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PHXII03:CURRENT ELECTRICITY

357286 The meter-bridge wire $A B$ shown in figure is $50 c m$ long. When $A D = 30 c m$ , no deflection occurs in the galvanometer. Find $R$ .
supporting img

1

1 Ω

2

2 Ω

3

3 Ω

4

4 Ω

Explanation:

$\frac{6}{R} = \frac{ℓ}{x - ℓ} \frac{6}{R} = \frac{30}{20} \Rightarrow R = 4 Ω$ .

PHXII03:CURRENT ELECTRICITY

357287 In the circuit shown, a metre bridge is in its balanced state. The meter bridge wire has a resistance 0.1 $o h m$ / $c m$ . The value of unknown resistance $X$ and the current drawn from the battery of negligible resistance is
supporting img

1

12 Ω, 0.5 a m p

2

6 Ω, 5 a m p

3

10 Ω, 0.1 a m p

4

4 Ω, 1.0 a m p

Explanation:

Resistance of the part $A C$
$R_{A C} = 0.1 \times 40 = 4 Ω$ and $R_{C B} = 0.1 \times 60 = 6 Ω$
In balanced condition $\frac{X}{6} = \frac{4}{6} \Rightarrow X = 4 Ω$
Equivalent resistance $R_{e q} = 5 Ω$ so current drawn from battery is 1 $A$
supporting img

PHXII03:CURRENT ELECTRICITY

357288 With resistance $P$ and $Q$ in the left and the right gap respectively of a metre bridge, the null point divides the wire in the ratio $3 : 4$ . When $P$ and $Q$ are increased by $20 Ω$ each, the null point divides the wire in the ratio $5 : 6$ . The values of $P$ and $Q$ are

1

30 Ω, 40 Ω

2

20 Ω, 40 Ω

3

30 Ω, 80 Ω

4

20 Ω, 20 Ω

Explanation:

For balanced bridge, $\frac{X}{R} = \frac{l_{x}}{l_{r}}$
$∴ \frac{P}{Q} = \frac{3}{4}$
$∴ 3 Q = 4 P (1)$
Now, $\frac{P + 20}{Q + 20} = \frac{5}{6}$
$∴ 5 Q - 6 P = 20 (2)$
From equation (1) and (2), we have,
$P = 30 Ω and Q = 40 Ω$

PHXII03:CURRENT ELECTRICITY

357289 In the given arrangement of experiment on metre bridge, if $A D$ corresponding to null deflection of the galvanometer is $X$ , what would be its value if the radius of the wire $A B$ is doubled ?
supporting img

1

X

2

\frac{X}{4}

3

4 X

4

2 X

KCET - 2021

PHXII03:CURRENT ELECTRICITY

357286 The meter-bridge wire $A B$ shown in figure is $50 c m$ long. When $A D = 30 c m$ , no deflection occurs in the galvanometer. Find $R$ .
supporting img

1

1 Ω

2

2 Ω

3

3 Ω

4

4 Ω

Explanation:

$\frac{6}{R} = \frac{ℓ}{x - ℓ} \frac{6}{R} = \frac{30}{20} \Rightarrow R = 4 Ω$ .

PHXII03:CURRENT ELECTRICITY

357287 In the circuit shown, a metre bridge is in its balanced state. The meter bridge wire has a resistance 0.1 $o h m$ / $c m$ . The value of unknown resistance $X$ and the current drawn from the battery of negligible resistance is
supporting img

1

12 Ω, 0.5 a m p

2

6 Ω, 5 a m p

3

10 Ω, 0.1 a m p

4

4 Ω, 1.0 a m p

Explanation:

Resistance of the part $A C$
$R_{A C} = 0.1 \times 40 = 4 Ω$ and $R_{C B} = 0.1 \times 60 = 6 Ω$
In balanced condition $\frac{X}{6} = \frac{4}{6} \Rightarrow X = 4 Ω$
Equivalent resistance $R_{e q} = 5 Ω$ so current drawn from battery is 1 $A$
supporting img

PHXII03:CURRENT ELECTRICITY

357288 With resistance $P$ and $Q$ in the left and the right gap respectively of a metre bridge, the null point divides the wire in the ratio $3 : 4$ . When $P$ and $Q$ are increased by $20 Ω$ each, the null point divides the wire in the ratio $5 : 6$ . The values of $P$ and $Q$ are

1

30 Ω, 40 Ω

2

20 Ω, 40 Ω

3

30 Ω, 80 Ω

4

20 Ω, 20 Ω

Explanation:

For balanced bridge, $\frac{X}{R} = \frac{l_{x}}{l_{r}}$
$∴ \frac{P}{Q} = \frac{3}{4}$
$∴ 3 Q = 4 P (1)$
Now, $\frac{P + 20}{Q + 20} = \frac{5}{6}$
$∴ 5 Q - 6 P = 20 (2)$
From equation (1) and (2), we have,
$P = 30 Ω and Q = 40 Ω$

PHXII03:CURRENT ELECTRICITY

357289 In the given arrangement of experiment on metre bridge, if $A D$ corresponding to null deflection of the galvanometer is $X$ , what would be its value if the radius of the wire $A B$ is doubled ?
supporting img

1

X

2

\frac{X}{4}

3

4 X

4

2 X

KCET - 2021

PHXII03:CURRENT ELECTRICITY

357286 The meter-bridge wire $A B$ shown in figure is $50 c m$ long. When $A D = 30 c m$ , no deflection occurs in the galvanometer. Find $R$ .
supporting img

1

1 Ω

2

2 Ω

3

3 Ω

4

4 Ω

Explanation:

$\frac{6}{R} = \frac{ℓ}{x - ℓ} \frac{6}{R} = \frac{30}{20} \Rightarrow R = 4 Ω$ .

PHXII03:CURRENT ELECTRICITY

357287 In the circuit shown, a metre bridge is in its balanced state. The meter bridge wire has a resistance 0.1 $o h m$ / $c m$ . The value of unknown resistance $X$ and the current drawn from the battery of negligible resistance is
supporting img

1

12 Ω, 0.5 a m p

2

6 Ω, 5 a m p

3

10 Ω, 0.1 a m p

4

4 Ω, 1.0 a m p

Explanation:

Resistance of the part $A C$
$R_{A C} = 0.1 \times 40 = 4 Ω$ and $R_{C B} = 0.1 \times 60 = 6 Ω$
In balanced condition $\frac{X}{6} = \frac{4}{6} \Rightarrow X = 4 Ω$
Equivalent resistance $R_{e q} = 5 Ω$ so current drawn from battery is 1 $A$
supporting img

PHXII03:CURRENT ELECTRICITY

357288 With resistance $P$ and $Q$ in the left and the right gap respectively of a metre bridge, the null point divides the wire in the ratio $3 : 4$ . When $P$ and $Q$ are increased by $20 Ω$ each, the null point divides the wire in the ratio $5 : 6$ . The values of $P$ and $Q$ are

1

30 Ω, 40 Ω

2

20 Ω, 40 Ω

3

30 Ω, 80 Ω

4

20 Ω, 20 Ω

Explanation:

For balanced bridge, $\frac{X}{R} = \frac{l_{x}}{l_{r}}$
$∴ \frac{P}{Q} = \frac{3}{4}$
$∴ 3 Q = 4 P (1)$
Now, $\frac{P + 20}{Q + 20} = \frac{5}{6}$
$∴ 5 Q - 6 P = 20 (2)$
From equation (1) and (2), we have,
$P = 30 Ω and Q = 40 Ω$

PHXII03:CURRENT ELECTRICITY

357289 In the given arrangement of experiment on metre bridge, if $A D$ corresponding to null deflection of the galvanometer is $X$ , what would be its value if the radius of the wire $A B$ is doubled ?
supporting img

1

X

2

\frac{X}{4}

3

4 X

4

2 X

KCET - 2021

PHXII03:CURRENT ELECTRICITY

357286 The meter-bridge wire $A B$ shown in figure is $50 c m$ long. When $A D = 30 c m$ , no deflection occurs in the galvanometer. Find $R$ .
supporting img

1

1 Ω

2

2 Ω

3

3 Ω

4

4 Ω

Explanation:

$\frac{6}{R} = \frac{ℓ}{x - ℓ} \frac{6}{R} = \frac{30}{20} \Rightarrow R = 4 Ω$ .

PHXII03:CURRENT ELECTRICITY

357287 In the circuit shown, a metre bridge is in its balanced state. The meter bridge wire has a resistance 0.1 $o h m$ / $c m$ . The value of unknown resistance $X$ and the current drawn from the battery of negligible resistance is
supporting img

1

12 Ω, 0.5 a m p

2

6 Ω, 5 a m p

3

10 Ω, 0.1 a m p

4

4 Ω, 1.0 a m p

Explanation:

Resistance of the part $A C$
$R_{A C} = 0.1 \times 40 = 4 Ω$ and $R_{C B} = 0.1 \times 60 = 6 Ω$
In balanced condition $\frac{X}{6} = \frac{4}{6} \Rightarrow X = 4 Ω$
Equivalent resistance $R_{e q} = 5 Ω$ so current drawn from battery is 1 $A$
supporting img

PHXII03:CURRENT ELECTRICITY

357288 With resistance $P$ and $Q$ in the left and the right gap respectively of a metre bridge, the null point divides the wire in the ratio $3 : 4$ . When $P$ and $Q$ are increased by $20 Ω$ each, the null point divides the wire in the ratio $5 : 6$ . The values of $P$ and $Q$ are

1

30 Ω, 40 Ω

2

20 Ω, 40 Ω

3

30 Ω, 80 Ω

4

20 Ω, 20 Ω

Explanation:

For balanced bridge, $\frac{X}{R} = \frac{l_{x}}{l_{r}}$
$∴ \frac{P}{Q} = \frac{3}{4}$
$∴ 3 Q = 4 P (1)$
Now, $\frac{P + 20}{Q + 20} = \frac{5}{6}$
$∴ 5 Q - 6 P = 20 (2)$
From equation (1) and (2), we have,
$P = 30 Ω and Q = 40 Ω$

PHXII03:CURRENT ELECTRICITY

357289 In the given arrangement of experiment on metre bridge, if $A D$ corresponding to null deflection of the galvanometer is $X$ , what would be its value if the radius of the wire $A B$ is doubled ?
supporting img

1

X

2

\frac{X}{4}

3

4 X

4

2 X

KCET - 2021

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