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PHXII03:CURRENT ELECTRICITY

357048 The figure shows three conductors I, II and III of same material, different lengths, $l, 2 l$ and $3 l$ and of different areas of cross-section 3 $A$ , $A$ and 2 $A$ respectively. Arrange them in the increasing order of current drawn from battery.
supporting img

1

i_{3} < i_{2} < i_{1}

2

i_{1} < i_{2} < i_{3}

3

i_{2} < i_{3} < i_{1}

4

i_{2} < i_{1} < i_{3}

Explanation:

Let $ε$ is the $e m f$ of the battery.
$i = \frac{ε}{R} = \frac{ε A}{ρ l}$
$i_{1} = \frac{ε (3 A)}{ρ l}, i_{2} = \frac{ε A}{ρ 2 l}, i_{3} = \frac{ε 2 A}{ρ 3 l}$
$i_{1} > i_{3} > i_{2}$
Correct option is (3)

PHXII03:CURRENT ELECTRICITY

357049 A conductor with rectangular cross section has dimension $(a \times 2 a \times 4 a)$ as shown in figure. Resistance across $A B$ is $x$ , across $C D$ is $y$ and across $E F$ is $z$ . Then
supporting img

1

x = y = z

2

x > y > z

3

y > z > x

4

x > z > y

Explanation:

$x = \frac{ρ \times 4 a}{a \times 2 a} = 2 \frac{ρ}{a}$
$\begin{aligned} y = \frac{ρ \times a}{4 a \times 2 a} = \frac{1}{8} \frac{ρ}{a} \\ z = \frac{ρ \times 2 a}{4 a \times a} = \frac{1}{2} \frac{ρ}{a} \end{aligned}$
so, $x > z > y$

PHXII03:CURRENT ELECTRICITY

357050 A metal wire of circular cross-section has a resistance $R_{1}$ . The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes $R_{2}$ then $R_{2} : R_{1}$ is

1

1 : 4

2

1 : 1

3

4 : 1

4

10 : 2

Explanation:

$R_{1} = ρ \frac{l_{1}}{A_{1}}$
As volume remains same
$l_{1} A_{1} = l_{2} A_{2} = (2 l_{1}) A_{2}$
$A_{2} = \frac{A_{1}}{2} \Rightarrow R_{2} = ρ \frac{(2 l_{1})}{A_{1} / 2} = 4 ρ \frac{l_{1}}{A_{1}} = 4 R$

PHXII03:CURRENT ELECTRICITY

357051 Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter 1 $m m$ . Conductor $B$ is a hollow tube of outer diameter 2 $m m$ and inner diameter 1 $m m$ . What is the ratio of resistance $R_{A} t o R_{B}$ ?

1

3 : 2

2

3 : 1

3

1 : 3

4

2 : 3

Explanation:

We have $R_{A} = \frac{ρ l}{π {(10^{- 3} \times 0.5)}^{2}}$
$R_{B} = \frac{ρ l}{π [(10^{- 3})^{2} - {(0.5 \times 10^{- 3})}^{2}]}$
$\Rightarrow \frac{R_{A}}{R_{B}} = \frac{{(10^{- 3})}^{2} - {(0.5 \times 10^{- 3})}^{2}}{{(10^{- 3} \times 0.5)}^{2}} = 3$

PHXII03:CURRENT ELECTRICITY

357048 The figure shows three conductors I, II and III of same material, different lengths, $l, 2 l$ and $3 l$ and of different areas of cross-section 3 $A$ , $A$ and 2 $A$ respectively. Arrange them in the increasing order of current drawn from battery.
supporting img

1

i_{3} < i_{2} < i_{1}

2

i_{1} < i_{2} < i_{3}

3

i_{2} < i_{3} < i_{1}

4

i_{2} < i_{1} < i_{3}

Explanation:

Let $ε$ is the $e m f$ of the battery.
$i = \frac{ε}{R} = \frac{ε A}{ρ l}$
$i_{1} = \frac{ε (3 A)}{ρ l}, i_{2} = \frac{ε A}{ρ 2 l}, i_{3} = \frac{ε 2 A}{ρ 3 l}$
$i_{1} > i_{3} > i_{2}$
Correct option is (3)

PHXII03:CURRENT ELECTRICITY

357049 A conductor with rectangular cross section has dimension $(a \times 2 a \times 4 a)$ as shown in figure. Resistance across $A B$ is $x$ , across $C D$ is $y$ and across $E F$ is $z$ . Then
supporting img

1

x = y = z

2

x > y > z

3

y > z > x

4

x > z > y

Explanation:

$x = \frac{ρ \times 4 a}{a \times 2 a} = 2 \frac{ρ}{a}$
$\begin{aligned} y = \frac{ρ \times a}{4 a \times 2 a} = \frac{1}{8} \frac{ρ}{a} \\ z = \frac{ρ \times 2 a}{4 a \times a} = \frac{1}{2} \frac{ρ}{a} \end{aligned}$
so, $x > z > y$

PHXII03:CURRENT ELECTRICITY

357050 A metal wire of circular cross-section has a resistance $R_{1}$ . The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes $R_{2}$ then $R_{2} : R_{1}$ is

1

1 : 4

2

1 : 1

3

4 : 1

4

10 : 2

Explanation:

$R_{1} = ρ \frac{l_{1}}{A_{1}}$
As volume remains same
$l_{1} A_{1} = l_{2} A_{2} = (2 l_{1}) A_{2}$
$A_{2} = \frac{A_{1}}{2} \Rightarrow R_{2} = ρ \frac{(2 l_{1})}{A_{1} / 2} = 4 ρ \frac{l_{1}}{A_{1}} = 4 R$

PHXII03:CURRENT ELECTRICITY

357051 Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter 1 $m m$ . Conductor $B$ is a hollow tube of outer diameter 2 $m m$ and inner diameter 1 $m m$ . What is the ratio of resistance $R_{A} t o R_{B}$ ?

1

3 : 2

2

3 : 1

3

1 : 3

4

2 : 3

Explanation:

We have $R_{A} = \frac{ρ l}{π {(10^{- 3} \times 0.5)}^{2}}$
$R_{B} = \frac{ρ l}{π [(10^{- 3})^{2} - {(0.5 \times 10^{- 3})}^{2}]}$
$\Rightarrow \frac{R_{A}}{R_{B}} = \frac{{(10^{- 3})}^{2} - {(0.5 \times 10^{- 3})}^{2}}{{(10^{- 3} \times 0.5)}^{2}} = 3$

PHXII03:CURRENT ELECTRICITY

357048 The figure shows three conductors I, II and III of same material, different lengths, $l, 2 l$ and $3 l$ and of different areas of cross-section 3 $A$ , $A$ and 2 $A$ respectively. Arrange them in the increasing order of current drawn from battery.
supporting img

1

i_{3} < i_{2} < i_{1}

2

i_{1} < i_{2} < i_{3}

3

i_{2} < i_{3} < i_{1}

4

i_{2} < i_{1} < i_{3}

Explanation:

Let $ε$ is the $e m f$ of the battery.
$i = \frac{ε}{R} = \frac{ε A}{ρ l}$
$i_{1} = \frac{ε (3 A)}{ρ l}, i_{2} = \frac{ε A}{ρ 2 l}, i_{3} = \frac{ε 2 A}{ρ 3 l}$
$i_{1} > i_{3} > i_{2}$
Correct option is (3)

PHXII03:CURRENT ELECTRICITY

357049 A conductor with rectangular cross section has dimension $(a \times 2 a \times 4 a)$ as shown in figure. Resistance across $A B$ is $x$ , across $C D$ is $y$ and across $E F$ is $z$ . Then
supporting img

1

x = y = z

2

x > y > z

3

y > z > x

4

x > z > y

Explanation:

$x = \frac{ρ \times 4 a}{a \times 2 a} = 2 \frac{ρ}{a}$
$\begin{aligned} y = \frac{ρ \times a}{4 a \times 2 a} = \frac{1}{8} \frac{ρ}{a} \\ z = \frac{ρ \times 2 a}{4 a \times a} = \frac{1}{2} \frac{ρ}{a} \end{aligned}$
so, $x > z > y$

PHXII03:CURRENT ELECTRICITY

357050 A metal wire of circular cross-section has a resistance $R_{1}$ . The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes $R_{2}$ then $R_{2} : R_{1}$ is

1

1 : 4

2

1 : 1

3

4 : 1

4

10 : 2

Explanation:

$R_{1} = ρ \frac{l_{1}}{A_{1}}$
As volume remains same
$l_{1} A_{1} = l_{2} A_{2} = (2 l_{1}) A_{2}$
$A_{2} = \frac{A_{1}}{2} \Rightarrow R_{2} = ρ \frac{(2 l_{1})}{A_{1} / 2} = 4 ρ \frac{l_{1}}{A_{1}} = 4 R$

PHXII03:CURRENT ELECTRICITY

357051 Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter 1 $m m$ . Conductor $B$ is a hollow tube of outer diameter 2 $m m$ and inner diameter 1 $m m$ . What is the ratio of resistance $R_{A} t o R_{B}$ ?

1

3 : 2

2

3 : 1

3

1 : 3

4

2 : 3

Explanation:

We have $R_{A} = \frac{ρ l}{π {(10^{- 3} \times 0.5)}^{2}}$
$R_{B} = \frac{ρ l}{π [(10^{- 3})^{2} - {(0.5 \times 10^{- 3})}^{2}]}$
$\Rightarrow \frac{R_{A}}{R_{B}} = \frac{{(10^{- 3})}^{2} - {(0.5 \times 10^{- 3})}^{2}}{{(10^{- 3} \times 0.5)}^{2}} = 3$

PHXII03:CURRENT ELECTRICITY

357048 The figure shows three conductors I, II and III of same material, different lengths, $l, 2 l$ and $3 l$ and of different areas of cross-section 3 $A$ , $A$ and 2 $A$ respectively. Arrange them in the increasing order of current drawn from battery.
supporting img

1

i_{3} < i_{2} < i_{1}

2

i_{1} < i_{2} < i_{3}

3

i_{2} < i_{3} < i_{1}

4

i_{2} < i_{1} < i_{3}

Explanation:

Let $ε$ is the $e m f$ of the battery.
$i = \frac{ε}{R} = \frac{ε A}{ρ l}$
$i_{1} = \frac{ε (3 A)}{ρ l}, i_{2} = \frac{ε A}{ρ 2 l}, i_{3} = \frac{ε 2 A}{ρ 3 l}$
$i_{1} > i_{3} > i_{2}$
Correct option is (3)

PHXII03:CURRENT ELECTRICITY

357049 A conductor with rectangular cross section has dimension $(a \times 2 a \times 4 a)$ as shown in figure. Resistance across $A B$ is $x$ , across $C D$ is $y$ and across $E F$ is $z$ . Then
supporting img

1

x = y = z

2

x > y > z

3

y > z > x

4

x > z > y

Explanation:

$x = \frac{ρ \times 4 a}{a \times 2 a} = 2 \frac{ρ}{a}$
$\begin{aligned} y = \frac{ρ \times a}{4 a \times 2 a} = \frac{1}{8} \frac{ρ}{a} \\ z = \frac{ρ \times 2 a}{4 a \times a} = \frac{1}{2} \frac{ρ}{a} \end{aligned}$
so, $x > z > y$

PHXII03:CURRENT ELECTRICITY

357050 A metal wire of circular cross-section has a resistance $R_{1}$ . The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes $R_{2}$ then $R_{2} : R_{1}$ is

1

1 : 4

2

1 : 1

3

4 : 1

4

10 : 2

Explanation:

$R_{1} = ρ \frac{l_{1}}{A_{1}}$
As volume remains same
$l_{1} A_{1} = l_{2} A_{2} = (2 l_{1}) A_{2}$
$A_{2} = \frac{A_{1}}{2} \Rightarrow R_{2} = ρ \frac{(2 l_{1})}{A_{1} / 2} = 4 ρ \frac{l_{1}}{A_{1}} = 4 R$

PHXII03:CURRENT ELECTRICITY

357051 Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter 1 $m m$ . Conductor $B$ is a hollow tube of outer diameter 2 $m m$ and inner diameter 1 $m m$ . What is the ratio of resistance $R_{A} t o R_{B}$ ?

1

3 : 2

2

3 : 1

3

1 : 3

4

2 : 3

Explanation:

We have $R_{A} = \frac{ρ l}{π {(10^{- 3} \times 0.5)}^{2}}$
$R_{B} = \frac{ρ l}{π [(10^{- 3})^{2} - {(0.5 \times 10^{- 3})}^{2}]}$
$\Rightarrow \frac{R_{A}}{R_{B}} = \frac{{(10^{- 3})}^{2} - {(0.5 \times 10^{- 3})}^{2}}{{(10^{- 3} \times 0.5)}^{2}} = 3$

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