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On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ. How much was the resistance on the left slot before interchanging the resistances ?

  • Option 1)

    910 Ω
     

  • Option 2)

     990 Ω
     

  • Option 3)

    505 Ω
     

  • Option 4)

    550 Ω
     

 

Answers (2)

 

As we learnt that

 

Meter bridge -

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

- wherein

 

 Lets say resistances are R and 1000-R

 

Case I : \frac{R}{l}=\frac{1000-R}{100-l}    eq 1

Case II: \frac{1000-R}{l-10}= \frac{R}{110-l}     eq 2

Multiply both equation

\frac{R(1000-R)}{l(l-10)}= \frac{R(1000-R)}{(100-l)(110-l)} \Rightarrow l^{2}-10l=11000+l^{2}-210l

\Rightarrow 200l=11000

or l=55cm

\Rightarrow \frac{R}{55}=\frac{1000-R}{45}

or 45 R=55000-55R

or R=550\Omega

 

 


Option 1)

910 Ω
 

This is incorrect

Option 2)

 990 Ω
 

This is incorrect

Option 3)

505 Ω
 

This is incorrect

Option 4)

550 Ω
 

This is correct

Posted by

Vakul

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