Current Electricity Test

Result

Current Electricity Test - 59

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Weekly Quiz Competition

Question 1
4 / -1

For comparing the e.m.f.\'s of two cells with a potentiometer, a standard cell is used to develop a potential gradient along the wires. Which of the following possibilities would make the experiment unsuccessful?

A
The e.m.f. of the standard cell is larger than the e.m.f.'s of the two cells

B
The diameter of the wires is the same and uniform throughout

C
The number of wires is ten

D
The e.m.f. of the standard cell is smaller than the e.m.f.'s of the two cells
Question 2
4 / -1

A potentiometer wire of length L and resistance 10 Ω is connected in series with a battery of e.m.f. 2.5 V and a resistance in its primary circuit. The null point corresponding to a cell of e.m.f. 1 V is obtained at a distance L/2. If the resistance in the primary circuit is doubled then the position of new null point will be

A
0.4 L

B
0.5 L

C
0.6 L

D
0.8 L

Solution
Question 3
4 / -1

An ammeter with internal resistance 90 Ω reads 1.85 A when connected in a circuit containing a battery and two resistors 700 Ω and 410 Ω in series. Actual current will be

A
1.85 A

B
Greater than 1.85 A

C
Less than 1.85 A

D
None of these
Question 4
4 / -1

The circuit shown here is used to compare the e.m.f.\'s of two cells E₁ and E₂ (E₁> E₂). The null point is at C when the galvanometer is connected to E₁. When the galvanometer is connected to E₂, the null point will be

A
To the left of C

B
To the right of C

C
At C itself

D
No where on AB

Solution
Question 5
4 / -1

In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2m when the cell is shunted by a 5 Ω resistance; and is at a length of 3m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell is, then

A
1.5 Ω

B
10 Ω

C
15 Ω

D
1 Ω

Solution
Question 6
4 / -1

A potentiometer circuit shown in the figure is set up to measure e.m.f. of a cell E. As the point P moves from X to V the galvanometer G shows deflection always in one direction, but the deflection decreases continuously until Y is reached. In order to obtain balance point between X and Y it is necessary to

A
Decrease the resistance R

B
Increase the resistance R

C
Reverse the terminals of battery V

D
Reverse the terminals of cell E
Question 7
4 / -1

In the Wheatstone\'s bridge (shown in figure) X = Y and A >B. The direction of the current between ab will be

A
From a to b

B
From b to a

C
From b to a through c

D
From a to b through c

Solution
Question 8
4 / -1

A
The galvanometer shows a deflection when switch S is closed

B
The galvanometer shows a deflection when switch S is open

C
The galvanometer shows no change in deflection whether S is open or closed

D
The galvanometer shows no deflection
Question 9
4 / -1

In the Wheatstone bridge shown below, in order to balance the bridge, we must have

A
R₁ = 3 Ω ; R₂ = 3 Ω

B
R₁ = 6 Ω ; R₂ =15 Ω

C
R₁ = 1.5 Ω ; R₂ = any finite value

D
R₁ = 3 Ω ; R₂ = any finite value

Solution
Question 10
4 / -1

A resistance of 4 Ω and a wire of length 5 metres and resistance 5 Ω are joined in series and connected to a cell of e.m.f. 10 V and internal resistance 1 Ω. A parallel combination of two identical cells is balanced across 300 cm of the wire. The e.m.f. E of each cell is

A
1.5 V

B
3.0 V

C
0.67 V

D
1.33 V

Solution