - 2020

PHXII03:CURRENT ELECTRICITY

357469 $\vec{E}$ is the electric field inside a conductor whose material has conductivity $σ$ and resistivity $ρ$ . The current density inside the conductor is $\vec{j}$ . The correct form of $O h m^{'} s$ law is

1

\vec{E} = σ \vec{j}

2

\vec{j} = ρ \vec{E}

3

\vec{E} = ρ \vec{j}

4

\vec{E} \cdot \vec{j} = ρ

Explanation:

Ohm's law
$V = i R$ , and $R = ρ \frac{l}{A}$
$∴ V = i ρ \frac{l}{A}$
$\Rightarrow \frac{V}{l} = \frac{i}{A} ρ$
$∴ \vec{E} = ρ \vec{j}$ .

KCET - 2024