Class 12th Physics Physics Current Electricity Overview

In a meter bridge experiment, the galvanometer shows a null point when jockey is pressed at 40.0cmusing a standard resistance of 100 $Ω$ as per the diagram. The least count of the scale used is 1 mm. The unknown resistance is nearly

Option 1 - <p>  <span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>(</mo> <mrow> <mn>6</mn> <mn>6</mn> <mo>.</mo> <mn>7</mn> <mo>±</mo> <mn>0</mn> <mo>.</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> <mi>Ω</mi> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>(</mo> <mrow> <mn>6</mn> <mn>6</mn> <mo>.</mo> <mn>7</mn> <mo>±</mo> <mn>0</mn> <mo>.</mo> <mn>3</mn> </mrow> <mo>)</mo> </mrow> <mi>Ω</mi> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>(</mo> <mrow> <mn>3</mn> <mn>3</mn> <mo>.</mo> <mn>3</mn> <mo>±</mo> <mn>0</mn> <mo>.</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> <mi>Ω</mi> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>(</mo> <mrow> <mn>3</mn> <mn>3</mn> <mo>.</mo> <mn>3</mn> <mo>±</mo> <mn>0</mn> <mo>.</mo> <mn>3</mn> </mrow> <mo>)</mo> </mrow> <mi>Ω</mi> </mrow> </math> </span></p>

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Vishal Baghel | Contributor-Level 10

4 months ago

Correct Option - 2

Detailed Solution:

$\Rightarrow \frac{R}{100} = \frac{40}{60}$

$\Rightarrow Δ R = (\frac{Δ l}{l} + \frac{Δ l}{(100 - l)}) R$

$= (\frac{0.1}{40} + \frac{0.1}{60}) * 66.7 = 0.27 = 0.3$

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