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The resistance in the two arms of the meter bridge  are 5\Omega and R\Omega, respectivilty. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6\ l_{1} . The resistance 'R' is:

  • Option 1)

    10\Omega

  • Option 2)

    15\Omega

  • Option 3)

    20\Omega

  • Option 4)

    25\Omega

 

Answers (1)

best_answer

 

Meter bridge -

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

- wherein

 

 

\frac{5}{l_{1}}=\frac{R}{100-l_{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(1)

\frac{5}{1.6\:l_{1}}=\frac{\frac{R}{2}}{100-1.6\:l_{1}}\:\:\:\:\:\:(2)

 

Divide (1) and (2)

\Rightarrow 1.6=\frac{100-1.6\:l_{1}}{100-l_{1}}\times 2

\Rightarrow 1.6l_{1}=40

Or, l_{1}=25 \:cm

From equation (1)

\frac{5}{25}=\frac{R}{75}\:\:\:\Rightarrow R=15 \Omega


Option 1)

10\Omega

This option is incorrect.

Option 2)

15\Omega

This option is correct.

Option 3)

20\Omega

This option is incorrect.

Option 4)

25\Omega

This option is incorrect.

Posted by

divya.saini

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