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Current Electricity

268337 When an unknown resistance and a resistance of $4 Ω$ areconnected in theleft and right gaps of a Meterbridge, the balance point is obtained at $50 cm$ . The shift in the balance point if a $4 Ω$ resistance is now connected in parallel to the resistance in the right gap is

1

66.7 cm

2

16.7 cm

3

34.6 cm

4

14.6 cm

Explanation:

$\frac{x}{y} = \frac{50}{50} - \dots (1); \frac{4}{2} = \frac{I}{(100 - I)}$
$I - 50 = 16.7$

Current Electricity

268338 In a meter bridge, the gaps are closed by resistances 2 and 3 ohms. The value of shunt to be added to 3 ohm resistor to shift the balancing point by $22.5 cm$ is

1

1_{Ω}

2

2 Ω

3

2.5 Ω

4

5 Ω

Explanation:

$\frac{2}{3} = \frac{I}{100 - I} \Rightarrow I = 40 cm; \frac{\frac{2}{3 r}}{3 + r} = \frac{62.5}{100 - 62.5}$

Current Electricity

268339 Two equal resistance are connected in the gaps of a metre bridge. If the resistance in the left gap is increased by $10 %$ , the balancing point shift

1

10 %

to right

2

10 %

to left

3

9.6 %

to right

4

4.8 %

to right

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$

Current Electricity

268403 The point in a $M$ etre bridge is at $35.6 cm$ . If the resistancesin thegaps are interchanged, thenew balancepoint is

1

64.4 cm 2

2

56 cm

3

41.2 cm

4

56.7 cm

Explanation:

$\frac{P}{Q} = \frac{35.6}{64.4}, \frac{Q}{R} = \frac{64.4}{35.6}$

Current Electricity

268404 In a metre bridge expt, when the resistances in the gaps are interchanged the balance point is increases by $10 cm$ . The ratio of the resistances is

1

\frac{15}{5}

2

\frac{12}{8}

3

\frac{11}{9}

4

\frac{10}{9}

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$

Current Electricity

268337 When an unknown resistance and a resistance of $4 Ω$ areconnected in theleft and right gaps of a Meterbridge, the balance point is obtained at $50 cm$ . The shift in the balance point if a $4 Ω$ resistance is now connected in parallel to the resistance in the right gap is

1

66.7 cm

2

16.7 cm

3

34.6 cm

4

14.6 cm

Explanation:

$\frac{x}{y} = \frac{50}{50} - \dots (1); \frac{4}{2} = \frac{I}{(100 - I)}$
$I - 50 = 16.7$

Current Electricity

268338 In a meter bridge, the gaps are closed by resistances 2 and 3 ohms. The value of shunt to be added to 3 ohm resistor to shift the balancing point by $22.5 cm$ is

1

1_{Ω}

2

2 Ω

3

2.5 Ω

4

5 Ω

Explanation:

$\frac{2}{3} = \frac{I}{100 - I} \Rightarrow I = 40 cm; \frac{\frac{2}{3 r}}{3 + r} = \frac{62.5}{100 - 62.5}$

Current Electricity

268339 Two equal resistance are connected in the gaps of a metre bridge. If the resistance in the left gap is increased by $10 %$ , the balancing point shift

1

10 %

to right

2

10 %

to left

3

9.6 %

to right

4

4.8 %

to right

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$

Current Electricity

268403 The point in a $M$ etre bridge is at $35.6 cm$ . If the resistancesin thegaps are interchanged, thenew balancepoint is

1

64.4 cm 2

2

56 cm

3

41.2 cm

4

56.7 cm

Explanation:

$\frac{P}{Q} = \frac{35.6}{64.4}, \frac{Q}{R} = \frac{64.4}{35.6}$

Current Electricity

268404 In a metre bridge expt, when the resistances in the gaps are interchanged the balance point is increases by $10 cm$ . The ratio of the resistances is

1

\frac{15}{5}

2

\frac{12}{8}

3

\frac{11}{9}

4

\frac{10}{9}

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$

Current Electricity

268337 When an unknown resistance and a resistance of $4 Ω$ areconnected in theleft and right gaps of a Meterbridge, the balance point is obtained at $50 cm$ . The shift in the balance point if a $4 Ω$ resistance is now connected in parallel to the resistance in the right gap is

1

66.7 cm

2

16.7 cm

3

34.6 cm

4

14.6 cm

Explanation:

$\frac{x}{y} = \frac{50}{50} - \dots (1); \frac{4}{2} = \frac{I}{(100 - I)}$
$I - 50 = 16.7$

Current Electricity

268338 In a meter bridge, the gaps are closed by resistances 2 and 3 ohms. The value of shunt to be added to 3 ohm resistor to shift the balancing point by $22.5 cm$ is

1

1_{Ω}

2

2 Ω

3

2.5 Ω

4

5 Ω

Explanation:

$\frac{2}{3} = \frac{I}{100 - I} \Rightarrow I = 40 cm; \frac{\frac{2}{3 r}}{3 + r} = \frac{62.5}{100 - 62.5}$

Current Electricity

268339 Two equal resistance are connected in the gaps of a metre bridge. If the resistance in the left gap is increased by $10 %$ , the balancing point shift

1

10 %

to right

2

10 %

to left

3

9.6 %

to right

4

4.8 %

to right

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$

Current Electricity

268403 The point in a $M$ etre bridge is at $35.6 cm$ . If the resistancesin thegaps are interchanged, thenew balancepoint is

1

64.4 cm 2

2

56 cm

3

41.2 cm

4

56.7 cm

Explanation:

$\frac{P}{Q} = \frac{35.6}{64.4}, \frac{Q}{R} = \frac{64.4}{35.6}$

Current Electricity

268404 In a metre bridge expt, when the resistances in the gaps are interchanged the balance point is increases by $10 cm$ . The ratio of the resistances is

1

\frac{15}{5}

2

\frac{12}{8}

3

\frac{11}{9}

4

\frac{10}{9}

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$

Current Electricity

268337 When an unknown resistance and a resistance of $4 Ω$ areconnected in theleft and right gaps of a Meterbridge, the balance point is obtained at $50 cm$ . The shift in the balance point if a $4 Ω$ resistance is now connected in parallel to the resistance in the right gap is

1

66.7 cm

2

16.7 cm

3

34.6 cm

4

14.6 cm

Explanation:

$\frac{x}{y} = \frac{50}{50} - \dots (1); \frac{4}{2} = \frac{I}{(100 - I)}$
$I - 50 = 16.7$

Current Electricity

268338 In a meter bridge, the gaps are closed by resistances 2 and 3 ohms. The value of shunt to be added to 3 ohm resistor to shift the balancing point by $22.5 cm$ is

1

1_{Ω}

2

2 Ω

3

2.5 Ω

4

5 Ω

Explanation:

$\frac{2}{3} = \frac{I}{100 - I} \Rightarrow I = 40 cm; \frac{\frac{2}{3 r}}{3 + r} = \frac{62.5}{100 - 62.5}$

Current Electricity

268339 Two equal resistance are connected in the gaps of a metre bridge. If the resistance in the left gap is increased by $10 %$ , the balancing point shift

1

10 %

to right

2

10 %

to left

3

9.6 %

to right

4

4.8 %

to right

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$

Current Electricity

268403 The point in a $M$ etre bridge is at $35.6 cm$ . If the resistancesin thegaps are interchanged, thenew balancepoint is

1

64.4 cm 2

2

56 cm

3

41.2 cm

4

56.7 cm

Explanation:

$\frac{P}{Q} = \frac{35.6}{64.4}, \frac{Q}{R} = \frac{64.4}{35.6}$

Current Electricity

268404 In a metre bridge expt, when the resistances in the gaps are interchanged the balance point is increases by $10 cm$ . The ratio of the resistances is

1

\frac{15}{5}

2

\frac{12}{8}

3

\frac{11}{9}

4

\frac{10}{9}

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$

Current Electricity

268337 When an unknown resistance and a resistance of $4 Ω$ areconnected in theleft and right gaps of a Meterbridge, the balance point is obtained at $50 cm$ . The shift in the balance point if a $4 Ω$ resistance is now connected in parallel to the resistance in the right gap is

1

66.7 cm

2

16.7 cm

3

34.6 cm

4

14.6 cm

Explanation:

$\frac{x}{y} = \frac{50}{50} - \dots (1); \frac{4}{2} = \frac{I}{(100 - I)}$
$I - 50 = 16.7$

Current Electricity

268338 In a meter bridge, the gaps are closed by resistances 2 and 3 ohms. The value of shunt to be added to 3 ohm resistor to shift the balancing point by $22.5 cm$ is

1

1_{Ω}

2

2 Ω

3

2.5 Ω

4

5 Ω

Explanation:

$\frac{2}{3} = \frac{I}{100 - I} \Rightarrow I = 40 cm; \frac{\frac{2}{3 r}}{3 + r} = \frac{62.5}{100 - 62.5}$

Current Electricity

268339 Two equal resistance are connected in the gaps of a metre bridge. If the resistance in the left gap is increased by $10 %$ , the balancing point shift

1

10 %

to right

2

10 %

to left

3

9.6 %

to right

4

4.8 %

to right

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$

Current Electricity

268403 The point in a $M$ etre bridge is at $35.6 cm$ . If the resistancesin thegaps are interchanged, thenew balancepoint is

1

64.4 cm 2

2

56 cm

3

41.2 cm

4

56.7 cm

Explanation:

$\frac{P}{Q} = \frac{35.6}{64.4}, \frac{Q}{R} = \frac{64.4}{35.6}$

Current Electricity

268404 In a metre bridge expt, when the resistances in the gaps are interchanged the balance point is increases by $10 cm$ . The ratio of the resistances is

1

\frac{15}{5}

2

\frac{12}{8}

3

\frac{11}{9}

4

\frac{10}{9}

Explanation:

$\frac{X}{R} = \frac{1}{100 - 1}$