Current Electricity Test

Result

Current Electricity Test - 8

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
4 / -1

In a potentiometer of 10 wires, the balance point is obtained on the 7^th wire. To shift the balance point to 9^th wire, we should

A
decrease resistance in the main circuit

B
increase resistance in the main circuit

C
decrease resistance in senes with the cell whose emf is to be measured

D
increase resistance in senes with the cell whose emf is to be determined

Solution

To shift the balance point on higher length, the potential gradient of the wire is to be decreased. The same can be obtained by decreasing the current of the main circuit, which is possible by increasing the resistance in series of potentiometer wire.
Question 2
4 / -1

AB is a wire of potentiometer with the increase in the value of resistance R, the shift in the balance point J will be

A
towards B

B
towards A

C
remains constant

D
first towards B then back towards A

Solution

Due to increase in resistance R the current through the wire will decrease and hence the potential gradient also decreases, which results in increase in balancing length. So / will shift towards B.
Question 3
4 / -1

In a potentiometer a cell of emf 1.5 V gives a balanced point at 32 cm length of the wire. If the cell is replaced by another cell then the balance point shifts to 65.0 cm then the emf of second cell is

A
3.05V

B
2.05V

C
4.05V

D
6.05V

Solution

Here, in the balance condition of potentiometer
ε₁/ε₂ = l₁/l₂
or ε₁= 1.5V , l₁= 32cm, l₂= 65cm
∴ ε₂ = ε₁ × l₂/l₁
= 1.5 × 65/32 = 3.05V
Question 4
4 / -1

3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell is

A
2.5Ω

B
2.25Ω

C
1.12Ω

D
3.2Ω

Solution

Internal resistance of cell

= 3.05 V
l₂ = 68.3cm or
= 1.12Ω
Question 5
4 / -1

In a potentiometer the balancing with a cell is at length of 220cm. On shunting the cell with a resistance of 3Ω balance length becomes 130cm. What is the internal resistance of this cell?

A
4.5Ω

B
7.8Ω

C
6.3Ω

D
2.08Ω

Solution

Here, l₁ = 220cm, l₂ = 130cm, R = 3Ω
∴ Internal resistance,
r =
= 2.08Ω