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PHXII03:CURRENT ELECTRICITY

357030 The relaxation time in conductors

1 Decrease with the increase of temperature

2 Increase with the increase of temperature

3 It does not depend on temperature

4 All of sudden changes at 400

K

PHXII03:CURRENT ELECTRICITY

357031 A steady current is set up in a metallic wire of non-uniform cross-section. How is the rate of flow of free electrons related to the area of cross-section $A$ ?

1

K

is independent of

A

2

K \propto A^{- 1}

3

K \propto A

4

K \propto A^{2}

Explanation:

As, $i =$ Anev $_{d}$
$\Rightarrow \frac{d q}{d t} =$ Anev $_{d} (∵ i = \frac{d q}{d t})$
$∴ \frac{d q}{d t} \propto A \Rightarrow K \propto A$
So, correct option is (3)

AIIMS - 2009

PHXII03:CURRENT ELECTRICITY

357032 Consider a copper wire of length $L$ , cross-section area $A$ . It has n number of free electrons per unit volume. Which of the following is the correct expression of drift velocity of the electrons when the wire carries a steady current I?

1

\frac{I}{n^{2} e L}

2

\frac{I}{n e L}

3

\frac{I}{n e A}

4

\frac{I}{n e^{2} L A}

Explanation:

$I$ is the current, $n$ is the number of free electrons per unit volume, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $V_{d} = \frac{I}{n e A}$ ,
The denominator represents the product of the number of free electrons per unit volume $(n)$ , the charge of an electron $(e)$ , and the cross-sectional area $(A)$ . So correct option is (3).

PHXII03:CURRENT ELECTRICITY

357033 A copper wire of length 1 $m$ and uniform cross sectional area $5 \times 10^{- 7} m^{2}$ carries a current of 1 $A$ . Assuming that there are $8 \times 10^{28}$ free electrons per $m^{3}$ of copper, how long will an electron take to drift from one end of the wire to the other ?

1

0.8 \times 10^{3} s

2

1.6 \times 10^{3} s

3

3.2 \times 10^{3} s

4

6.4 \times 10^{3} s

Explanation:

Length of the wire $= l = 1 m$
Cross section area of the wire
$= A = 5 \times 10^{- 7} m^{2}$
Number density of electrons in the wire
$= n = 8 \times 10^{28} p e r m^{3}$
Charge of an electron $= e = 1.6 \times 10^{- 19} C$
Drift velocity of electrons in the wire $= v_{d}$
Current in the wire $= I = n e v_{d} A = 1 A$
$\Rightarrow v_{d} = \frac{I}{n e A}$
$∴$ Time taken by an electron to drift from one end of the wire to the other end $= t = \frac{l}{v_{d}} = \frac{l n e A}{I}$
$\Rightarrow t =$
$\frac{1 \times 8 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 7}}{1} = 6.4 \times 10^{3} s$

KCET - 2021

PHXII03:CURRENT ELECTRICITY

357034 An electric cell of e.m.f. $ε$ is connected across a copper wire of diameter $d$ and length $l$ . The drift velocity of electrons in the wire is $v_{d}$ . If the length of the wire is changed to 2 $l$ , the new drift velocity of electrons in the copper wire will be

1

2 v_{d}

2

v_{d} / 4

3

v_{d} / 2

4

v_{d}

Explanation:

$v_{d} = \frac{i}{n e A} = \frac{ε}{R \times n e A} = \frac{ε}{ρ \times l \times n \times e}$
$v_{d}^{'} = \frac{ε}{ρ \times 2 l \times n \times e} \Rightarrow \frac{v_{d}^{'}}{v_{d}} = \frac{1}{2}$

PHXII03:CURRENT ELECTRICITY

357030 The relaxation time in conductors

1 Decrease with the increase of temperature

2 Increase with the increase of temperature

3 It does not depend on temperature

4 All of sudden changes at 400

K

PHXII03:CURRENT ELECTRICITY

357031 A steady current is set up in a metallic wire of non-uniform cross-section. How is the rate of flow of free electrons related to the area of cross-section $A$ ?

1

K

is independent of

A

2

K \propto A^{- 1}

3

K \propto A

4

K \propto A^{2}

Explanation:

As, $i =$ Anev $_{d}$
$\Rightarrow \frac{d q}{d t} =$ Anev $_{d} (∵ i = \frac{d q}{d t})$
$∴ \frac{d q}{d t} \propto A \Rightarrow K \propto A$
So, correct option is (3)

AIIMS - 2009

PHXII03:CURRENT ELECTRICITY

357032 Consider a copper wire of length $L$ , cross-section area $A$ . It has n number of free electrons per unit volume. Which of the following is the correct expression of drift velocity of the electrons when the wire carries a steady current I?

1

\frac{I}{n^{2} e L}

2

\frac{I}{n e L}

3

\frac{I}{n e A}

4

\frac{I}{n e^{2} L A}

Explanation:

$I$ is the current, $n$ is the number of free electrons per unit volume, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $V_{d} = \frac{I}{n e A}$ ,
The denominator represents the product of the number of free electrons per unit volume $(n)$ , the charge of an electron $(e)$ , and the cross-sectional area $(A)$ . So correct option is (3).

PHXII03:CURRENT ELECTRICITY

357033 A copper wire of length 1 $m$ and uniform cross sectional area $5 \times 10^{- 7} m^{2}$ carries a current of 1 $A$ . Assuming that there are $8 \times 10^{28}$ free electrons per $m^{3}$ of copper, how long will an electron take to drift from one end of the wire to the other ?

1

0.8 \times 10^{3} s

2

1.6 \times 10^{3} s

3

3.2 \times 10^{3} s

4

6.4 \times 10^{3} s

Explanation:

Length of the wire $= l = 1 m$
Cross section area of the wire
$= A = 5 \times 10^{- 7} m^{2}$
Number density of electrons in the wire
$= n = 8 \times 10^{28} p e r m^{3}$
Charge of an electron $= e = 1.6 \times 10^{- 19} C$
Drift velocity of electrons in the wire $= v_{d}$
Current in the wire $= I = n e v_{d} A = 1 A$
$\Rightarrow v_{d} = \frac{I}{n e A}$
$∴$ Time taken by an electron to drift from one end of the wire to the other end $= t = \frac{l}{v_{d}} = \frac{l n e A}{I}$
$\Rightarrow t =$
$\frac{1 \times 8 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 7}}{1} = 6.4 \times 10^{3} s$

KCET - 2021

PHXII03:CURRENT ELECTRICITY

357034 An electric cell of e.m.f. $ε$ is connected across a copper wire of diameter $d$ and length $l$ . The drift velocity of electrons in the wire is $v_{d}$ . If the length of the wire is changed to 2 $l$ , the new drift velocity of electrons in the copper wire will be

1

2 v_{d}

2

v_{d} / 4

3

v_{d} / 2

4

v_{d}

Explanation:

$v_{d} = \frac{i}{n e A} = \frac{ε}{R \times n e A} = \frac{ε}{ρ \times l \times n \times e}$
$v_{d}^{'} = \frac{ε}{ρ \times 2 l \times n \times e} \Rightarrow \frac{v_{d}^{'}}{v_{d}} = \frac{1}{2}$

PHXII03:CURRENT ELECTRICITY

357030 The relaxation time in conductors

1 Decrease with the increase of temperature

2 Increase with the increase of temperature

3 It does not depend on temperature

4 All of sudden changes at 400

K

PHXII03:CURRENT ELECTRICITY

357031 A steady current is set up in a metallic wire of non-uniform cross-section. How is the rate of flow of free electrons related to the area of cross-section $A$ ?

1

K

is independent of

A

2

K \propto A^{- 1}

3

K \propto A

4

K \propto A^{2}

Explanation:

As, $i =$ Anev $_{d}$
$\Rightarrow \frac{d q}{d t} =$ Anev $_{d} (∵ i = \frac{d q}{d t})$
$∴ \frac{d q}{d t} \propto A \Rightarrow K \propto A$
So, correct option is (3)

AIIMS - 2009

PHXII03:CURRENT ELECTRICITY

357032 Consider a copper wire of length $L$ , cross-section area $A$ . It has n number of free electrons per unit volume. Which of the following is the correct expression of drift velocity of the electrons when the wire carries a steady current I?

1

\frac{I}{n^{2} e L}

2

\frac{I}{n e L}

3

\frac{I}{n e A}

4

\frac{I}{n e^{2} L A}

Explanation:

$I$ is the current, $n$ is the number of free electrons per unit volume, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $V_{d} = \frac{I}{n e A}$ ,
The denominator represents the product of the number of free electrons per unit volume $(n)$ , the charge of an electron $(e)$ , and the cross-sectional area $(A)$ . So correct option is (3).

PHXII03:CURRENT ELECTRICITY

357033 A copper wire of length 1 $m$ and uniform cross sectional area $5 \times 10^{- 7} m^{2}$ carries a current of 1 $A$ . Assuming that there are $8 \times 10^{28}$ free electrons per $m^{3}$ of copper, how long will an electron take to drift from one end of the wire to the other ?

1

0.8 \times 10^{3} s

2

1.6 \times 10^{3} s

3

3.2 \times 10^{3} s

4

6.4 \times 10^{3} s

Explanation:

Length of the wire $= l = 1 m$
Cross section area of the wire
$= A = 5 \times 10^{- 7} m^{2}$
Number density of electrons in the wire
$= n = 8 \times 10^{28} p e r m^{3}$
Charge of an electron $= e = 1.6 \times 10^{- 19} C$
Drift velocity of electrons in the wire $= v_{d}$
Current in the wire $= I = n e v_{d} A = 1 A$
$\Rightarrow v_{d} = \frac{I}{n e A}$
$∴$ Time taken by an electron to drift from one end of the wire to the other end $= t = \frac{l}{v_{d}} = \frac{l n e A}{I}$
$\Rightarrow t =$
$\frac{1 \times 8 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 7}}{1} = 6.4 \times 10^{3} s$

KCET - 2021

PHXII03:CURRENT ELECTRICITY

357034 An electric cell of e.m.f. $ε$ is connected across a copper wire of diameter $d$ and length $l$ . The drift velocity of electrons in the wire is $v_{d}$ . If the length of the wire is changed to 2 $l$ , the new drift velocity of electrons in the copper wire will be

1

2 v_{d}

2

v_{d} / 4

3

v_{d} / 2

4

v_{d}

Explanation:

$v_{d} = \frac{i}{n e A} = \frac{ε}{R \times n e A} = \frac{ε}{ρ \times l \times n \times e}$
$v_{d}^{'} = \frac{ε}{ρ \times 2 l \times n \times e} \Rightarrow \frac{v_{d}^{'}}{v_{d}} = \frac{1}{2}$

PHXII03:CURRENT ELECTRICITY

357030 The relaxation time in conductors

1 Decrease with the increase of temperature

2 Increase with the increase of temperature

3 It does not depend on temperature

4 All of sudden changes at 400

K

PHXII03:CURRENT ELECTRICITY

357031 A steady current is set up in a metallic wire of non-uniform cross-section. How is the rate of flow of free electrons related to the area of cross-section $A$ ?

1

K

is independent of

A

2

K \propto A^{- 1}

3

K \propto A

4

K \propto A^{2}

Explanation:

As, $i =$ Anev $_{d}$
$\Rightarrow \frac{d q}{d t} =$ Anev $_{d} (∵ i = \frac{d q}{d t})$
$∴ \frac{d q}{d t} \propto A \Rightarrow K \propto A$
So, correct option is (3)

AIIMS - 2009

PHXII03:CURRENT ELECTRICITY

357032 Consider a copper wire of length $L$ , cross-section area $A$ . It has n number of free electrons per unit volume. Which of the following is the correct expression of drift velocity of the electrons when the wire carries a steady current I?

1

\frac{I}{n^{2} e L}

2

\frac{I}{n e L}

3

\frac{I}{n e A}

4

\frac{I}{n e^{2} L A}

Explanation:

$I$ is the current, $n$ is the number of free electrons per unit volume, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $V_{d} = \frac{I}{n e A}$ ,
The denominator represents the product of the number of free electrons per unit volume $(n)$ , the charge of an electron $(e)$ , and the cross-sectional area $(A)$ . So correct option is (3).

PHXII03:CURRENT ELECTRICITY

357033 A copper wire of length 1 $m$ and uniform cross sectional area $5 \times 10^{- 7} m^{2}$ carries a current of 1 $A$ . Assuming that there are $8 \times 10^{28}$ free electrons per $m^{3}$ of copper, how long will an electron take to drift from one end of the wire to the other ?

1

0.8 \times 10^{3} s

2

1.6 \times 10^{3} s

3

3.2 \times 10^{3} s

4

6.4 \times 10^{3} s

Explanation:

Length of the wire $= l = 1 m$
Cross section area of the wire
$= A = 5 \times 10^{- 7} m^{2}$
Number density of electrons in the wire
$= n = 8 \times 10^{28} p e r m^{3}$
Charge of an electron $= e = 1.6 \times 10^{- 19} C$
Drift velocity of electrons in the wire $= v_{d}$
Current in the wire $= I = n e v_{d} A = 1 A$
$\Rightarrow v_{d} = \frac{I}{n e A}$
$∴$ Time taken by an electron to drift from one end of the wire to the other end $= t = \frac{l}{v_{d}} = \frac{l n e A}{I}$
$\Rightarrow t =$
$\frac{1 \times 8 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 7}}{1} = 6.4 \times 10^{3} s$

KCET - 2021

PHXII03:CURRENT ELECTRICITY

357034 An electric cell of e.m.f. $ε$ is connected across a copper wire of diameter $d$ and length $l$ . The drift velocity of electrons in the wire is $v_{d}$ . If the length of the wire is changed to 2 $l$ , the new drift velocity of electrons in the copper wire will be

1

2 v_{d}

2

v_{d} / 4

3

v_{d} / 2

4

v_{d}

Explanation:

$v_{d} = \frac{i}{n e A} = \frac{ε}{R \times n e A} = \frac{ε}{ρ \times l \times n \times e}$
$v_{d}^{'} = \frac{ε}{ρ \times 2 l \times n \times e} \Rightarrow \frac{v_{d}^{'}}{v_{d}} = \frac{1}{2}$

PHXII03:CURRENT ELECTRICITY

357030 The relaxation time in conductors

1 Decrease with the increase of temperature

2 Increase with the increase of temperature

3 It does not depend on temperature

4 All of sudden changes at 400

K

PHXII03:CURRENT ELECTRICITY

357031 A steady current is set up in a metallic wire of non-uniform cross-section. How is the rate of flow of free electrons related to the area of cross-section $A$ ?

1

K

is independent of

A

2

K \propto A^{- 1}

3

K \propto A

4

K \propto A^{2}

Explanation:

As, $i =$ Anev $_{d}$
$\Rightarrow \frac{d q}{d t} =$ Anev $_{d} (∵ i = \frac{d q}{d t})$
$∴ \frac{d q}{d t} \propto A \Rightarrow K \propto A$
So, correct option is (3)

AIIMS - 2009

PHXII03:CURRENT ELECTRICITY

357032 Consider a copper wire of length $L$ , cross-section area $A$ . It has n number of free electrons per unit volume. Which of the following is the correct expression of drift velocity of the electrons when the wire carries a steady current I?

1

\frac{I}{n^{2} e L}

2

\frac{I}{n e L}

3

\frac{I}{n e A}

4

\frac{I}{n e^{2} L A}

Explanation:

$I$ is the current, $n$ is the number of free electrons per unit volume, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $V_{d} = \frac{I}{n e A}$ ,
The denominator represents the product of the number of free electrons per unit volume $(n)$ , the charge of an electron $(e)$ , and the cross-sectional area $(A)$ . So correct option is (3).

PHXII03:CURRENT ELECTRICITY

357033 A copper wire of length 1 $m$ and uniform cross sectional area $5 \times 10^{- 7} m^{2}$ carries a current of 1 $A$ . Assuming that there are $8 \times 10^{28}$ free electrons per $m^{3}$ of copper, how long will an electron take to drift from one end of the wire to the other ?

1

0.8 \times 10^{3} s

2

1.6 \times 10^{3} s

3

3.2 \times 10^{3} s

4

6.4 \times 10^{3} s

Explanation:

Length of the wire $= l = 1 m$
Cross section area of the wire
$= A = 5 \times 10^{- 7} m^{2}$
Number density of electrons in the wire
$= n = 8 \times 10^{28} p e r m^{3}$
Charge of an electron $= e = 1.6 \times 10^{- 19} C$
Drift velocity of electrons in the wire $= v_{d}$
Current in the wire $= I = n e v_{d} A = 1 A$
$\Rightarrow v_{d} = \frac{I}{n e A}$
$∴$ Time taken by an electron to drift from one end of the wire to the other end $= t = \frac{l}{v_{d}} = \frac{l n e A}{I}$
$\Rightarrow t =$
$\frac{1 \times 8 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 7}}{1} = 6.4 \times 10^{3} s$

KCET - 2021

PHXII03:CURRENT ELECTRICITY

357034 An electric cell of e.m.f. $ε$ is connected across a copper wire of diameter $d$ and length $l$ . The drift velocity of electrons in the wire is $v_{d}$ . If the length of the wire is changed to 2 $l$ , the new drift velocity of electrons in the copper wire will be

1

2 v_{d}

2

v_{d} / 4

3

v_{d} / 2

4

v_{d}

Explanation:

$v_{d} = \frac{i}{n e A} = \frac{ε}{R \times n e A} = \frac{ε}{ρ \times l \times n \times e}$
$v_{d}^{'} = \frac{ε}{ρ \times 2 l \times n \times e} \Rightarrow \frac{v_{d}^{'}}{v_{d}} = \frac{1}{2}$