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PHXII03:CURRENT ELECTRICITY

357311 A potentiometer is connected across $A$ and $B$ and a balance is obtained at $64.0 c m$ . When potentiometer lead to $B$ is moved to $C$ , a balance is found at $8.0 c m$ . If the potentiometer is now connected across $B$ and $C$ , a balance will be found at
supporting img

1

8.0 c m

2

56.0 c m

3

64.0 c m

4

72.0 c m

Explanation:

supporting img
$V_{A B} = ε_{1} \propto 64$
$V_{A C} = (ε_{1} - ε_{2}) \propto 8$
$V_{B C} = ε_{2} \propto l$
$∴ 64 - l = 8$
or $l = 64 - 8 = 56 c m$

PHXII03:CURRENT ELECTRICITY

357312 In a potentiometer experiment two cells of e.m.f.’s $E_{1}$ and $E_{2}$ are used in series and the balancing length is found to be 58 $c m$ of the wire. If the polarity of $E_{2}$ is reversed, then the balancing length becomes 29 $c m$ . The ratio $\frac{E_{1}}{E_{2}}$ of the e.m.f. of the two cells is $(E_{1} > E_{2})$

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

Let $K$ be the potential gradient of the rod.
Given that
$E_{1} + E_{2} = (K) 58 c m$
$E_{1} - E_{2} = (K) 29 c m$
By dividing the equations
$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = 2 \Rightarrow \frac{E_{1}}{E_{2}} = 3$

PHXII03:CURRENT ELECTRICITY

357313 A potentiometer of one metre length, an unknown, e.m.f voltage source is balanced at 60 $c m$ length of potentiometer wire, while a 3 volt battery is balanced at 45 $c m$ length. Then the e.m.f. of the unknown voltage source is

1

4 V

2

4.5 V

3

3 V

4

2.25 V

Explanation:

$l_{1} = 60 c m; E_{2} = 3 V, l_{2} = 45 c m$
In balanced condition
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow \frac{E_{1}}{3} = \frac{60}{45} \Rightarrow E_{1} = 4 V .$

PHXII03:CURRENT ELECTRICITY

357314 In a potentiometer the balance length with standard cadmium cell is 509 $c m$ . The emf of a cell which when connected in place of the standard cell gave a balance length of 750 $c m$ is (emf of standard cell is 1.018 $V$ )

1

1.5 V

2

0.5 V

3

1.08 V

4

1.2 V

Explanation:

Given that $ε_{1} = 1.018, l_{1} = 509 c m, l_{2} = 750 c m$
$\frac{ε_{1}}{ε_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow ε_{2} = 1.5 V$

PHXII03:CURRENT ELECTRICITY

357315 Figure shows a 200 $c m$ potentiometer wire $A B$ with resistance $15 Ω$ . Find the balancing length for a 3 $V$ cell.
supporting img

1

20 c m

2

10 c m

3

40 c m

4

60 c m

Explanation:

Current in primary circuit of above potentiometer, $i = \frac{20}{20} = 1 A$
Potential difference across wire $A B, V_{A B} = i R_{A B} = 15 V$
Potential gradient on wire, $λ = \frac{V_{A B}}{L} = \frac{15}{200} = 0.075 V / c m$
If $l_{l}$ is the balancing length then, $3 = λ l_{1} \Rightarrow l_{1} = \frac{3}{0.075} = 40 c m$

PHXII03:CURRENT ELECTRICITY

357311 A potentiometer is connected across $A$ and $B$ and a balance is obtained at $64.0 c m$ . When potentiometer lead to $B$ is moved to $C$ , a balance is found at $8.0 c m$ . If the potentiometer is now connected across $B$ and $C$ , a balance will be found at
supporting img

1

8.0 c m

2

56.0 c m

3

64.0 c m

4

72.0 c m

Explanation:

supporting img
$V_{A B} = ε_{1} \propto 64$
$V_{A C} = (ε_{1} - ε_{2}) \propto 8$
$V_{B C} = ε_{2} \propto l$
$∴ 64 - l = 8$
or $l = 64 - 8 = 56 c m$

PHXII03:CURRENT ELECTRICITY

357312 In a potentiometer experiment two cells of e.m.f.’s $E_{1}$ and $E_{2}$ are used in series and the balancing length is found to be 58 $c m$ of the wire. If the polarity of $E_{2}$ is reversed, then the balancing length becomes 29 $c m$ . The ratio $\frac{E_{1}}{E_{2}}$ of the e.m.f. of the two cells is $(E_{1} > E_{2})$

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

Let $K$ be the potential gradient of the rod.
Given that
$E_{1} + E_{2} = (K) 58 c m$
$E_{1} - E_{2} = (K) 29 c m$
By dividing the equations
$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = 2 \Rightarrow \frac{E_{1}}{E_{2}} = 3$

PHXII03:CURRENT ELECTRICITY

357313 A potentiometer of one metre length, an unknown, e.m.f voltage source is balanced at 60 $c m$ length of potentiometer wire, while a 3 volt battery is balanced at 45 $c m$ length. Then the e.m.f. of the unknown voltage source is

1

4 V

2

4.5 V

3

3 V

4

2.25 V

Explanation:

$l_{1} = 60 c m; E_{2} = 3 V, l_{2} = 45 c m$
In balanced condition
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow \frac{E_{1}}{3} = \frac{60}{45} \Rightarrow E_{1} = 4 V .$

PHXII03:CURRENT ELECTRICITY

357314 In a potentiometer the balance length with standard cadmium cell is 509 $c m$ . The emf of a cell which when connected in place of the standard cell gave a balance length of 750 $c m$ is (emf of standard cell is 1.018 $V$ )

1

1.5 V

2

0.5 V

3

1.08 V

4

1.2 V

Explanation:

Given that $ε_{1} = 1.018, l_{1} = 509 c m, l_{2} = 750 c m$
$\frac{ε_{1}}{ε_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow ε_{2} = 1.5 V$

PHXII03:CURRENT ELECTRICITY

357315 Figure shows a 200 $c m$ potentiometer wire $A B$ with resistance $15 Ω$ . Find the balancing length for a 3 $V$ cell.
supporting img

1

20 c m

2

10 c m

3

40 c m

4

60 c m

Explanation:

Current in primary circuit of above potentiometer, $i = \frac{20}{20} = 1 A$
Potential difference across wire $A B, V_{A B} = i R_{A B} = 15 V$
Potential gradient on wire, $λ = \frac{V_{A B}}{L} = \frac{15}{200} = 0.075 V / c m$
If $l_{l}$ is the balancing length then, $3 = λ l_{1} \Rightarrow l_{1} = \frac{3}{0.075} = 40 c m$

PHXII03:CURRENT ELECTRICITY

357311 A potentiometer is connected across $A$ and $B$ and a balance is obtained at $64.0 c m$ . When potentiometer lead to $B$ is moved to $C$ , a balance is found at $8.0 c m$ . If the potentiometer is now connected across $B$ and $C$ , a balance will be found at
supporting img

1

8.0 c m

2

56.0 c m

3

64.0 c m

4

72.0 c m

Explanation:

supporting img
$V_{A B} = ε_{1} \propto 64$
$V_{A C} = (ε_{1} - ε_{2}) \propto 8$
$V_{B C} = ε_{2} \propto l$
$∴ 64 - l = 8$
or $l = 64 - 8 = 56 c m$

PHXII03:CURRENT ELECTRICITY

357312 In a potentiometer experiment two cells of e.m.f.’s $E_{1}$ and $E_{2}$ are used in series and the balancing length is found to be 58 $c m$ of the wire. If the polarity of $E_{2}$ is reversed, then the balancing length becomes 29 $c m$ . The ratio $\frac{E_{1}}{E_{2}}$ of the e.m.f. of the two cells is $(E_{1} > E_{2})$

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

Let $K$ be the potential gradient of the rod.
Given that
$E_{1} + E_{2} = (K) 58 c m$
$E_{1} - E_{2} = (K) 29 c m$
By dividing the equations
$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = 2 \Rightarrow \frac{E_{1}}{E_{2}} = 3$

PHXII03:CURRENT ELECTRICITY

357313 A potentiometer of one metre length, an unknown, e.m.f voltage source is balanced at 60 $c m$ length of potentiometer wire, while a 3 volt battery is balanced at 45 $c m$ length. Then the e.m.f. of the unknown voltage source is

1

4 V

2

4.5 V

3

3 V

4

2.25 V

Explanation:

$l_{1} = 60 c m; E_{2} = 3 V, l_{2} = 45 c m$
In balanced condition
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow \frac{E_{1}}{3} = \frac{60}{45} \Rightarrow E_{1} = 4 V .$

PHXII03:CURRENT ELECTRICITY

357314 In a potentiometer the balance length with standard cadmium cell is 509 $c m$ . The emf of a cell which when connected in place of the standard cell gave a balance length of 750 $c m$ is (emf of standard cell is 1.018 $V$ )

1

1.5 V

2

0.5 V

3

1.08 V

4

1.2 V

Explanation:

Given that $ε_{1} = 1.018, l_{1} = 509 c m, l_{2} = 750 c m$
$\frac{ε_{1}}{ε_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow ε_{2} = 1.5 V$

PHXII03:CURRENT ELECTRICITY

357315 Figure shows a 200 $c m$ potentiometer wire $A B$ with resistance $15 Ω$ . Find the balancing length for a 3 $V$ cell.
supporting img

1

20 c m

2

10 c m

3

40 c m

4

60 c m

Explanation:

Current in primary circuit of above potentiometer, $i = \frac{20}{20} = 1 A$
Potential difference across wire $A B, V_{A B} = i R_{A B} = 15 V$
Potential gradient on wire, $λ = \frac{V_{A B}}{L} = \frac{15}{200} = 0.075 V / c m$
If $l_{l}$ is the balancing length then, $3 = λ l_{1} \Rightarrow l_{1} = \frac{3}{0.075} = 40 c m$

PHXII03:CURRENT ELECTRICITY

357311 A potentiometer is connected across $A$ and $B$ and a balance is obtained at $64.0 c m$ . When potentiometer lead to $B$ is moved to $C$ , a balance is found at $8.0 c m$ . If the potentiometer is now connected across $B$ and $C$ , a balance will be found at
supporting img

1

8.0 c m

2

56.0 c m

3

64.0 c m

4

72.0 c m

Explanation:

supporting img
$V_{A B} = ε_{1} \propto 64$
$V_{A C} = (ε_{1} - ε_{2}) \propto 8$
$V_{B C} = ε_{2} \propto l$
$∴ 64 - l = 8$
or $l = 64 - 8 = 56 c m$

PHXII03:CURRENT ELECTRICITY

357312 In a potentiometer experiment two cells of e.m.f.’s $E_{1}$ and $E_{2}$ are used in series and the balancing length is found to be 58 $c m$ of the wire. If the polarity of $E_{2}$ is reversed, then the balancing length becomes 29 $c m$ . The ratio $\frac{E_{1}}{E_{2}}$ of the e.m.f. of the two cells is $(E_{1} > E_{2})$

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

Let $K$ be the potential gradient of the rod.
Given that
$E_{1} + E_{2} = (K) 58 c m$
$E_{1} - E_{2} = (K) 29 c m$
By dividing the equations
$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = 2 \Rightarrow \frac{E_{1}}{E_{2}} = 3$

PHXII03:CURRENT ELECTRICITY

357313 A potentiometer of one metre length, an unknown, e.m.f voltage source is balanced at 60 $c m$ length of potentiometer wire, while a 3 volt battery is balanced at 45 $c m$ length. Then the e.m.f. of the unknown voltage source is

1

4 V

2

4.5 V

3

3 V

4

2.25 V

Explanation:

$l_{1} = 60 c m; E_{2} = 3 V, l_{2} = 45 c m$
In balanced condition
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow \frac{E_{1}}{3} = \frac{60}{45} \Rightarrow E_{1} = 4 V .$

PHXII03:CURRENT ELECTRICITY

357314 In a potentiometer the balance length with standard cadmium cell is 509 $c m$ . The emf of a cell which when connected in place of the standard cell gave a balance length of 750 $c m$ is (emf of standard cell is 1.018 $V$ )

1

1.5 V

2

0.5 V

3

1.08 V

4

1.2 V

Explanation:

Given that $ε_{1} = 1.018, l_{1} = 509 c m, l_{2} = 750 c m$
$\frac{ε_{1}}{ε_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow ε_{2} = 1.5 V$

PHXII03:CURRENT ELECTRICITY

357315 Figure shows a 200 $c m$ potentiometer wire $A B$ with resistance $15 Ω$ . Find the balancing length for a 3 $V$ cell.
supporting img

1

20 c m

2

10 c m

3

40 c m

4

60 c m

Explanation:

Current in primary circuit of above potentiometer, $i = \frac{20}{20} = 1 A$
Potential difference across wire $A B, V_{A B} = i R_{A B} = 15 V$
Potential gradient on wire, $λ = \frac{V_{A B}}{L} = \frac{15}{200} = 0.075 V / c m$
If $l_{l}$ is the balancing length then, $3 = λ l_{1} \Rightarrow l_{1} = \frac{3}{0.075} = 40 c m$

PHXII03:CURRENT ELECTRICITY

357311 A potentiometer is connected across $A$ and $B$ and a balance is obtained at $64.0 c m$ . When potentiometer lead to $B$ is moved to $C$ , a balance is found at $8.0 c m$ . If the potentiometer is now connected across $B$ and $C$ , a balance will be found at
supporting img

1

8.0 c m

2

56.0 c m

3

64.0 c m

4

72.0 c m

Explanation:

supporting img
$V_{A B} = ε_{1} \propto 64$
$V_{A C} = (ε_{1} - ε_{2}) \propto 8$
$V_{B C} = ε_{2} \propto l$
$∴ 64 - l = 8$
or $l = 64 - 8 = 56 c m$

PHXII03:CURRENT ELECTRICITY

357312 In a potentiometer experiment two cells of e.m.f.’s $E_{1}$ and $E_{2}$ are used in series and the balancing length is found to be 58 $c m$ of the wire. If the polarity of $E_{2}$ is reversed, then the balancing length becomes 29 $c m$ . The ratio $\frac{E_{1}}{E_{2}}$ of the e.m.f. of the two cells is $(E_{1} > E_{2})$

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

Let $K$ be the potential gradient of the rod.
Given that
$E_{1} + E_{2} = (K) 58 c m$
$E_{1} - E_{2} = (K) 29 c m$
By dividing the equations
$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = 2 \Rightarrow \frac{E_{1}}{E_{2}} = 3$

PHXII03:CURRENT ELECTRICITY

357313 A potentiometer of one metre length, an unknown, e.m.f voltage source is balanced at 60 $c m$ length of potentiometer wire, while a 3 volt battery is balanced at 45 $c m$ length. Then the e.m.f. of the unknown voltage source is

1

4 V

2

4.5 V

3

3 V

4

2.25 V

Explanation:

$l_{1} = 60 c m; E_{2} = 3 V, l_{2} = 45 c m$
In balanced condition
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow \frac{E_{1}}{3} = \frac{60}{45} \Rightarrow E_{1} = 4 V .$

PHXII03:CURRENT ELECTRICITY

357314 In a potentiometer the balance length with standard cadmium cell is 509 $c m$ . The emf of a cell which when connected in place of the standard cell gave a balance length of 750 $c m$ is (emf of standard cell is 1.018 $V$ )

1

1.5 V

2

0.5 V

3

1.08 V

4

1.2 V

Explanation:

Given that $ε_{1} = 1.018, l_{1} = 509 c m, l_{2} = 750 c m$
$\frac{ε_{1}}{ε_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow ε_{2} = 1.5 V$

PHXII03:CURRENT ELECTRICITY

357315 Figure shows a 200 $c m$ potentiometer wire $A B$ with resistance $15 Ω$ . Find the balancing length for a 3 $V$ cell.
supporting img

1

20 c m

2

10 c m

3

40 c m

4

60 c m

Explanation:

Current in primary circuit of above potentiometer, $i = \frac{20}{20} = 1 A$
Potential difference across wire $A B, V_{A B} = i R_{A B} = 15 V$
Potential gradient on wire, $λ = \frac{V_{A B}}{L} = \frac{15}{200} = 0.075 V / c m$
If $l_{l}$ is the balancing length then, $3 = λ l_{1} \Rightarrow l_{1} = \frac{3}{0.075} = 40 c m$