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Current Electricity

268520 Three ammeters $P, Q$ and $R$ with internal resistances $r, 1.5 r$ ,3r respectively . $Q$ and $R$ parallel and this combination is in series with $P$ , The whole combination concted between $X$ and $Y$ . When the battery connected between $X$ and $Y$ , the ratio of the readings of $P, Q$ and $R$ is

1

2 : 1 : 1

2

3 : 2 : 1

3

3 : 1 : 2

4

1 : 1 : 1

Explanation:

$i = \frac{V}{R}$

Current Electricity

268521 The potential difference between the points
$A$ and $B$ is

1

1.50 V

2

2.50 V

3

1.00 V

4

0.50 V

Explanation:

For the first loop $12 = 5 i + i_{1}$
For thesecond loop $0 = 7 (i - i_{1}) - i_{1}$
or $8 i_{1} = 7 i$ or, $i_{1} = (7 / 8) i$
Therefore, we obtain $12 = 5 i + \frac{7}{8} i = \frac{47 i}{8}$
or, $i = \frac{12 \times 8}{47} A = 2.04 A$
$i_{1} = \frac{7}{8} \times 2.04 A = 1.79 A$
Thus, the p.d across $A$ and $B$ is
$V_{A} - V_{B} = (i - i_{1}) \times 2 = 0.25 \times 2 = 0.50 V$

Current Electricity

268522 The resistance of a semicircle shown in fig. betweenitstwo end faces is (G iven that radial thickness $= 3 cm$ , axial thickness $= 4 cm$ , inner radius $= 6 cm$ and resistivity $= 4 \times 10^{- 6} Ω cm$ )

1

24.15 \times 10^{- 6} Ω

2

7.85 \times 10^{- 7} Ω

3

7.85 \times 10^{- 6} Ω

4

7.85 \times 10^{- 5} Ω

Explanation:

Here, $A = 4 cm \times 3 cm = 12 {cm}^{2}$
$I = π r = π (6 + 3 / 2) = π \times 7.5 cm$
$R = \frac{ρ l}{A} = \frac{4 \times 10^{- 6} \times π \times 7.5}{12}$
$= 7.85 \times 10^{- 6}$ ohm

Current Electricity

268523 ABCD is a square where each side is a uniform wire of resistance12 . A point $E$ lies on $C D$ such that if a uniform wire of resistance $1 Ω$ is connected across AE and constant potential difference is applied across $A$ and $C$ , then $B$ and $E$ are equipotential .

1

\frac{C E}{E D} = 1

2

\frac{C E}{E D} = \frac{1}{\sqrt{2}}

3

\frac{C E}{E D} = \frac{1}{2}

4

\frac{C E}{E D} = \sqrt{2}

Explanation:

Equivalent resistance between $A$ and $E$ is
$y = \frac{(x + 1)}{x + 2}$
For $B$ and $E$ to be at equal potential, we get
$\frac{R_{A E}}{R_{A B}} = \frac{R_{E C}}{R_{B C}} \Rightarrow \frac{x + 1}{(x + 2) 1} = \frac{1 - x}{1}$
Solving $x = \sqrt{2} - 1$
Now $\frac{C E}{E D} = \frac{1 - x}{x} = \sqrt{2}$

Current Electricity

268520 Three ammeters $P, Q$ and $R$ with internal resistances $r, 1.5 r$ ,3r respectively . $Q$ and $R$ parallel and this combination is in series with $P$ , The whole combination concted between $X$ and $Y$ . When the battery connected between $X$ and $Y$ , the ratio of the readings of $P, Q$ and $R$ is

1

2 : 1 : 1

2

3 : 2 : 1

3

3 : 1 : 2

4

1 : 1 : 1

Explanation:

$i = \frac{V}{R}$

Current Electricity

268521 The potential difference between the points
$A$ and $B$ is

1

1.50 V

2

2.50 V

3

1.00 V

4

0.50 V

Explanation:

For the first loop $12 = 5 i + i_{1}$
For thesecond loop $0 = 7 (i - i_{1}) - i_{1}$
or $8 i_{1} = 7 i$ or, $i_{1} = (7 / 8) i$
Therefore, we obtain $12 = 5 i + \frac{7}{8} i = \frac{47 i}{8}$
or, $i = \frac{12 \times 8}{47} A = 2.04 A$
$i_{1} = \frac{7}{8} \times 2.04 A = 1.79 A$
Thus, the p.d across $A$ and $B$ is
$V_{A} - V_{B} = (i - i_{1}) \times 2 = 0.25 \times 2 = 0.50 V$

Current Electricity

268522 The resistance of a semicircle shown in fig. betweenitstwo end faces is (G iven that radial thickness $= 3 cm$ , axial thickness $= 4 cm$ , inner radius $= 6 cm$ and resistivity $= 4 \times 10^{- 6} Ω cm$ )

1

24.15 \times 10^{- 6} Ω

2

7.85 \times 10^{- 7} Ω

3

7.85 \times 10^{- 6} Ω

4

7.85 \times 10^{- 5} Ω

Explanation:

Here, $A = 4 cm \times 3 cm = 12 {cm}^{2}$
$I = π r = π (6 + 3 / 2) = π \times 7.5 cm$
$R = \frac{ρ l}{A} = \frac{4 \times 10^{- 6} \times π \times 7.5}{12}$
$= 7.85 \times 10^{- 6}$ ohm

Current Electricity

268523 ABCD is a square where each side is a uniform wire of resistance12 . A point $E$ lies on $C D$ such that if a uniform wire of resistance $1 Ω$ is connected across AE and constant potential difference is applied across $A$ and $C$ , then $B$ and $E$ are equipotential .

1

\frac{C E}{E D} = 1

2

\frac{C E}{E D} = \frac{1}{\sqrt{2}}

3

\frac{C E}{E D} = \frac{1}{2}

4

\frac{C E}{E D} = \sqrt{2}

Explanation:

Equivalent resistance between $A$ and $E$ is
$y = \frac{(x + 1)}{x + 2}$
For $B$ and $E$ to be at equal potential, we get
$\frac{R_{A E}}{R_{A B}} = \frac{R_{E C}}{R_{B C}} \Rightarrow \frac{x + 1}{(x + 2) 1} = \frac{1 - x}{1}$
Solving $x = \sqrt{2} - 1$
Now $\frac{C E}{E D} = \frac{1 - x}{x} = \sqrt{2}$

Current Electricity

268520 Three ammeters $P, Q$ and $R$ with internal resistances $r, 1.5 r$ ,3r respectively . $Q$ and $R$ parallel and this combination is in series with $P$ , The whole combination concted between $X$ and $Y$ . When the battery connected between $X$ and $Y$ , the ratio of the readings of $P, Q$ and $R$ is

1

2 : 1 : 1

2

3 : 2 : 1

3

3 : 1 : 2

4

1 : 1 : 1

Explanation:

$i = \frac{V}{R}$

Current Electricity

268521 The potential difference between the points
$A$ and $B$ is

1

1.50 V

2

2.50 V

3

1.00 V

4

0.50 V

Explanation:

For the first loop $12 = 5 i + i_{1}$
For thesecond loop $0 = 7 (i - i_{1}) - i_{1}$
or $8 i_{1} = 7 i$ or, $i_{1} = (7 / 8) i$
Therefore, we obtain $12 = 5 i + \frac{7}{8} i = \frac{47 i}{8}$
or, $i = \frac{12 \times 8}{47} A = 2.04 A$
$i_{1} = \frac{7}{8} \times 2.04 A = 1.79 A$
Thus, the p.d across $A$ and $B$ is
$V_{A} - V_{B} = (i - i_{1}) \times 2 = 0.25 \times 2 = 0.50 V$

Current Electricity

268522 The resistance of a semicircle shown in fig. betweenitstwo end faces is (G iven that radial thickness $= 3 cm$ , axial thickness $= 4 cm$ , inner radius $= 6 cm$ and resistivity $= 4 \times 10^{- 6} Ω cm$ )

1

24.15 \times 10^{- 6} Ω

2

7.85 \times 10^{- 7} Ω

3

7.85 \times 10^{- 6} Ω

4

7.85 \times 10^{- 5} Ω

Explanation:

Here, $A = 4 cm \times 3 cm = 12 {cm}^{2}$
$I = π r = π (6 + 3 / 2) = π \times 7.5 cm$
$R = \frac{ρ l}{A} = \frac{4 \times 10^{- 6} \times π \times 7.5}{12}$
$= 7.85 \times 10^{- 6}$ ohm

Current Electricity

268523 ABCD is a square where each side is a uniform wire of resistance12 . A point $E$ lies on $C D$ such that if a uniform wire of resistance $1 Ω$ is connected across AE and constant potential difference is applied across $A$ and $C$ , then $B$ and $E$ are equipotential .

1

\frac{C E}{E D} = 1

2

\frac{C E}{E D} = \frac{1}{\sqrt{2}}

3

\frac{C E}{E D} = \frac{1}{2}

4

\frac{C E}{E D} = \sqrt{2}

Explanation:

Equivalent resistance between $A$ and $E$ is
$y = \frac{(x + 1)}{x + 2}$
For $B$ and $E$ to be at equal potential, we get
$\frac{R_{A E}}{R_{A B}} = \frac{R_{E C}}{R_{B C}} \Rightarrow \frac{x + 1}{(x + 2) 1} = \frac{1 - x}{1}$
Solving $x = \sqrt{2} - 1$
Now $\frac{C E}{E D} = \frac{1 - x}{x} = \sqrt{2}$

Current Electricity

268520 Three ammeters $P, Q$ and $R$ with internal resistances $r, 1.5 r$ ,3r respectively . $Q$ and $R$ parallel and this combination is in series with $P$ , The whole combination concted between $X$ and $Y$ . When the battery connected between $X$ and $Y$ , the ratio of the readings of $P, Q$ and $R$ is

1

2 : 1 : 1

2

3 : 2 : 1

3

3 : 1 : 2

4

1 : 1 : 1

Explanation:

$i = \frac{V}{R}$

Current Electricity

268521 The potential difference between the points
$A$ and $B$ is

1

1.50 V

2

2.50 V

3

1.00 V

4

0.50 V

Explanation:

For the first loop $12 = 5 i + i_{1}$
For thesecond loop $0 = 7 (i - i_{1}) - i_{1}$
or $8 i_{1} = 7 i$ or, $i_{1} = (7 / 8) i$
Therefore, we obtain $12 = 5 i + \frac{7}{8} i = \frac{47 i}{8}$
or, $i = \frac{12 \times 8}{47} A = 2.04 A$
$i_{1} = \frac{7}{8} \times 2.04 A = 1.79 A$
Thus, the p.d across $A$ and $B$ is
$V_{A} - V_{B} = (i - i_{1}) \times 2 = 0.25 \times 2 = 0.50 V$

Current Electricity

268522 The resistance of a semicircle shown in fig. betweenitstwo end faces is (G iven that radial thickness $= 3 cm$ , axial thickness $= 4 cm$ , inner radius $= 6 cm$ and resistivity $= 4 \times 10^{- 6} Ω cm$ )

1

24.15 \times 10^{- 6} Ω

2

7.85 \times 10^{- 7} Ω

3

7.85 \times 10^{- 6} Ω

4

7.85 \times 10^{- 5} Ω

Explanation:

Here, $A = 4 cm \times 3 cm = 12 {cm}^{2}$
$I = π r = π (6 + 3 / 2) = π \times 7.5 cm$
$R = \frac{ρ l}{A} = \frac{4 \times 10^{- 6} \times π \times 7.5}{12}$
$= 7.85 \times 10^{- 6}$ ohm

Current Electricity

268523 ABCD is a square where each side is a uniform wire of resistance12 . A point $E$ lies on $C D$ such that if a uniform wire of resistance $1 Ω$ is connected across AE and constant potential difference is applied across $A$ and $C$ , then $B$ and $E$ are equipotential .

1

\frac{C E}{E D} = 1

2

\frac{C E}{E D} = \frac{1}{\sqrt{2}}

3

\frac{C E}{E D} = \frac{1}{2}

4

\frac{C E}{E D} = \sqrt{2}

Explanation:

Equivalent resistance between $A$ and $E$ is
$y = \frac{(x + 1)}{x + 2}$
For $B$ and $E$ to be at equal potential, we get
$\frac{R_{A E}}{R_{A B}} = \frac{R_{E C}}{R_{B C}} \Rightarrow \frac{x + 1}{(x + 2) 1} = \frac{1 - x}{1}$
Solving $x = \sqrt{2} - 1$
Now $\frac{C E}{E D} = \frac{1 - x}{x} = \sqrt{2}$

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