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PHXII03:CURRENT ELECTRICITY

357035 The electron drift speed is small and the charge of the electron is also small but still, we obtain large current in a conductor. This is due to

1 the conducting property of the conductor

2 the resistance of the conductor is small

3 the electron number density of the conductor is small

4 the electron number density of the conductor is enormous.

Explanation:

It is because electron number density, i.e. number of electrons per unit volume of a conductor is very large $(\approx 10^{28} m^{- 3})$

KCET - 2006

PHXII03:CURRENT ELECTRICITY

357036 Two cylindrical rods of uniform cross-section area $A$ and $2 A$ , having free electrons per unit volume $2 n$ and $n$ , respectively, are joined in series. A current $I$ flows through them in steady state. The ratio of drift velocity of free electron in left rod to drift velocity of electron in the right $rod, (\frac{v_{L}}{v_{R}})$ , is
supporting img

1

\frac{1}{2}

2 1

3 2

4 4

Explanation:

Since current $I = n e A v_{d}$ through the both the rods is same
$2 (n) e A v_{L} = n e (2 A) v_{R}$
$\Rightarrow \frac{v_{L}}{v_{R}} = 1$

PHXII03:CURRENT ELECTRICITY

357037 The drift velocity of electrons for a conductor connected in an electrical circuit is $V_{d}$ . The conductor is now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be

1

2 V_{d}

2

\frac{V_{d}}{4}

3

\frac{V_{d}}{2}

4

V_{d}

Explanation:

Drift velocity of electron, $v_{d} = \frac{- e E τ}{m}$
As $E = \frac{V}{l}$
We can write, $v_{d} = \frac{- e V τ}{m l}$
where, $l =$ length of conductor,
$V$ is applied voltage
As drift velocity does not depends upon area, so it will remain same, even after changing area of conductor. So, correct option is (4).

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357038 For a given electric current the drift velocity of conduction electrons in a copper wire is $v_{d}$ and their mobility is $μ$ . When the current is increased at constant temperature

1

v_{d}

decreases,

μ

remains the same

2

v_{d}

remains the same,

μ

decreases

3

v_{d}

increases,

μ

remains the same

4

v_{d}

remains the same,

μ

increases

Explanation:

$I = n q A v_{d}$
$\Rightarrow$ As $I$ increases, $v_{d}$ also increases.
But $μ = \frac{v_{d}}{E}; J = \frac{I}{A}$ and $J \propto E$ .
As both $v_{d}$ and $E$ change, $μ$ remains same.
Correct option is (3).

KECT - 2023

PHXII03:CURRENT ELECTRICITY

357035 The electron drift speed is small and the charge of the electron is also small but still, we obtain large current in a conductor. This is due to

1 the conducting property of the conductor

2 the resistance of the conductor is small

3 the electron number density of the conductor is small

4 the electron number density of the conductor is enormous.

Explanation:

It is because electron number density, i.e. number of electrons per unit volume of a conductor is very large $(\approx 10^{28} m^{- 3})$

KCET - 2006

PHXII03:CURRENT ELECTRICITY

357036 Two cylindrical rods of uniform cross-section area $A$ and $2 A$ , having free electrons per unit volume $2 n$ and $n$ , respectively, are joined in series. A current $I$ flows through them in steady state. The ratio of drift velocity of free electron in left rod to drift velocity of electron in the right $rod, (\frac{v_{L}}{v_{R}})$ , is
supporting img

1

\frac{1}{2}

2 1

3 2

4 4

Explanation:

Since current $I = n e A v_{d}$ through the both the rods is same
$2 (n) e A v_{L} = n e (2 A) v_{R}$
$\Rightarrow \frac{v_{L}}{v_{R}} = 1$

PHXII03:CURRENT ELECTRICITY

357037 The drift velocity of electrons for a conductor connected in an electrical circuit is $V_{d}$ . The conductor is now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be

1

2 V_{d}

2

\frac{V_{d}}{4}

3

\frac{V_{d}}{2}

4

V_{d}

Explanation:

Drift velocity of electron, $v_{d} = \frac{- e E τ}{m}$
As $E = \frac{V}{l}$
We can write, $v_{d} = \frac{- e V τ}{m l}$
where, $l =$ length of conductor,
$V$ is applied voltage
As drift velocity does not depends upon area, so it will remain same, even after changing area of conductor. So, correct option is (4).

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357038 For a given electric current the drift velocity of conduction electrons in a copper wire is $v_{d}$ and their mobility is $μ$ . When the current is increased at constant temperature

1

v_{d}

decreases,

μ

remains the same

2

v_{d}

remains the same,

μ

decreases

3

v_{d}

increases,

μ

remains the same

4

v_{d}

remains the same,

μ

increases

Explanation:

$I = n q A v_{d}$
$\Rightarrow$ As $I$ increases, $v_{d}$ also increases.
But $μ = \frac{v_{d}}{E}; J = \frac{I}{A}$ and $J \propto E$ .
As both $v_{d}$ and $E$ change, $μ$ remains same.
Correct option is (3).

KECT - 2023

PHXII03:CURRENT ELECTRICITY

357035 The electron drift speed is small and the charge of the electron is also small but still, we obtain large current in a conductor. This is due to

1 the conducting property of the conductor

2 the resistance of the conductor is small

3 the electron number density of the conductor is small

4 the electron number density of the conductor is enormous.

Explanation:

It is because electron number density, i.e. number of electrons per unit volume of a conductor is very large $(\approx 10^{28} m^{- 3})$

KCET - 2006

PHXII03:CURRENT ELECTRICITY

357036 Two cylindrical rods of uniform cross-section area $A$ and $2 A$ , having free electrons per unit volume $2 n$ and $n$ , respectively, are joined in series. A current $I$ flows through them in steady state. The ratio of drift velocity of free electron in left rod to drift velocity of electron in the right $rod, (\frac{v_{L}}{v_{R}})$ , is
supporting img

1

\frac{1}{2}

2 1

3 2

4 4

Explanation:

Since current $I = n e A v_{d}$ through the both the rods is same
$2 (n) e A v_{L} = n e (2 A) v_{R}$
$\Rightarrow \frac{v_{L}}{v_{R}} = 1$

PHXII03:CURRENT ELECTRICITY

357037 The drift velocity of electrons for a conductor connected in an electrical circuit is $V_{d}$ . The conductor is now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be

1

2 V_{d}

2

\frac{V_{d}}{4}

3

\frac{V_{d}}{2}

4

V_{d}

Explanation:

Drift velocity of electron, $v_{d} = \frac{- e E τ}{m}$
As $E = \frac{V}{l}$
We can write, $v_{d} = \frac{- e V τ}{m l}$
where, $l =$ length of conductor,
$V$ is applied voltage
As drift velocity does not depends upon area, so it will remain same, even after changing area of conductor. So, correct option is (4).

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357038 For a given electric current the drift velocity of conduction electrons in a copper wire is $v_{d}$ and their mobility is $μ$ . When the current is increased at constant temperature

1

v_{d}

decreases,

μ

remains the same

2

v_{d}

remains the same,

μ

decreases

3

v_{d}

increases,

μ

remains the same

4

v_{d}

remains the same,

μ

increases

Explanation:

$I = n q A v_{d}$
$\Rightarrow$ As $I$ increases, $v_{d}$ also increases.
But $μ = \frac{v_{d}}{E}; J = \frac{I}{A}$ and $J \propto E$ .
As both $v_{d}$ and $E$ change, $μ$ remains same.
Correct option is (3).

KECT - 2023

PHXII03:CURRENT ELECTRICITY

357035 The electron drift speed is small and the charge of the electron is also small but still, we obtain large current in a conductor. This is due to

1 the conducting property of the conductor

2 the resistance of the conductor is small

3 the electron number density of the conductor is small

4 the electron number density of the conductor is enormous.

Explanation:

It is because electron number density, i.e. number of electrons per unit volume of a conductor is very large $(\approx 10^{28} m^{- 3})$

KCET - 2006

PHXII03:CURRENT ELECTRICITY

357036 Two cylindrical rods of uniform cross-section area $A$ and $2 A$ , having free electrons per unit volume $2 n$ and $n$ , respectively, are joined in series. A current $I$ flows through them in steady state. The ratio of drift velocity of free electron in left rod to drift velocity of electron in the right $rod, (\frac{v_{L}}{v_{R}})$ , is
supporting img

1

\frac{1}{2}

2 1

3 2

4 4

Explanation:

Since current $I = n e A v_{d}$ through the both the rods is same
$2 (n) e A v_{L} = n e (2 A) v_{R}$
$\Rightarrow \frac{v_{L}}{v_{R}} = 1$

PHXII03:CURRENT ELECTRICITY

357037 The drift velocity of electrons for a conductor connected in an electrical circuit is $V_{d}$ . The conductor is now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be

1

2 V_{d}

2

\frac{V_{d}}{4}

3

\frac{V_{d}}{2}

4

V_{d}

Explanation:

Drift velocity of electron, $v_{d} = \frac{- e E τ}{m}$
As $E = \frac{V}{l}$
We can write, $v_{d} = \frac{- e V τ}{m l}$
where, $l =$ length of conductor,
$V$ is applied voltage
As drift velocity does not depends upon area, so it will remain same, even after changing area of conductor. So, correct option is (4).

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357038 For a given electric current the drift velocity of conduction electrons in a copper wire is $v_{d}$ and their mobility is $μ$ . When the current is increased at constant temperature

1

v_{d}

decreases,

μ

remains the same

2

v_{d}

remains the same,

μ

decreases

3

v_{d}

increases,

μ

remains the same

4

v_{d}

remains the same,

μ

increases

Explanation:

$I = n q A v_{d}$
$\Rightarrow$ As $I$ increases, $v_{d}$ also increases.
But $μ = \frac{v_{d}}{E}; J = \frac{I}{A}$ and $J \propto E$ .
As both $v_{d}$ and $E$ change, $μ$ remains same.
Correct option is (3).

KECT - 2023

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