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Byju's Answer

Standard XII

Physics

Meter Bridge

A meter bridg...

Question

A meter bridge is set-up as shown, to determine an unknown resistance $X$ using a standard resistor of 10 $Ω$ . The galvanometer shows null point when tapping-key is at 52 $c m$ mark. The end-corrections are 1 $c m$ and 2 $c m$ respectively for the ends A and B. Then the determined value of $X$ is

10.2

Ω

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10.6

Ω

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10.8

Ω

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11.1

Ω

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Solution

The correct option is B 10.6 $Ω$
From given conditions, $l_{1} = 52 + 1 = 53 c m$
$l_{2} = 48 + 2 = 50 c m$
so by balancing condition
$\Rightarrow \frac{l_{1}}{l_{2}} = \frac{X}{R}$
$\Rightarrow \frac{53}{50} = \frac{X}{10}$
$\Rightarrow X = 10.6 Ω$

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A meter bridge is set-up as shown, to determine an unknown resistance X using a standard resistor of 10 Ω. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. Then the determined value of X is

The correct option is B 10.6 ΩFrom given conditions,l1=52+1=53 cm l2=48+2=50 cm so by balancing condition ⇒l1l2=XR ⇒5350=X10 ⇒X=10.6 Ω

A meter bridge is set-up as shown, to determine an unknown resistance $X$ using a standard resistor of 10 $Ω$ . The galvanometer shows null point when tapping-key is at 52 $c m$ mark. The end-corrections are 1 $c m$ and 2 $c m$ respectively for the ends A and B. Then the determined value of $X$ is

The correct option is B 10.6 $Ω$
From given conditions, $l_{1} = 52 + 1 = 53 c m$
$l_{2} = 48 + 2 = 50 c m$
so by balancing condition
$\Rightarrow \frac{l_{1}}{l_{2}} = \frac{X}{R}$
$\Rightarrow \frac{53}{50} = \frac{X}{10}$
$\Rightarrow X = 10.6 Ω$