# 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$\Omega$

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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