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)
D Divya Saini

 

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.

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