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PHXII03:CURRENT ELECTRICITY

357060 Assertion :
Bending of a wire does not affect resistivity of a material.
Reason :
Resistance of wire is proportional to resistivity of material.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The formula for resistance is $R = ρ \frac{l}{A}$ , where $R$ is resistance, $ρ$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).

PHXII03:CURRENT ELECTRICITY

357061 The electric intensity $E$ , current density $j$ and specific resistance $k$ are related to each other by the relation

1

k = j E

2

E = j k

3

E = k / j

4

E = j / k

PHXII03:CURRENT ELECTRICITY

357062 A wire of resistance $160 Ω$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

1

40 Ω

2

16 Ω

3

10 Ω

4

640 Ω

Explanation:

Volume of the wire before being melted and after will remain same therefore.
Volume of wire before melting = Volume of wire after melting $A^{'} l^{'} = A l$
$π r^{' 2} l^{'} = π r^{2} l$
As $l = \frac{l^{'}}{4}; π r^{' 2} l^{'} = π r^{2} \frac{l^{'}}{4} r^{2}$
$r = 2 r^{'} \frac{R}{R^{'}} = \frac{ρ l}{A} \times \frac{A^{'}}{ρ l^{'}} = \frac{l}{l^{'}} \times \frac{π {r^{'}}^{2}}{π r^{2}} = \frac{l}{l^{'}} \frac{{r^{'}}^{2}}{r^{2}}$
$R = R^{'} (\frac{l^{'}}{4 l^{'}} \times \frac{r^{' 2}}{{(2 r^{'})}^{2}}) = \frac{160}{16} = 10 Ω$

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357063 Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are

1

1 : 15 : 25

2

1 : 3 : 5

3

125 : 15 : 1

4

5 : 3 : 1

Explanation:

As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes.
$∴ V_{1} : V_{2} : V_{3} = 1 : 3 : 5$
Also $l_{1} : l_{2} : l_{3} = 5 : 3 : 1$ (given)
As resistance
$R = (ρ l) / A = (ρ l^{2}) / A l = (ρ l^{2}) / V$
$∴ R_{1} : R_{2} : R_{3} = (l_{1}^{2} / V_{1}) : (l_{2}^{2} / V_{2}) : (l_{3}^{2} / V_{3})$
$= (5^{2} / 1) : (3^{2} / 3) : (1^{2} / 5) = 125 : 15 : 1$

KCET - 2019

PHXII03:CURRENT ELECTRICITY

357064 The length of the wire is doubled. Its conductance will be

1 remain unchanged

2 halved

3 doubled

4 quadrupled

Explanation:

Conductance is given by $σ = \frac{1}{R}$ $Substituting R = ρ \frac{l}{A}, we get σ = \frac{A}{ρ l}$
If $A$ is constant, then we have $σ \propto \frac{1}{l}$
$\Rightarrow \frac{σ_{2}}{σ_{1}} = \frac{l_{1}}{l_{2}}$
Substituting, $l_{2} = 2 l_{1}$ , we get $σ_{2} = \frac{σ_{1}}{2}$
We see that, conductance will be halved.

PHXII03:CURRENT ELECTRICITY

357060 Assertion :
Bending of a wire does not affect resistivity of a material.
Reason :
Resistance of wire is proportional to resistivity of material.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The formula for resistance is $R = ρ \frac{l}{A}$ , where $R$ is resistance, $ρ$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).

PHXII03:CURRENT ELECTRICITY

357061 The electric intensity $E$ , current density $j$ and specific resistance $k$ are related to each other by the relation

1

k = j E

2

E = j k

3

E = k / j

4

E = j / k

PHXII03:CURRENT ELECTRICITY

357062 A wire of resistance $160 Ω$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

1

40 Ω

2

16 Ω

3

10 Ω

4

640 Ω

Explanation:

Volume of the wire before being melted and after will remain same therefore.
Volume of wire before melting = Volume of wire after melting $A^{'} l^{'} = A l$
$π r^{' 2} l^{'} = π r^{2} l$
As $l = \frac{l^{'}}{4}; π r^{' 2} l^{'} = π r^{2} \frac{l^{'}}{4} r^{2}$
$r = 2 r^{'} \frac{R}{R^{'}} = \frac{ρ l}{A} \times \frac{A^{'}}{ρ l^{'}} = \frac{l}{l^{'}} \times \frac{π {r^{'}}^{2}}{π r^{2}} = \frac{l}{l^{'}} \frac{{r^{'}}^{2}}{r^{2}}$
$R = R^{'} (\frac{l^{'}}{4 l^{'}} \times \frac{r^{' 2}}{{(2 r^{'})}^{2}}) = \frac{160}{16} = 10 Ω$

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357063 Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are

1

1 : 15 : 25

2

1 : 3 : 5

3

125 : 15 : 1

4

5 : 3 : 1

Explanation:

As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes.
$∴ V_{1} : V_{2} : V_{3} = 1 : 3 : 5$
Also $l_{1} : l_{2} : l_{3} = 5 : 3 : 1$ (given)
As resistance
$R = (ρ l) / A = (ρ l^{2}) / A l = (ρ l^{2}) / V$
$∴ R_{1} : R_{2} : R_{3} = (l_{1}^{2} / V_{1}) : (l_{2}^{2} / V_{2}) : (l_{3}^{2} / V_{3})$
$= (5^{2} / 1) : (3^{2} / 3) : (1^{2} / 5) = 125 : 15 : 1$

KCET - 2019

PHXII03:CURRENT ELECTRICITY

357064 The length of the wire is doubled. Its conductance will be

1 remain unchanged

2 halved

3 doubled

4 quadrupled

Explanation:

Conductance is given by $σ = \frac{1}{R}$ $Substituting R = ρ \frac{l}{A}, we get σ = \frac{A}{ρ l}$
If $A$ is constant, then we have $σ \propto \frac{1}{l}$
$\Rightarrow \frac{σ_{2}}{σ_{1}} = \frac{l_{1}}{l_{2}}$
Substituting, $l_{2} = 2 l_{1}$ , we get $σ_{2} = \frac{σ_{1}}{2}$
We see that, conductance will be halved.

PHXII03:CURRENT ELECTRICITY

357060 Assertion :
Bending of a wire does not affect resistivity of a material.
Reason :
Resistance of wire is proportional to resistivity of material.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The formula for resistance is $R = ρ \frac{l}{A}$ , where $R$ is resistance, $ρ$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).

PHXII03:CURRENT ELECTRICITY

357061 The electric intensity $E$ , current density $j$ and specific resistance $k$ are related to each other by the relation

1

k = j E

2

E = j k

3

E = k / j

4

E = j / k

PHXII03:CURRENT ELECTRICITY

357062 A wire of resistance $160 Ω$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

1

40 Ω

2

16 Ω

3

10 Ω

4

640 Ω

Explanation:

Volume of the wire before being melted and after will remain same therefore.
Volume of wire before melting = Volume of wire after melting $A^{'} l^{'} = A l$
$π r^{' 2} l^{'} = π r^{2} l$
As $l = \frac{l^{'}}{4}; π r^{' 2} l^{'} = π r^{2} \frac{l^{'}}{4} r^{2}$
$r = 2 r^{'} \frac{R}{R^{'}} = \frac{ρ l}{A} \times \frac{A^{'}}{ρ l^{'}} = \frac{l}{l^{'}} \times \frac{π {r^{'}}^{2}}{π r^{2}} = \frac{l}{l^{'}} \frac{{r^{'}}^{2}}{r^{2}}$
$R = R^{'} (\frac{l^{'}}{4 l^{'}} \times \frac{r^{' 2}}{{(2 r^{'})}^{2}}) = \frac{160}{16} = 10 Ω$

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357063 Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are

1

1 : 15 : 25

2

1 : 3 : 5

3

125 : 15 : 1

4

5 : 3 : 1

Explanation:

As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes.
$∴ V_{1} : V_{2} : V_{3} = 1 : 3 : 5$
Also $l_{1} : l_{2} : l_{3} = 5 : 3 : 1$ (given)
As resistance
$R = (ρ l) / A = (ρ l^{2}) / A l = (ρ l^{2}) / V$
$∴ R_{1} : R_{2} : R_{3} = (l_{1}^{2} / V_{1}) : (l_{2}^{2} / V_{2}) : (l_{3}^{2} / V_{3})$
$= (5^{2} / 1) : (3^{2} / 3) : (1^{2} / 5) = 125 : 15 : 1$

KCET - 2019

PHXII03:CURRENT ELECTRICITY

357064 The length of the wire is doubled. Its conductance will be

1 remain unchanged

2 halved

3 doubled

4 quadrupled

Explanation:

Conductance is given by $σ = \frac{1}{R}$ $Substituting R = ρ \frac{l}{A}, we get σ = \frac{A}{ρ l}$
If $A$ is constant, then we have $σ \propto \frac{1}{l}$
$\Rightarrow \frac{σ_{2}}{σ_{1}} = \frac{l_{1}}{l_{2}}$
Substituting, $l_{2} = 2 l_{1}$ , we get $σ_{2} = \frac{σ_{1}}{2}$
We see that, conductance will be halved.

PHXII03:CURRENT ELECTRICITY

357060 Assertion :
Bending of a wire does not affect resistivity of a material.
Reason :
Resistance of wire is proportional to resistivity of material.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The formula for resistance is $R = ρ \frac{l}{A}$ , where $R$ is resistance, $ρ$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).

PHXII03:CURRENT ELECTRICITY

357061 The electric intensity $E$ , current density $j$ and specific resistance $k$ are related to each other by the relation

1

k = j E

2

E = j k

3

E = k / j

4

E = j / k

PHXII03:CURRENT ELECTRICITY

357062 A wire of resistance $160 Ω$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

1

40 Ω

2

16 Ω

3

10 Ω

4

640 Ω

Explanation:

Volume of the wire before being melted and after will remain same therefore.
Volume of wire before melting = Volume of wire after melting $A^{'} l^{'} = A l$
$π r^{' 2} l^{'} = π r^{2} l$
As $l = \frac{l^{'}}{4}; π r^{' 2} l^{'} = π r^{2} \frac{l^{'}}{4} r^{2}$
$r = 2 r^{'} \frac{R}{R^{'}} = \frac{ρ l}{A} \times \frac{A^{'}}{ρ l^{'}} = \frac{l}{l^{'}} \times \frac{π {r^{'}}^{2}}{π r^{2}} = \frac{l}{l^{'}} \frac{{r^{'}}^{2}}{r^{2}}$
$R = R^{'} (\frac{l^{'}}{4 l^{'}} \times \frac{r^{' 2}}{{(2 r^{'})}^{2}}) = \frac{160}{16} = 10 Ω$

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357063 Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are

1

1 : 15 : 25

2

1 : 3 : 5

3

125 : 15 : 1

4

5 : 3 : 1

Explanation:

As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes.
$∴ V_{1} : V_{2} : V_{3} = 1 : 3 : 5$
Also $l_{1} : l_{2} : l_{3} = 5 : 3 : 1$ (given)
As resistance
$R = (ρ l) / A = (ρ l^{2}) / A l = (ρ l^{2}) / V$
$∴ R_{1} : R_{2} : R_{3} = (l_{1}^{2} / V_{1}) : (l_{2}^{2} / V_{2}) : (l_{3}^{2} / V_{3})$
$= (5^{2} / 1) : (3^{2} / 3) : (1^{2} / 5) = 125 : 15 : 1$

KCET - 2019

PHXII03:CURRENT ELECTRICITY

357064 The length of the wire is doubled. Its conductance will be

1 remain unchanged

2 halved

3 doubled

4 quadrupled

Explanation:

Conductance is given by $σ = \frac{1}{R}$ $Substituting R = ρ \frac{l}{A}, we get σ = \frac{A}{ρ l}$
If $A$ is constant, then we have $σ \propto \frac{1}{l}$
$\Rightarrow \frac{σ_{2}}{σ_{1}} = \frac{l_{1}}{l_{2}}$
Substituting, $l_{2} = 2 l_{1}$ , we get $σ_{2} = \frac{σ_{1}}{2}$
We see that, conductance will be halved.

PHXII03:CURRENT ELECTRICITY

357060 Assertion :
Bending of a wire does not affect resistivity of a material.
Reason :
Resistance of wire is proportional to resistivity of material.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The formula for resistance is $R = ρ \frac{l}{A}$ , where $R$ is resistance, $ρ$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).

PHXII03:CURRENT ELECTRICITY

357061 The electric intensity $E$ , current density $j$ and specific resistance $k$ are related to each other by the relation

1

k = j E

2

E = j k

3

E = k / j

4

E = j / k

PHXII03:CURRENT ELECTRICITY

357062 A wire of resistance $160 Ω$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

1

40 Ω

2

16 Ω

3

10 Ω

4

640 Ω

Explanation:

Volume of the wire before being melted and after will remain same therefore.
Volume of wire before melting = Volume of wire after melting $A^{'} l^{'} = A l$
$π r^{' 2} l^{'} = π r^{2} l$
As $l = \frac{l^{'}}{4}; π r^{' 2} l^{'} = π r^{2} \frac{l^{'}}{4} r^{2}$
$r = 2 r^{'} \frac{R}{R^{'}} = \frac{ρ l}{A} \times \frac{A^{'}}{ρ l^{'}} = \frac{l}{l^{'}} \times \frac{π {r^{'}}^{2}}{π r^{2}} = \frac{l}{l^{'}} \frac{{r^{'}}^{2}}{r^{2}}$
$R = R^{'} (\frac{l^{'}}{4 l^{'}} \times \frac{r^{' 2}}{{(2 r^{'})}^{2}}) = \frac{160}{16} = 10 Ω$

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357063 Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are

1

1 : 15 : 25

2

1 : 3 : 5

3

125 : 15 : 1

4

5 : 3 : 1

Explanation:

As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes.
$∴ V_{1} : V_{2} : V_{3} = 1 : 3 : 5$
Also $l_{1} : l_{2} : l_{3} = 5 : 3 : 1$ (given)
As resistance
$R = (ρ l) / A = (ρ l^{2}) / A l = (ρ l^{2}) / V$
$∴ R_{1} : R_{2} : R_{3} = (l_{1}^{2} / V_{1}) : (l_{2}^{2} / V_{2}) : (l_{3}^{2} / V_{3})$
$= (5^{2} / 1) : (3^{2} / 3) : (1^{2} / 5) = 125 : 15 : 1$

KCET - 2019

PHXII03:CURRENT ELECTRICITY

357064 The length of the wire is doubled. Its conductance will be

1 remain unchanged

2 halved

3 doubled

4 quadrupled

Explanation:

Conductance is given by $σ = \frac{1}{R}$ $Substituting R = ρ \frac{l}{A}, we get σ = \frac{A}{ρ l}$
If $A$ is constant, then we have $σ \propto \frac{1}{l}$
$\Rightarrow \frac{σ_{2}}{σ_{1}} = \frac{l_{1}}{l_{2}}$
Substituting, $l_{2} = 2 l_{1}$ , we get $σ_{2} = \frac{σ_{1}}{2}$
We see that, conductance will be halved.

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