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A potentiometer PQ is set up to compare two resistances as shown in the figure.  The ammeter A in the circuit reads 1.0 A when two way key K3 is open.  The balance point is at a length l1 cm from P when two way key K3 is plugged in between 2 and 1, while the balance point is at a length l2 cm from P when key K3 is plugged in between 3 and 1.

  The ratio of two resistances \frac{R_{1}}{R_{2}}    is  found to be :

 

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

Answers (2)

best_answer

As we learnt in

Meter bridge -

To find the resistance of a given wire using a meter bridge and hence determine the specific resistance of its materials

- wherein

 

 When key is at point (1)

V_{1}=iR_{1}=Xl_{1}

When key is at 3 

V_{2}=i(R_{1}+R_{2})=Xl_{2}

\frac{R_{1}}{R_{2}}=\frac{l_{i}}{l_{2}}=\frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}-l_{1}}


Option 1)

This option is incorrect 

Option 2)

This option is incorrect 

Option 3)

This option is incorrect 

Option 4)

This option is correct 

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Aadil

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