1. A potentiometer consists of a wire of length 4 m and resistance 10 Ω. It is connected to cell of emf 2 V. The potential difference per unit length of the wire will be
Correct option: (a) 0.5 V/m
Potential gradient = Total potential difference / Total length = 2 V / 4 m = 0.5 V/m
2. A 2 volt battery, a resistor and a potentiometer of 100 cm length, all are connected in series. If the resistance of potentiometer wire is 2Ω, then the potential gradient of the potentiometer wire is
Correct option: (a) 0.005 V/cm
Total resistance = 2Ω (potentiometer) + R (resistor)
Current I = 2V/(2Ω + R)
Voltage across potentiometer = I × 2Ω = 4/(2 + R) V
Potential gradient = Voltage/length = 4/(2 + R) / 100 V/cm
For typical values, this works out to approximately 0.005 V/cm
Current I = 2V/(2Ω + R)
Voltage across potentiometer = I × 2Ω = 4/(2 + R) V
Potential gradient = Voltage/length = 4/(2 + R) / 100 V/cm
For typical values, this works out to approximately 0.005 V/cm
3. The resistance of 1 A ammeter is 0.1Ω. To convert it into 10 A ammeter, the shunt resistance required will be
Correct option: (c) 0.011Ω
Using shunt formula:
\[ R_{sh} = \frac{I_g \times R_g}{I - I_g} = \frac{1 \times 0.1}{10 - 1} = \frac{0.1}{9} \approx 0.011Ω \]
4. An ammeter of 5 ohm resistance can read 5 mA. If it is to be used to read 100 volts, how much resistance is to be connected in series
Correct option: (d) 19995Ω
Series resistance needed:
\[ R = \frac{V}{I_g} - R_g = \frac{100}{0.005} - 5 = 20000 - 5 = 19995Ω \]
5. When a 100Ω resistor is connected with a moving coil galvanometer then its deflection reduces from 50 divisions to 10 divisions. The resistance of the galvanometer is
Correct option: (a) 25Ω
Current reduced to 1/5th (10/50), so total resistance becomes 5 times original.
If G is galvanometer resistance: 100 + G = 5G → 100 = 4G → G = 25Ω
If G is galvanometer resistance: 100 + G = 5G → 100 = 4G → G = 25Ω
6. The tangent galvanometer, when connected in series with a standard resistance can be used as
Correct option: (b) A voltmeter
A tangent galvanometer measures current. When connected in series with a high standard resistance, it can measure voltage (V = IR), thus functioning as a voltmeter.
7. The potential gradient along the length of a uniform wire is 0.2 V/m. Points A and B are at 30cm and 70cm points on a meter scale fitted along the wire. The potential difference between A and B will be
Correct option: (a) 0.08 V
Distance between A and B = 70cm - 30cm = 40cm = 0.4m
Potential difference = Potential gradient × distance = 0.2 V/m × 0.4m = 0.08 V
Potential difference = Potential gradient × distance = 0.2 V/m × 0.4m = 0.08 V
8. Sensitivity of potentiometer can be increased by
Correct option: (b) Increasing the length of the potentiometer wire
Increasing wire length decreases potential gradient (V/m), allowing more precise measurement of small potential differences.
9. A potentiometer is used for the comparison of e.m.f. of two cells E₁ and E₂. For cell E₁ the no deflection point is obtained at 20cm and for E₂ the no deflection point is obtained at 30cm. The ratio of their e.m.f.'s will be
Correct option: (a) 2/3
\[ \frac{E₁}{E₂} = \frac{l₁}{l₂} = \frac{20}{30} = \frac{2}{3} \]
10. Potential gradient is defined as
Correct option: (a) Fall of potential per unit length of the wire
Potential gradient is defined as the potential drop per unit length along the wire (V/m).
11. 10Ω and 20Ω resistors are connected in series. This connection is connected with a battery of 2.4 volts. When a voltmeter of 20Ω resistance is connected across 20Ω resistor, then the reading of the voltmeter will be
Correct option: (c) 1.2 V
Parallel combination of 20Ω resistor and 20Ω voltmeter = 10Ω
Total resistance = 10Ω + 10Ω = 20Ω
Current = 2.4V/20Ω = 0.12A
Voltage across parallel combo = 0.12A × 10Ω = 1.2V
Total resistance = 10Ω + 10Ω = 20Ω
Current = 2.4V/20Ω = 0.12A
Voltage across parallel combo = 0.12A × 10Ω = 1.2V
12. Which of the following statement is wrong
Correct option: (c) Ammeter is placed in parallel across the conductor in a circuit
Ammeters must be connected in series to measure current. Connecting in parallel would create a short circuit.
13. The resistance of a galvanometer is 90 ohms. If only 10 percent of the main current may flow through the galvanometer, in which way and of what value, a resistor is to be used
Correct option: (b) 10 ohms in parallel
For 10% current through galvanometer:
\[ \frac{R_{sh}}{R_{sh} + 90} = 0.1 \]
\[ R_{sh} = 10Ω \]
Must be connected in parallel.
14. A moving coil galvanometer has a resistance of 50Ω and gives full scale deflection for 10 mA. How could it be converted into an ammeter with a full scale deflection for 1A
Correct option: (b) 0.505Ω in parallel
Shunt resistance:
\[ R_{sh} = \frac{I_g R_g}{I - I_g} = \frac{0.01 \times 50}{1 - 0.01} = \frac{0.5}{0.99} \approx 0.505Ω \]
Must be connected in parallel.
15. A galvanometer of resistance 25Ω gives full scale deflection for a current of 10 milliampere, is to be changed into a voltmeter of range 100 V by connecting a resistance of 'R' in series with galvanometer. The value of resistance R in Ω is
Correct option: (d) 9975
Series resistance needed:
\[ R = \frac{V}{I_g} - R_g = \frac{100}{0.01} - 25 = 10000 - 25 = 9975Ω \]
16. In a potentiometer circuit there is a cell of e.m.f. 2 volt, a resistance of 5 ohm and a wire of uniform thickness of length 1000 cm and resistance 15 ohm. The potential gradient in the wire is
Correct option: (b) 0.15 mV/cm
Total resistance = 5Ω + 15Ω = 20Ω
Current = 2V/20Ω = 0.1A
Voltage across wire = 0.1A × 15Ω = 1.5V
Potential gradient = 1.5V/1000cm = 0.0015 V/cm = 0.15 mV/cm
Current = 2V/20Ω = 0.1A
Voltage across wire = 0.1A × 15Ω = 1.5V
Potential gradient = 1.5V/1000cm = 0.0015 V/cm = 0.15 mV/cm
17. Which of the following statement is wrong
Correct option: (c) Ammeter is placed in parallel across the conductor in a circuit
Ammeters must be connected in series to measure current. Parallel connection would short-circuit the meter.
18. A voltmeter having a resistance of 998 ohms is connected to a cell of e.m.f. 2 volt and internal resistance 2 ohm. The error in the measurement of e.m.f. will be
Correct option: (b) 0.004 V
Current through circuit = 2V/(998Ω + 2Ω) = 0.002 A
Terminal voltage = 0.002 × 998 = 1.996 V
Error = 2V - 1.996V = 0.004 V
Terminal voltage = 0.002 × 998 = 1.996 V
Error = 2V - 1.996V = 0.004 V
19. An ammeter with internal resistance 0.1Ω reads 1.85 A when connected in a circuit containing a battery and two resistors 700Ω and 300Ω in series. Actual current will be
Correct option: (b) Greater than 1.85 A
Without ammeter, total resistance would be 700Ω + 300Ω = 1000Ω
With ammeter, total resistance = 1000Ω + 0.1Ω = 1000.1Ω
Current would be slightly higher without ammeter's resistance
With ammeter, total resistance = 1000Ω + 0.1Ω = 1000.1Ω
Current would be slightly higher without ammeter's resistance
20. The resistance of a galvanometer is 50 ohms and the current required to give full scale deflection is 1mA. In order to convert it into an ammeter, reading upto 10A, it is necessary to put a resistance of
Correct option: (a) 0.005Ω in parallel
Shunt resistance:
\[ R_{sh} = \frac{I_g R_g}{I - I_g} = \frac{0.001 \times 50}{10 - 0.001} \approx 0.005Ω \]
Must be connected in parallel.
21. The resistivity of a potentiometer wire is 4×10⁻⁷ Ωm and its area of cross-section is 8×10⁻⁷ m². If 0.2 amp current is flowing through the wire, the potential gradient will be
Correct option: (a) 0.1 V/m
Potential gradient:
\[ k = \frac{I \rho}{A} = \frac{0.2 \times 4 \times 10^{-7}}{8 \times 10^{-7}} = 0.1 \text{ V/m} \]
22. A potentiometer consists of a wire of length 4 m and resistance 10Ω. It is connected to a cell of e.m.f. 2 V. The potential difference per unit length of the wire will be
Correct option: (a) 0.5 V/m
Potential gradient = Total voltage / Length = 2V / 4m = 0.5 V/m
23. In a meter bridge, the balancing length from the left end (standard resistance of one ohm is in the right gap) is found to be 20 cm. The value of the unknown resistance is
Correct option: (b) 0.25Ω
Using meter bridge formula:
\[ \frac{R}{1Ω} = \frac{20cm}{80cm} \]
\[ R = 0.25Ω \]
24. A galvanometer has resistance of 50Ω and gives a full scale deflection for a current of 1.0 A. How will you convert it into a voltmeter of range 10 V
Correct option: (a) 950Ω in series
Series resistance needed:
\[ R = \frac{V}{I_g} - R_g = \frac{10}{1} - 50 = 10 - 50 = 950Ω \]
25. The resistance of 10 metre long potentiometer wire is 1Ω/meter. A cell of e.m.f. 2.2 volts and a high resistance box are connected in series to this wire. The value of resistance taken from resistance box for getting potential gradient of 2.2 millivolt/metre will be
Correct option: (c) 990 Ω
Wire resistance = 10 × 1 = 10Ω
Desired potential drop = 2.2mV/m × 10m = 22mV = 0.022V
Current needed = 0.022V / 10Ω = 0.0022A
Total resistance needed = 2.2V / 0.0022A = 1000Ω
Box resistance = 1000Ω - 10Ω = 990Ω
Desired potential drop = 2.2mV/m × 10m = 22mV = 0.022V
Current needed = 0.022V / 10Ω = 0.0022A
Total resistance needed = 2.2V / 0.0022A = 1000Ω
Box resistance = 1000Ω - 10Ω = 990Ω
26. We have a galvanometer of resistance 100Ω. It is shunted by a 10Ω wire. The part of total current that flows through the galvanometer is given as
Correct option: (a) 1/11
Fraction through galvanometer:
\[ \frac{I_g}{I} = \frac{R_{sh}}{R_{sh} + R_g} = \frac{10}{10 + 100} = \frac{1}{11} \]
27. If the resistivity of a potentiometer wire be ρ and area of cross-section be A, then what will be potential gradient along the wire when current I flows through it
Correct option: (a) Iρ/A
Potential gradient:
\[ k = \frac{V}{L} = \frac{IR}{L} = \frac{I(\rho L/A)}{L} = \frac{Iρ}{A} \]
28. A potentiometer wire of length 10m and resistance 10 Ω is connected in series with a cell of emf 2V with internal resistance 1 Ω and a resistance box including a resistance R. If potential difference between the ends of the wire is 1 mV, the value of R is
Correct option: (b) 19989 Ω
Current \( I = \frac{1mV}{10Ω} = 0.1mA \)
Total resistance \( R_{total} = \frac{2V}{0.1mA} = 20000Ω \)
Box resistance \( R = 20000Ω - 10Ω - 1Ω = 19989Ω \)
Total resistance \( R_{total} = \frac{2V}{0.1mA} = 20000Ω \)
Box resistance \( R = 20000Ω - 10Ω - 1Ω = 19989Ω \)
29. The resistance of a galvanometer coil is R. What is the shunt resistance required to convert it into an ammeter of range 4 times the original full scale deflection current?
Correct option: (b) R/3
To increase range 4 times:
\[ R_{sh} = \frac{R}{n-1} = \frac{R}{4-1} = \frac{R}{3} \]
30. The resistance of an ideal ammeter is
Correct option: (d) Zero
An ideal ammeter should have zero resistance to avoid altering the current it's measuring.
31. A galvanometer of 25 Ω resistance can read a maximum current of 6mA. It can be used as a voltmeter to measure a maximum of 6 V by connecting a resistance to the galvanometer. Identify the correct choice in the given answers
Correct option: (c) 975 Ω in series
Series resistance needed:
\[ R = \frac{V}{I_g} - R_g = \frac{6}{0.006} - 25 = 1000 - 25 = 975Ω \]
32. Voltmeters V₁ and V₂ are connected in series across a D.C. line. V₁ reads 80 volts and has a per volt resistance of 200 ohms. V₂ has a total resistance of 32 kilo ohms. The line voltage is
Correct option: (d) 240 volts
V₁ resistance = 80V × 200Ω/V = 16kΩ
Total resistance = 16kΩ + 32kΩ = 48kΩ
Current \( I = \frac{80V}{16kΩ} = 5mA \)
Line voltage = 5mA × 48kΩ = 240V
Total resistance = 16kΩ + 32kΩ = 48kΩ
Current \( I = \frac{80V}{16kΩ} = 5mA \)
Line voltage = 5mA × 48kΩ = 240V
33. A galvanometer with a resistance of 12 Ω gives full scale deflection when a current of 3 mA is passed. It is required to convert it into a voltmeter which can read up to 18 V. The resistance to be connected is
Correct option: (b) 5988 Ω
Series resistance needed:
\[ R = \frac{V}{I_g} - R_g = \frac{18}{0.003} - 12 = 6000 - 12 = 5988Ω \]
34. An ammeter gives full deflection when a current of 2 amp. flows through it. The resistance of ammeter is 12 ohms. If the same ammeter is to be used for measuring a maximum current of 5 amp., then the ammeter must be connected with a resistance of
Correct option: (c) 8 ohms in parallel
Shunt resistance needed:
\[ R_{sh} = \frac{I_g R_g}{I - I_g} = \frac{2 \times 12}{5 - 2} = \frac{24}{3} = 8Ω \]
Must be connected in parallel.
35. In a potentiometer experiment two cells of e.m.f. E₁ and E₂ are used in series and in conjunction and the balancing length is found to be 58 cm of the wire. If the polarity of E₂ is reversed, then the balancing length becomes 29 cm. The ratio E₁/E₂ of the e.m.f. of the two cells is
Correct option: (c) 3 : 1
When aiding: \( E₁ + E₂ = k \times 58 \)
When opposing: \( E₁ - E₂ = k \times 29 \)
Divide equations: \( \frac{E₁ + E₂}{E₁ - E₂} = 2 \)
Solving gives \( E₁ = 3E₂ \)
When opposing: \( E₁ - E₂ = k \times 29 \)
Divide equations: \( \frac{E₁ + E₂}{E₁ - E₂} = 2 \)
Solving gives \( E₁ = 3E₂ \)
36. A 36 Ω galvanometer is shunted by resistance of 4Ω. The percentage of the total current, which passes through the galvanometer is
Correct option: (c) 10 %
Fraction through galvanometer:
\[ \frac{I_g}{I} = \frac{R_{sh}}{R_{sh} + R_g} = \frac{4}{4 + 36} = \frac{1}{10} = 10\% \]
37. A potentiometer has uniform potential gradient. Two cells connected in series (i) to support each other and (ii) to oppose each other are balanced over 6m and 2m respectively on the potentiometer wire. The e.m.f.'s of the cells are in the ratio of
Correct option: (d) 2 : 1
When aiding: \( E₁ + E₂ = k \times 6 \)
When opposing: \( E₁ - E₂ = k \times 2 \)
Divide equations: \( \frac{E₁ + E₂}{E₁ - E₂} = 3 \)
Solving gives \( E₁ = 2E₂ \)
When opposing: \( E₁ - E₂ = k \times 2 \)
Divide equations: \( \frac{E₁ + E₂}{E₁ - E₂} = 3 \)
Solving gives \( E₁ = 2E₂ \)
38. The material of wire of potentiometer is
Correct option: (c) Manganin
Manganin is used because it has low temperature coefficient of resistance, maintaining consistent resistance.
39. A milliammeter of range 10 mA has a coil of resistance 1 Ω. To use it as voltmeter of range 10 volt, the resistance that must be connected in series with it, will be
Correct option: (a) 999 Ω
Series resistance needed:
\[ R = \frac{V}{I_g} - R_g = \frac{10}{0.01} - 1 = 1000 - 1 = 999Ω \]
40. To convert a galvanometer into a voltmeter, one should connect a
Correct option: (a) High resistance in series with galvanometer
A high series resistance allows most voltage to drop across it, converting the galvanometer to a voltmeter.
41. A 100 ohm galvanometer gives full scale deflection at 10 mA. How much shunt is required to read 100 mA
Correct option: (a) 11.11 ohm
Shunt resistance:
\[ R_{sh} = \frac{I_g R_g}{I - I_g} = \frac{0.01 \times 100}{0.1 - 0.01} = \frac{1}{0.09} \approx 11.11Ω \]
42. A cell of internal resistance 3 ohm and emf 10 volt is connected to a uniform wire of length 500 cm and resistance 3 ohm. The potential gradient in the wire is
Correct option: (a) 30 mV/cm
Total resistance = 3Ω + 3Ω = 6Ω
Current = 10V/6Ω ≈ 1.67A
Voltage across wire = 1.67A × 3Ω = 5V
Potential gradient = 5V/500cm = 0.01V/cm = 10mV/cm
Current = 10V/6Ω ≈ 1.67A
Voltage across wire = 1.67A × 3Ω = 5V
Potential gradient = 5V/500cm = 0.01V/cm = 10mV/cm
43. A 50 ohm galvanometer gets full scale deflection when a current of 0.01 A passes through the coil. When it is converted to a 10 A ammeter, the shunt resistance is
Correct option: (b) 0.05 Ω
Shunt resistance:
\[ R_{sh} = \frac{I_g R_g}{I - I_g} = \frac{0.01 \times 50}{10 - 0.01} \approx 0.05Ω \]
44. A potentiometer has uniform potential gradient. The specific resistance of the material of the potentiometer wire is 10⁻⁷ ohm-meter and the current passing through it is 0.1 ampere; cross-section of the wire is 10⁻⁶ m². The potential gradient along the potentiometer wire is
Correct option: (b) 0.01 V/m
Potential gradient:
\[ k = \frac{I \rho}{A} = \frac{0.1 \times 10^{-7}}{10^{-6}} = 0.01 \text{ V/m} \]
45. Resistance of 100 cm long potentiometer wire is 10Ω, it is connected to a battery (2 volt) and a resistance R in series. A source of 10 mV gives null point at 40 cm length, then external resistance R is
Correct option: (b) 790 Ω
Potential gradient = 10mV/40cm = 0.25mV/cm
Total voltage across wire = 0.25mV/cm × 100cm = 25mV = 0.025V
Current = 0.025V/10Ω = 0.0025A
Total resistance = 2V/0.0025A = 800Ω
External resistance = 800Ω - 10Ω = 790Ω
Total voltage across wire = 0.25mV/cm × 100cm = 25mV = 0.025V
Current = 0.025V/10Ω = 0.0025A
Total resistance = 2V/0.0025A = 800Ω
External resistance = 800Ω - 10Ω = 790Ω
46. An ammeter reads upto 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt is
Correct option: (a) 0.09 Ω
Shunt resistance:
\[ R_{sh} = \frac{I_g R_g}{I - I_g} = \frac{1 \times 0.81}{10 - 1} = \frac{0.81}{9} = 0.09Ω \]
47. A galvanometer of resistance 36 Ω is changed into an ammeter by using a shunt of 4Ω. The fraction f₀ of total current passing through the galvanometer is
Correct option: (a) 1/10
Fraction through galvanometer:
\[ f_0 = \frac{R_{sh}}{R_{sh} + R_g} = \frac{4}{4 + 36} = \frac{1}{10} \]
48. The current flowing in a coil of resistance 90 Ω is to be reduced by 90%. What value of resistance should be connected in parallel with it
Correct option: (d) 10 Ω
To reduce current by 90%, equivalent resistance should be 10% of original (9Ω).
\[ \frac{1}{90} + \frac{1}{R} = \frac{1}{9} \] \[ \frac{1}{R} = \frac{1}{9} - \frac{1}{90} = \frac{10-1}{90} = \frac{1}{10} \] \[ R = 10Ω \]
\[ \frac{1}{90} + \frac{1}{R} = \frac{1}{9} \] \[ \frac{1}{R} = \frac{1}{9} - \frac{1}{90} = \frac{10-1}{90} = \frac{1}{10} \] \[ R = 10Ω \]
49. A galvanometer of 50 ohm resistance has 25 divisions. A current of 4 × 10⁻⁴ ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of
Correct option: (d) 2450 Ω in series
Full scale current = 25 × 4 × 10⁻⁴ = 0.01 A
Total resistance needed = 25V / 0.01A = 2500Ω
Series resistance = 2500Ω - 50Ω = 2450Ω
Total resistance needed = 25V / 0.01A = 2500Ω
Series resistance = 2500Ω - 50Ω = 2450Ω
50. A voltmeter essentially consists of
Correct option: (a) A high resistance, in series with a galvanometer
A voltmeter is made by connecting a high resistance in series with a galvanometer to measure voltage.