Question

Q.11 A steady current flows through the potentiometer
wire. When an unknown emf is measured the
balance point is found to be at the position of 45cm
from positive end of the potentiometer wire. When
a standard cell with emf 1.018 Visused, the jockey
position is 30 cm at balance. What is the unknown
emf?
(A) 1.018 V
(B) 0.679 V
(C) 1.527 V
(D) none of the above.​

Answer 1

(a) Constant emf of the given standard cell, E  

1

​  

=1.02V

Balance point on the wire, l  

1

​  

=67.3cm

A cell of unknown emf, ε,replaced the standard cell. Therefore, new balance point on the wire, l=82.3cm

The relation connecting emf and balance point is,

l  

1

​  

 

E  

1

​  

 

​  

=  

l

ε

​  

 

ε=  

l  

1

​  

 

l

​  

×E  

1

​  

 

  =  

67.3

82.3

​  

×1.02=1.247V

The value of unknown emfis 1.247 V.

(b) The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

(c) The balance point is not affected by the presence of high resistance.  

(d) The point is not affected by the internal resistance of the driver cell.

(e) The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.

(f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.(a) Constant emf of the given standard cell, E  

1

​  

=1.02V

Balance point on the wire, l  

1

​  

=67.3cm

A cell of unknown emf, ε,replaced the standard cell. Therefore, new balance point on the wire, l=82.3cm

The relation connecting emf and balance point is,

l  

1

​  

 

E  

1

​  

 

​  

=  

l

ε

​  

 

ε=  

l  

1

​  

 

l

​  

×E  

1

​  

 

  =  

67.3

82.3

​  

×1.02=1.247V

The value of unknown emfis 1.247 V.

(b) The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

(c) The balance point is not affected by the presence of high resistance.  

(d) The point is not affected by the internal resistance of the driver cell.

(e) The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.

(f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

Answer 2

The unknown emf is (C) 1.527 V.

Given:

The unknown emf is measured the balance point is found to be at the position of 45cm from the positive end of the potentiometer wire. When

a standard cell with emf 1.018 Visused, the jockey the position is 30 cm at balance.

To Find:

The unknown emf.

Solution:

To find the unknown emf we will follow the following steps:

As we know,

In potentiometer measured emf is directly proportional to length balanced.

Balancing length across unknown EMF = 45cm.

Balancing length across standard EMF of 1.018V = 30cm.

So,

Emf and length relation is-

$\frac{E1}{E2}$

Here, E1 is unknown emf while E2 is standard emf.

l1 and l2 are balancing lengths across unknown and standard emf respectively.

$\frac{E1}{E2} = \frac{l1}{l2}$

Now,

$\frac{E1}{1.018} = \frac{45}{30}$

$E1 = \frac{45}{30} \times 1.018 = 1.527volt$

Henceforth, the unknown emf is (C)1.527 volt.

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