QBManager

Current Electricity

152166 Two wires of equal diameters of resistivities $ρ_{1}$ and $ρ_{2}$ and lengths $x_{1}$ and $x_{2}$ respectively are joined in series. The equivalent resistivity of the combination is

1

\frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} + x_{2}}

2

\frac{ρ_{1} x_{2} - ρ_{2} x_{1}}{x_{1} - x_{2}}

3

\frac{ρ_{1} x_{2} + ρ_{2} x_{1}}{x_{1} + x_{2}}

4

\frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} - x_{2}}

Explanation:

A Given,
First wire resistivity $= ρ_{1}$
Second wire resistivity $= ρ_{2}$
Length of first wire $= x_{1}$
Length of second wire $= x_{2}$
Diameter of both wire are equal
So, $A_{1} = A_{2} = A$
We know that,
$R = ρ \frac{l}{A}$
For first wire -
$R_{1} = \frac{ρ_{1} x_{1}}{A}$
For second wire -
$R_{2} = \frac{ρ_{2} x_{2}}{A}$
Since, both resistance are joined in series, Hence,
$R_{eq} = R_{1} + R_{2}$
$R_{eq} = \frac{1}{A} (ρ_{1} x_{1} + ρ_{2} x_{2})$
$ρ_{eq} \cdot \frac{(x_{1} + x_{2})}{A} = \frac{1}{A} (ρ_{1} x_{1} + ρ_{2} x_{2})$
$ρ_{eq} = \frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} + x_{2}}$

EAMCET-2000

Current Electricity

152167 A nichrome wire $50 cm$ long and $1 {mm}^{2}$ crosssection carries a current of 4 A. When connected to $2 V$ battery. The resistivity of nichrome wire in $Ω - m$ is

1

1 \times 10^{- 6}

2

4 \times 10^{- 7}

3

3 \times 10^{- 7}

4

2 \times 10^{- 7}

Explanation:

A Given,
Length of wire $(l) = 50 cm = 0.5 m$
Area of wire $(A) = 1 {mm}^{2} = 1 \times 10^{- 6} m^{2}$
Voltage $(V) = 2 V$
Current $(I) = 4 A$
We know that,
$V = IR$
$R = \frac{V}{I} = \frac{2}{4}$
$R = 0.5 Ω$
$R = ρ \frac{l}{A}$
$ρ = \frac{AR}{l}$
$ρ = \frac{1 \times 10^{- 6} \times 0.5}{0.5}$
$ρ = 1 \times 10^{- 6} Ω m .$
$∵ R = ρ \frac{l}{A}$

EAMCET-2001

Current Electricity

152168 n conducting wires of same dimensions but having resistivities $1, 2, 3$ , connected in series. The equivalent resistivity of the combination is

1

\frac{n (n + 1)}{2}

2

\frac{n + 1}{2}

3

\frac{n + 1}{2 n}

4

\frac{2 n}{n + 1}

Explanation:

A Given,
Number of wire $= n$
Resistivity are given as,
$ρ_{1} = 1, ρ_{2} = 2, ρ_{3} = 3, ρ_{4} = 4 \dots \dots \dots . ρ_{n} = n$
We know that,
$R = \frac{ρ l}{A}$
$∴ R_{1} = \frac{ρ_{l} l}{A}$
$R_{2} = \frac{ρ_{1} l}{A}, R_{3} = \frac{ρ_{3} l}{A}, R_{n} = \frac{ρ_{n} l}{A}$
Since, all resistance are joined in series.
$R_{eq} = R_{1} + R_{2} + R_{3} + \dots \dots \dots \dots + R_{n}$
$ρ_{eq} \frac{l}{A} = \frac{l}{A} (ρ_{1} + ρ_{2} + ρ_{3} + \dots \dots \dots ρ_{n})$
$ρ_{eq} = (1 + 2 + 3 + 4 + \dots \dots \dots \dots + n)$
$ρ_{eq} = \frac{n (n + 1)}{2}$

EAMCET -2000

Current Electricity

152169 An aluminum (resistivity $ρ = 2.2 \times 10^{- 8} Ω - m$ ) wire of a diameter $1.4 mm$ is used to make a $4 Ω$ resistor. The length of the wire is

1

220 m

2

1000 m

3

280 m

4

1 m

Explanation:

C Given,
Resistivity $(ρ) = 2.2 \times 10^{- 8} Ω m$
Diameter $(d) = 1.4 mm = 1.4 \times 10^{- 3} m$
Resistance $(R) = 4 Ω$
We know that,
$R = \frac{ρ l}{A}$
$l = \frac{RA}{ρ}$
$l = \frac{4 \times \frac{π}{4} \times d^{2}}{2.2 \times 10^{- 8}}$
$l = \frac{π {(1.4 \times 10^{- 3})}^{2}}{2.2 \times 10^{- 8}}$
$l = \frac{π \times 1.96 \times 10^{- 6}}{2.2 \times 10^{- 8}}$
$l = 280 m$

EAMCET-2007

Current Electricity

152166 Two wires of equal diameters of resistivities $ρ_{1}$ and $ρ_{2}$ and lengths $x_{1}$ and $x_{2}$ respectively are joined in series. The equivalent resistivity of the combination is

1

\frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} + x_{2}}

2

\frac{ρ_{1} x_{2} - ρ_{2} x_{1}}{x_{1} - x_{2}}

3

\frac{ρ_{1} x_{2} + ρ_{2} x_{1}}{x_{1} + x_{2}}

4

\frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} - x_{2}}

Explanation:

A Given,
First wire resistivity $= ρ_{1}$
Second wire resistivity $= ρ_{2}$
Length of first wire $= x_{1}$
Length of second wire $= x_{2}$
Diameter of both wire are equal
So, $A_{1} = A_{2} = A$
We know that,
$R = ρ \frac{l}{A}$
For first wire -
$R_{1} = \frac{ρ_{1} x_{1}}{A}$
For second wire -
$R_{2} = \frac{ρ_{2} x_{2}}{A}$
Since, both resistance are joined in series, Hence,
$R_{eq} = R_{1} + R_{2}$
$R_{eq} = \frac{1}{A} (ρ_{1} x_{1} + ρ_{2} x_{2})$
$ρ_{eq} \cdot \frac{(x_{1} + x_{2})}{A} = \frac{1}{A} (ρ_{1} x_{1} + ρ_{2} x_{2})$
$ρ_{eq} = \frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} + x_{2}}$

EAMCET-2000

Current Electricity

152167 A nichrome wire $50 cm$ long and $1 {mm}^{2}$ crosssection carries a current of 4 A. When connected to $2 V$ battery. The resistivity of nichrome wire in $Ω - m$ is

1

1 \times 10^{- 6}

2

4 \times 10^{- 7}

3

3 \times 10^{- 7}

4

2 \times 10^{- 7}

Explanation:

A Given,
Length of wire $(l) = 50 cm = 0.5 m$
Area of wire $(A) = 1 {mm}^{2} = 1 \times 10^{- 6} m^{2}$
Voltage $(V) = 2 V$
Current $(I) = 4 A$
We know that,
$V = IR$
$R = \frac{V}{I} = \frac{2}{4}$
$R = 0.5 Ω$
$R = ρ \frac{l}{A}$
$ρ = \frac{AR}{l}$
$ρ = \frac{1 \times 10^{- 6} \times 0.5}{0.5}$
$ρ = 1 \times 10^{- 6} Ω m .$
$∵ R = ρ \frac{l}{A}$

EAMCET-2001

Current Electricity

152168 n conducting wires of same dimensions but having resistivities $1, 2, 3$ , connected in series. The equivalent resistivity of the combination is

1

\frac{n (n + 1)}{2}

2

\frac{n + 1}{2}

3

\frac{n + 1}{2 n}

4

\frac{2 n}{n + 1}

Explanation:

A Given,
Number of wire $= n$
Resistivity are given as,
$ρ_{1} = 1, ρ_{2} = 2, ρ_{3} = 3, ρ_{4} = 4 \dots \dots \dots . ρ_{n} = n$
We know that,
$R = \frac{ρ l}{A}$
$∴ R_{1} = \frac{ρ_{l} l}{A}$
$R_{2} = \frac{ρ_{1} l}{A}, R_{3} = \frac{ρ_{3} l}{A}, R_{n} = \frac{ρ_{n} l}{A}$
Since, all resistance are joined in series.
$R_{eq} = R_{1} + R_{2} + R_{3} + \dots \dots \dots \dots + R_{n}$
$ρ_{eq} \frac{l}{A} = \frac{l}{A} (ρ_{1} + ρ_{2} + ρ_{3} + \dots \dots \dots ρ_{n})$
$ρ_{eq} = (1 + 2 + 3 + 4 + \dots \dots \dots \dots + n)$
$ρ_{eq} = \frac{n (n + 1)}{2}$

EAMCET -2000

Current Electricity

152169 An aluminum (resistivity $ρ = 2.2 \times 10^{- 8} Ω - m$ ) wire of a diameter $1.4 mm$ is used to make a $4 Ω$ resistor. The length of the wire is

1

220 m

2

1000 m

3

280 m

4

1 m

Explanation:

C Given,
Resistivity $(ρ) = 2.2 \times 10^{- 8} Ω m$
Diameter $(d) = 1.4 mm = 1.4 \times 10^{- 3} m$
Resistance $(R) = 4 Ω$
We know that,
$R = \frac{ρ l}{A}$
$l = \frac{RA}{ρ}$
$l = \frac{4 \times \frac{π}{4} \times d^{2}}{2.2 \times 10^{- 8}}$
$l = \frac{π {(1.4 \times 10^{- 3})}^{2}}{2.2 \times 10^{- 8}}$
$l = \frac{π \times 1.96 \times 10^{- 6}}{2.2 \times 10^{- 8}}$
$l = 280 m$

EAMCET-2007

Current Electricity

152166 Two wires of equal diameters of resistivities $ρ_{1}$ and $ρ_{2}$ and lengths $x_{1}$ and $x_{2}$ respectively are joined in series. The equivalent resistivity of the combination is

1

\frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} + x_{2}}

2

\frac{ρ_{1} x_{2} - ρ_{2} x_{1}}{x_{1} - x_{2}}

3

\frac{ρ_{1} x_{2} + ρ_{2} x_{1}}{x_{1} + x_{2}}

4

\frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} - x_{2}}

Explanation:

A Given,
First wire resistivity $= ρ_{1}$
Second wire resistivity $= ρ_{2}$
Length of first wire $= x_{1}$
Length of second wire $= x_{2}$
Diameter of both wire are equal
So, $A_{1} = A_{2} = A$
We know that,
$R = ρ \frac{l}{A}$
For first wire -
$R_{1} = \frac{ρ_{1} x_{1}}{A}$
For second wire -
$R_{2} = \frac{ρ_{2} x_{2}}{A}$
Since, both resistance are joined in series, Hence,
$R_{eq} = R_{1} + R_{2}$
$R_{eq} = \frac{1}{A} (ρ_{1} x_{1} + ρ_{2} x_{2})$
$ρ_{eq} \cdot \frac{(x_{1} + x_{2})}{A} = \frac{1}{A} (ρ_{1} x_{1} + ρ_{2} x_{2})$
$ρ_{eq} = \frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} + x_{2}}$

EAMCET-2000

Current Electricity

152167 A nichrome wire $50 cm$ long and $1 {mm}^{2}$ crosssection carries a current of 4 A. When connected to $2 V$ battery. The resistivity of nichrome wire in $Ω - m$ is

1

1 \times 10^{- 6}

2

4 \times 10^{- 7}

3

3 \times 10^{- 7}

4

2 \times 10^{- 7}

Explanation:

A Given,
Length of wire $(l) = 50 cm = 0.5 m$
Area of wire $(A) = 1 {mm}^{2} = 1 \times 10^{- 6} m^{2}$
Voltage $(V) = 2 V$
Current $(I) = 4 A$
We know that,
$V = IR$
$R = \frac{V}{I} = \frac{2}{4}$
$R = 0.5 Ω$
$R = ρ \frac{l}{A}$
$ρ = \frac{AR}{l}$
$ρ = \frac{1 \times 10^{- 6} \times 0.5}{0.5}$
$ρ = 1 \times 10^{- 6} Ω m .$
$∵ R = ρ \frac{l}{A}$

EAMCET-2001

Current Electricity

152168 n conducting wires of same dimensions but having resistivities $1, 2, 3$ , connected in series. The equivalent resistivity of the combination is

1

\frac{n (n + 1)}{2}

2

\frac{n + 1}{2}

3

\frac{n + 1}{2 n}

4

\frac{2 n}{n + 1}

Explanation:

A Given,
Number of wire $= n$
Resistivity are given as,
$ρ_{1} = 1, ρ_{2} = 2, ρ_{3} = 3, ρ_{4} = 4 \dots \dots \dots . ρ_{n} = n$
We know that,
$R = \frac{ρ l}{A}$
$∴ R_{1} = \frac{ρ_{l} l}{A}$
$R_{2} = \frac{ρ_{1} l}{A}, R_{3} = \frac{ρ_{3} l}{A}, R_{n} = \frac{ρ_{n} l}{A}$
Since, all resistance are joined in series.
$R_{eq} = R_{1} + R_{2} + R_{3} + \dots \dots \dots \dots + R_{n}$
$ρ_{eq} \frac{l}{A} = \frac{l}{A} (ρ_{1} + ρ_{2} + ρ_{3} + \dots \dots \dots ρ_{n})$
$ρ_{eq} = (1 + 2 + 3 + 4 + \dots \dots \dots \dots + n)$
$ρ_{eq} = \frac{n (n + 1)}{2}$

EAMCET -2000

Current Electricity

152169 An aluminum (resistivity $ρ = 2.2 \times 10^{- 8} Ω - m$ ) wire of a diameter $1.4 mm$ is used to make a $4 Ω$ resistor. The length of the wire is

1

220 m

2

1000 m

3

280 m

4

1 m

Explanation:

C Given,
Resistivity $(ρ) = 2.2 \times 10^{- 8} Ω m$
Diameter $(d) = 1.4 mm = 1.4 \times 10^{- 3} m$
Resistance $(R) = 4 Ω$
We know that,
$R = \frac{ρ l}{A}$
$l = \frac{RA}{ρ}$
$l = \frac{4 \times \frac{π}{4} \times d^{2}}{2.2 \times 10^{- 8}}$
$l = \frac{π {(1.4 \times 10^{- 3})}^{2}}{2.2 \times 10^{- 8}}$
$l = \frac{π \times 1.96 \times 10^{- 6}}{2.2 \times 10^{- 8}}$
$l = 280 m$

EAMCET-2007

Current Electricity

152166 Two wires of equal diameters of resistivities $ρ_{1}$ and $ρ_{2}$ and lengths $x_{1}$ and $x_{2}$ respectively are joined in series. The equivalent resistivity of the combination is

1

\frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} + x_{2}}

2

\frac{ρ_{1} x_{2} - ρ_{2} x_{1}}{x_{1} - x_{2}}

3

\frac{ρ_{1} x_{2} + ρ_{2} x_{1}}{x_{1} + x_{2}}

4

\frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} - x_{2}}

Explanation:

A Given,
First wire resistivity $= ρ_{1}$
Second wire resistivity $= ρ_{2}$
Length of first wire $= x_{1}$
Length of second wire $= x_{2}$
Diameter of both wire are equal
So, $A_{1} = A_{2} = A$
We know that,
$R = ρ \frac{l}{A}$
For first wire -
$R_{1} = \frac{ρ_{1} x_{1}}{A}$
For second wire -
$R_{2} = \frac{ρ_{2} x_{2}}{A}$
Since, both resistance are joined in series, Hence,
$R_{eq} = R_{1} + R_{2}$
$R_{eq} = \frac{1}{A} (ρ_{1} x_{1} + ρ_{2} x_{2})$
$ρ_{eq} \cdot \frac{(x_{1} + x_{2})}{A} = \frac{1}{A} (ρ_{1} x_{1} + ρ_{2} x_{2})$
$ρ_{eq} = \frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} + x_{2}}$

EAMCET-2000

Current Electricity

152167 A nichrome wire $50 cm$ long and $1 {mm}^{2}$ crosssection carries a current of 4 A. When connected to $2 V$ battery. The resistivity of nichrome wire in $Ω - m$ is

1

1 \times 10^{- 6}

2

4 \times 10^{- 7}

3

3 \times 10^{- 7}

4

2 \times 10^{- 7}

Explanation:

A Given,
Length of wire $(l) = 50 cm = 0.5 m$
Area of wire $(A) = 1 {mm}^{2} = 1 \times 10^{- 6} m^{2}$
Voltage $(V) = 2 V$
Current $(I) = 4 A$
We know that,
$V = IR$
$R = \frac{V}{I} = \frac{2}{4}$
$R = 0.5 Ω$
$R = ρ \frac{l}{A}$
$ρ = \frac{AR}{l}$
$ρ = \frac{1 \times 10^{- 6} \times 0.5}{0.5}$
$ρ = 1 \times 10^{- 6} Ω m .$
$∵ R = ρ \frac{l}{A}$

EAMCET-2001

Current Electricity

152168 n conducting wires of same dimensions but having resistivities $1, 2, 3$ , connected in series. The equivalent resistivity of the combination is

1

\frac{n (n + 1)}{2}

2

\frac{n + 1}{2}

3

\frac{n + 1}{2 n}

4

\frac{2 n}{n + 1}

Explanation:

A Given,
Number of wire $= n$
Resistivity are given as,
$ρ_{1} = 1, ρ_{2} = 2, ρ_{3} = 3, ρ_{4} = 4 \dots \dots \dots . ρ_{n} = n$
We know that,
$R = \frac{ρ l}{A}$
$∴ R_{1} = \frac{ρ_{l} l}{A}$
$R_{2} = \frac{ρ_{1} l}{A}, R_{3} = \frac{ρ_{3} l}{A}, R_{n} = \frac{ρ_{n} l}{A}$
Since, all resistance are joined in series.
$R_{eq} = R_{1} + R_{2} + R_{3} + \dots \dots \dots \dots + R_{n}$
$ρ_{eq} \frac{l}{A} = \frac{l}{A} (ρ_{1} + ρ_{2} + ρ_{3} + \dots \dots \dots ρ_{n})$
$ρ_{eq} = (1 + 2 + 3 + 4 + \dots \dots \dots \dots + n)$
$ρ_{eq} = \frac{n (n + 1)}{2}$

EAMCET -2000

Current Electricity

152169 An aluminum (resistivity $ρ = 2.2 \times 10^{- 8} Ω - m$ ) wire of a diameter $1.4 mm$ is used to make a $4 Ω$ resistor. The length of the wire is

1

220 m

2

1000 m

3

280 m

4

1 m

Explanation:

C Given,
Resistivity $(ρ) = 2.2 \times 10^{- 8} Ω m$
Diameter $(d) = 1.4 mm = 1.4 \times 10^{- 3} m$
Resistance $(R) = 4 Ω$
We know that,
$R = \frac{ρ l}{A}$
$l = \frac{RA}{ρ}$
$l = \frac{4 \times \frac{π}{4} \times d^{2}}{2.2 \times 10^{- 8}}$
$l = \frac{π {(1.4 \times 10^{- 3})}^{2}}{2.2 \times 10^{- 8}}$
$l = \frac{π \times 1.96 \times 10^{- 6}}{2.2 \times 10^{- 8}}$
$l = 280 m$

EAMCET-2007