QBManager

Current Electricity

151743 Consider a horizontal sheet with charge density $5 \times 10^{- 6} C / m^{2}$ . A particle of mass $8 \times 10^{- 6} g$ is dropped from a certain height onto this sheet. The number of electrons that should be removed from this particle so that it stays close to the sheet without falling on it is $($ Assume $g = 10 m / s^{2}$ and $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2})$

1

\frac{2}{9 π} \times 10^{8}

2

\frac{2 π}{9} \times 10^{8}

3

\frac{π}{5} \times 10^{8}

4

\frac{1}{18 π} \times 10^{8}

Explanation:

D
Given that,
Charge density $(σ) = 5 \times 10^{- 6} C / m^{2}$
$mass (m) = 8 \times 10^{- 6} g = 8 \times 10^{- 9} kg$
For the particle do not fall down the electrostatic force on it should balance its weight.
$F = mg$
$qE = mg$
$q (\frac{σ}{2 ε_{0}}) = mg$
$q = \frac{2 mg ε_{0}}{σ}$
$q = \frac{2 \times 8 \times 10^{- 9} \times 10 \times 1}{5 \times 10^{- 6} \times 4 π \times 9 \times 10^{9}}$
No. of electron $(n) = \frac{q}{e}$
$= \frac{2 \times 8 \times 10^{- 9} \times 10}{5 \times 10^{- 6} \times 4 π \times 9 \times 10^{9} \times 1.6 \times 10^{- 19}} = \frac{1}{18 π} \times 10^{8}$

TS EAMCET 08.05.2019

Current Electricity

151744 Though the electron drift velocity is small and electron charge is very small, a conductor can carry an appreciably large current because:

1 Electron number density is very large

2 Drift velocity of electron is very large

3 Electron number density depends on temperature

4 Relaxation time is small

Explanation:

A The current in a metallic conductor is given by $I = {neAv}_{d}$
Where $n =$ number of charge carrier
$e = Charge of electron$
$I = Current flow$
$V_{d} = Drift velocity$
The conductors are abundant in the concentration of charge (electron) carrier; a conductor has a large carrier density $≃ 10^{28} m^{- 3}$ .

Karnataka CET-2019

Current Electricity

151746 A current of 10 amp is passing through a metallic wire of cross sectional area $4 \times 10^{- 6} m^{2}$ . If the density of the aluminum conductor is $2.7$ $gm / cc$ considering aluminum gives 1 electrons per atom for conduction find the drift speed of the electrons if molecular weight of aluminum is $27 g m$ .

1

1.6 \times 10^{- 4} m / s

2

3.6 \times 10^{- 4} m / s

3

2.6 \times 10^{- 4} m / s

4

1.5 \times 10^{- 4} m / s

Explanation:

C Given, $I = 10 amp, m_{w} = 27 \times 10^{- 3} kg$
$A = 4 \times 10^{- 6} m^{2}, ρ = 2.7 \times 10^{3} kg / m^{3}$
Number of electron $(n) = \frac{N_{A} ρ}{m_{w}}$
$= \frac{6 \times 10^{23} \times 2.7 \times 10^{3}}{27 \times 10^{- 3}} = 6 \times 10^{28}$
Drift velocity $(v_{d}) = \frac{I}{nAe}$
$= \frac{10}{6 \times 10^{28} \times 4 \times 10^{- 6} \times 1.6 \times 10^{- 19}}$
$= 2.6 \times 10^{- 4} m / s$

AIIMS-25.05.2019(M) Shift-1

Current Electricity

151748 Estimate the magnitude of current that passes through a wire, if $0.1 mol$ of electrons flow through it in $40 \min$ . (Assume, Avogadro's number $= 6.0 \times 10^{23}$ )

1

4 A

2

9 A

3

12 A

4

14 A

Explanation:

A Given,
Number of electrons $= 0.1 mol$
Time $= 40 \min = 60 \times 40 = 2400 \sec$
Avogadro's number $= 6.0 \times 10^{23}$
The magnitude of current,
$I = \frac{Number of electron \times electron charge}{Time}$
$I = \frac{0.1 \times 6 \times 10^{23} \times 1.6 \times 10^{- 19}}{2400}$
$I = \frac{96}{24}$
$I = 4 A$

TS- EAMCET-03.05.2019

Current Electricity

151743 Consider a horizontal sheet with charge density $5 \times 10^{- 6} C / m^{2}$ . A particle of mass $8 \times 10^{- 6} g$ is dropped from a certain height onto this sheet. The number of electrons that should be removed from this particle so that it stays close to the sheet without falling on it is $($ Assume $g = 10 m / s^{2}$ and $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2})$

1

\frac{2}{9 π} \times 10^{8}

2

\frac{2 π}{9} \times 10^{8}

3

\frac{π}{5} \times 10^{8}

4

\frac{1}{18 π} \times 10^{8}

Explanation:

D
Given that,
Charge density $(σ) = 5 \times 10^{- 6} C / m^{2}$
$mass (m) = 8 \times 10^{- 6} g = 8 \times 10^{- 9} kg$
For the particle do not fall down the electrostatic force on it should balance its weight.
$F = mg$
$qE = mg$
$q (\frac{σ}{2 ε_{0}}) = mg$
$q = \frac{2 mg ε_{0}}{σ}$
$q = \frac{2 \times 8 \times 10^{- 9} \times 10 \times 1}{5 \times 10^{- 6} \times 4 π \times 9 \times 10^{9}}$
No. of electron $(n) = \frac{q}{e}$
$= \frac{2 \times 8 \times 10^{- 9} \times 10}{5 \times 10^{- 6} \times 4 π \times 9 \times 10^{9} \times 1.6 \times 10^{- 19}} = \frac{1}{18 π} \times 10^{8}$

TS EAMCET 08.05.2019

Current Electricity

151744 Though the electron drift velocity is small and electron charge is very small, a conductor can carry an appreciably large current because:

1 Electron number density is very large

2 Drift velocity of electron is very large

3 Electron number density depends on temperature

4 Relaxation time is small

Explanation:

A The current in a metallic conductor is given by $I = {neAv}_{d}$
Where $n =$ number of charge carrier
$e = Charge of electron$
$I = Current flow$
$V_{d} = Drift velocity$
The conductors are abundant in the concentration of charge (electron) carrier; a conductor has a large carrier density $≃ 10^{28} m^{- 3}$ .

Karnataka CET-2019

Current Electricity

151746 A current of 10 amp is passing through a metallic wire of cross sectional area $4 \times 10^{- 6} m^{2}$ . If the density of the aluminum conductor is $2.7$ $gm / cc$ considering aluminum gives 1 electrons per atom for conduction find the drift speed of the electrons if molecular weight of aluminum is $27 g m$ .

1

1.6 \times 10^{- 4} m / s

2

3.6 \times 10^{- 4} m / s

3

2.6 \times 10^{- 4} m / s

4

1.5 \times 10^{- 4} m / s

Explanation:

C Given, $I = 10 amp, m_{w} = 27 \times 10^{- 3} kg$
$A = 4 \times 10^{- 6} m^{2}, ρ = 2.7 \times 10^{3} kg / m^{3}$
Number of electron $(n) = \frac{N_{A} ρ}{m_{w}}$
$= \frac{6 \times 10^{23} \times 2.7 \times 10^{3}}{27 \times 10^{- 3}} = 6 \times 10^{28}$
Drift velocity $(v_{d}) = \frac{I}{nAe}$
$= \frac{10}{6 \times 10^{28} \times 4 \times 10^{- 6} \times 1.6 \times 10^{- 19}}$
$= 2.6 \times 10^{- 4} m / s$

AIIMS-25.05.2019(M) Shift-1

Current Electricity

151748 Estimate the magnitude of current that passes through a wire, if $0.1 mol$ of electrons flow through it in $40 \min$ . (Assume, Avogadro's number $= 6.0 \times 10^{23}$ )

1

4 A

2

9 A

3

12 A

4

14 A

Explanation:

A Given,
Number of electrons $= 0.1 mol$
Time $= 40 \min = 60 \times 40 = 2400 \sec$
Avogadro's number $= 6.0 \times 10^{23}$
The magnitude of current,
$I = \frac{Number of electron \times electron charge}{Time}$
$I = \frac{0.1 \times 6 \times 10^{23} \times 1.6 \times 10^{- 19}}{2400}$
$I = \frac{96}{24}$
$I = 4 A$

TS- EAMCET-03.05.2019

Current Electricity

151743 Consider a horizontal sheet with charge density $5 \times 10^{- 6} C / m^{2}$ . A particle of mass $8 \times 10^{- 6} g$ is dropped from a certain height onto this sheet. The number of electrons that should be removed from this particle so that it stays close to the sheet without falling on it is $($ Assume $g = 10 m / s^{2}$ and $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2})$

1

\frac{2}{9 π} \times 10^{8}

2

\frac{2 π}{9} \times 10^{8}

3

\frac{π}{5} \times 10^{8}

4

\frac{1}{18 π} \times 10^{8}

Explanation:

D
Given that,
Charge density $(σ) = 5 \times 10^{- 6} C / m^{2}$
$mass (m) = 8 \times 10^{- 6} g = 8 \times 10^{- 9} kg$
For the particle do not fall down the electrostatic force on it should balance its weight.
$F = mg$
$qE = mg$
$q (\frac{σ}{2 ε_{0}}) = mg$
$q = \frac{2 mg ε_{0}}{σ}$
$q = \frac{2 \times 8 \times 10^{- 9} \times 10 \times 1}{5 \times 10^{- 6} \times 4 π \times 9 \times 10^{9}}$
No. of electron $(n) = \frac{q}{e}$
$= \frac{2 \times 8 \times 10^{- 9} \times 10}{5 \times 10^{- 6} \times 4 π \times 9 \times 10^{9} \times 1.6 \times 10^{- 19}} = \frac{1}{18 π} \times 10^{8}$

TS EAMCET 08.05.2019

Current Electricity

151744 Though the electron drift velocity is small and electron charge is very small, a conductor can carry an appreciably large current because:

1 Electron number density is very large

2 Drift velocity of electron is very large

3 Electron number density depends on temperature

4 Relaxation time is small

Explanation:

A The current in a metallic conductor is given by $I = {neAv}_{d}$
Where $n =$ number of charge carrier
$e = Charge of electron$
$I = Current flow$
$V_{d} = Drift velocity$
The conductors are abundant in the concentration of charge (electron) carrier; a conductor has a large carrier density $≃ 10^{28} m^{- 3}$ .

Karnataka CET-2019

Current Electricity

151746 A current of 10 amp is passing through a metallic wire of cross sectional area $4 \times 10^{- 6} m^{2}$ . If the density of the aluminum conductor is $2.7$ $gm / cc$ considering aluminum gives 1 electrons per atom for conduction find the drift speed of the electrons if molecular weight of aluminum is $27 g m$ .

1

1.6 \times 10^{- 4} m / s

2

3.6 \times 10^{- 4} m / s

3

2.6 \times 10^{- 4} m / s

4

1.5 \times 10^{- 4} m / s

Explanation:

C Given, $I = 10 amp, m_{w} = 27 \times 10^{- 3} kg$
$A = 4 \times 10^{- 6} m^{2}, ρ = 2.7 \times 10^{3} kg / m^{3}$
Number of electron $(n) = \frac{N_{A} ρ}{m_{w}}$
$= \frac{6 \times 10^{23} \times 2.7 \times 10^{3}}{27 \times 10^{- 3}} = 6 \times 10^{28}$
Drift velocity $(v_{d}) = \frac{I}{nAe}$
$= \frac{10}{6 \times 10^{28} \times 4 \times 10^{- 6} \times 1.6 \times 10^{- 19}}$
$= 2.6 \times 10^{- 4} m / s$

AIIMS-25.05.2019(M) Shift-1

Current Electricity

151748 Estimate the magnitude of current that passes through a wire, if $0.1 mol$ of electrons flow through it in $40 \min$ . (Assume, Avogadro's number $= 6.0 \times 10^{23}$ )

1

4 A

2

9 A

3

12 A

4

14 A

Explanation:

A Given,
Number of electrons $= 0.1 mol$
Time $= 40 \min = 60 \times 40 = 2400 \sec$
Avogadro's number $= 6.0 \times 10^{23}$
The magnitude of current,
$I = \frac{Number of electron \times electron charge}{Time}$
$I = \frac{0.1 \times 6 \times 10^{23} \times 1.6 \times 10^{- 19}}{2400}$
$I = \frac{96}{24}$
$I = 4 A$

TS- EAMCET-03.05.2019

Current Electricity

151743 Consider a horizontal sheet with charge density $5 \times 10^{- 6} C / m^{2}$ . A particle of mass $8 \times 10^{- 6} g$ is dropped from a certain height onto this sheet. The number of electrons that should be removed from this particle so that it stays close to the sheet without falling on it is $($ Assume $g = 10 m / s^{2}$ and $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2})$

1

\frac{2}{9 π} \times 10^{8}

2

\frac{2 π}{9} \times 10^{8}

3

\frac{π}{5} \times 10^{8}

4

\frac{1}{18 π} \times 10^{8}

Explanation:

D
Given that,
Charge density $(σ) = 5 \times 10^{- 6} C / m^{2}$
$mass (m) = 8 \times 10^{- 6} g = 8 \times 10^{- 9} kg$
For the particle do not fall down the electrostatic force on it should balance its weight.
$F = mg$
$qE = mg$
$q (\frac{σ}{2 ε_{0}}) = mg$
$q = \frac{2 mg ε_{0}}{σ}$
$q = \frac{2 \times 8 \times 10^{- 9} \times 10 \times 1}{5 \times 10^{- 6} \times 4 π \times 9 \times 10^{9}}$
No. of electron $(n) = \frac{q}{e}$
$= \frac{2 \times 8 \times 10^{- 9} \times 10}{5 \times 10^{- 6} \times 4 π \times 9 \times 10^{9} \times 1.6 \times 10^{- 19}} = \frac{1}{18 π} \times 10^{8}$

TS EAMCET 08.05.2019

Current Electricity

151744 Though the electron drift velocity is small and electron charge is very small, a conductor can carry an appreciably large current because:

1 Electron number density is very large

2 Drift velocity of electron is very large

3 Electron number density depends on temperature

4 Relaxation time is small

Explanation:

A The current in a metallic conductor is given by $I = {neAv}_{d}$
Where $n =$ number of charge carrier
$e = Charge of electron$
$I = Current flow$
$V_{d} = Drift velocity$
The conductors are abundant in the concentration of charge (electron) carrier; a conductor has a large carrier density $≃ 10^{28} m^{- 3}$ .

Karnataka CET-2019

Current Electricity

151746 A current of 10 amp is passing through a metallic wire of cross sectional area $4 \times 10^{- 6} m^{2}$ . If the density of the aluminum conductor is $2.7$ $gm / cc$ considering aluminum gives 1 electrons per atom for conduction find the drift speed of the electrons if molecular weight of aluminum is $27 g m$ .

1

1.6 \times 10^{- 4} m / s

2

3.6 \times 10^{- 4} m / s

3

2.6 \times 10^{- 4} m / s

4

1.5 \times 10^{- 4} m / s

Explanation:

C Given, $I = 10 amp, m_{w} = 27 \times 10^{- 3} kg$
$A = 4 \times 10^{- 6} m^{2}, ρ = 2.7 \times 10^{3} kg / m^{3}$
Number of electron $(n) = \frac{N_{A} ρ}{m_{w}}$
$= \frac{6 \times 10^{23} \times 2.7 \times 10^{3}}{27 \times 10^{- 3}} = 6 \times 10^{28}$
Drift velocity $(v_{d}) = \frac{I}{nAe}$
$= \frac{10}{6 \times 10^{28} \times 4 \times 10^{- 6} \times 1.6 \times 10^{- 19}}$
$= 2.6 \times 10^{- 4} m / s$

AIIMS-25.05.2019(M) Shift-1

Current Electricity

151748 Estimate the magnitude of current that passes through a wire, if $0.1 mol$ of electrons flow through it in $40 \min$ . (Assume, Avogadro's number $= 6.0 \times 10^{23}$ )

1

4 A

2

9 A

3

12 A

4

14 A

Explanation:

A Given,
Number of electrons $= 0.1 mol$
Time $= 40 \min = 60 \times 40 = 2400 \sec$
Avogadro's number $= 6.0 \times 10^{23}$
The magnitude of current,
$I = \frac{Number of electron \times electron charge}{Time}$
$I = \frac{0.1 \times 6 \times 10^{23} \times 1.6 \times 10^{- 19}}{2400}$
$I = \frac{96}{24}$
$I = 4 A$

TS- EAMCET-03.05.2019

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series